Good News on Race in the US

October 31, 2006

Two days ago, the Washington Post published an article, “Minority entrollment in college still lagging.” The actual numbers it gives show some underrepresentation of minorities in college, but not that much, and not for all minority groups.

Minority enrollments rose by 50.7% to 4.7 million between 1993 and 2003, while the number of white students increased 3.4%, to 10.5 million, the report says.

White high school graduates are more likely than black or Hispanic peers to enroll in college. The report says 47.3% of white high school graduates ages 18 to 24 attend college, vs. 41.1% of black and 35.2% of Hispanic high school graduates.

Among students who entered college in 1995-96, 36.4% of blacks and 42% of Hispanics earned a bachelor’s degree within six years, vs. 58% of whites and 62.3% of Asian-Americans.

By 2003, most minority groups’ representation among college students was nearly equal to their nationwide representation. The total number of college students of known race in the US is 15.2 million. Of these, 1.9 million are black, representing 12.5% of students compared with 13.5% of Americans; 1.6 million are Hispanic, representing 10.5% of students compared with 14.5% of Americans; just under a million are Asian, representing 6.6% of students compared with 4.3% of the population; and 163,000 are Native American, representing 1.1% of students compared with 0.8% of the population.
Of course, these numbers are somewhat skewed, especially for Hispanics. There are more than 14.5% Hispanics among Americans aged 18 to 24, obviously. So the underrepresentation of Hispanics in US universities is even more acute, though it’s also the one that’s getting better the fastest (between 1993 and 2003, white enrollment increased 3.4%, nonwhite enrollment increased 50.7%, and Hispanic enrollment increased 70%).

The overall minority increases are encouraging, “but we are also concerned by what still seems to be slow growth,” says Beverly Daniel Tatum, president of Spelman College, a historically black college for women in Atlanta, and chair of a commission that produces the annual report. “While we see forward movement, it is incremental and not transformational.”

That, she says, would require better preparation and encouragement in elementary and high schools. “Students of color often have limited access to the courses they need … (and) college guidance,” Tatum says. And a key reason some minority college students don’t persist is because “they’re simply running out of money.”

In other words, since black and Hispanic Americans go to underfunded public schools and have to deal with mounting tuition, they have lower rates of going to college and higher rates of attrition.


October 31, 2006

Dana writes about liberals and libertarians. Asking whether liberals can support libertarians, he offers the conventional issue-by-issue approach:

I’m certain that most of my liberal friends who frequent [Liberal Avenger] wouldn’t have much problem with many Libertarian positions on drugs, constitutional rights (though I’d note here that the emergence of campus speech codes were not the products of conservative thinking), immigration, and abortion. But one wonders how they would see the party’s position on gun rights, on taxes, on welfare, and a host of other issues.

If there is one great value to the Libertarian Party, it is that they understand that freedom and the nanny state are incompatible: if the government is going to take care of you, then the government has to have some say, a lot of say, in how you live your life. The Libertarians wouldn’t care if you smoked or ate a lot of fatty foods or used currently illegal drugs. But they also wouldn’t pay to care for you when you got sick or fat or FUBARed on drugs.

The problem with the issue-based thinking is that it presupposes a libertarian view of things. In that view, every government action that isn’t intended to enforce property rights is categorically immoral. Therefore, liberals are good on free speech, gay rights, immigration, abortion, and separation of church and state, but not on taxes, health care, education, welfare, or anti-discrimination laws; and conservatives are the other way around.

Although it’s possible to make a neutral issue-based approach, the libertarian viewpoint has two problems. First, it categorizes issues artificially. Libertarians say liberals agree with them on social issues and conservatives do on economic issues. But in the US, liberals have a better record on budget balancing, which libertarians support, and are not any more protectionist than conservatives (in other countries, the liberal party tends to be the least protectionist in politics); and libertarians agree with conservatives rather than liberals on anti-discrimination laws, which are a social issue.

But more importantly, the libertarian schema leaves out issue emphasis. My list of positions on issues would shed no light about my political tendencies without the accompanying priority scores. This is especially important for libertarians, who arose as a group in opposition to the New Deal rather than Prohibition, who often found themselves defending fascism in the Cold War, and who are now likelier to vote Republican than Democratic. By the way, it’s not just an American phenomenon: the German Free Democrats are now solidly allied with the Christian Democrats rather than the Social Democrats.

The most important thing to know about modern libertarianism is not that it’s capitalistic. Neo-liberalism is just as capitalistic and tends to come to different conclusions on a lot of issues. It’s that it’s anti-empirical. Austrian economics lionizes factless analysis of principles over empirical data; I’ve yet to meet a libertarian who can distinguish “but it works better” realism from totalitarianism.

When Hayek said in The Road to Serfdom that the growth of government spending was a threat to freedom, he had an excuse: at that time there was no evidence to the contrary. But now when the correlation between social and economic scores on the Political Compass is strongly positive and when the liberal approach to taxing and spending has produced higher individual freedom than the conservative approach, to say that is just plain wrong.

Libertarians may think that the nanny state and freedom are incompatible. But liberals realize that unregulated capitalism is incompatible even with the crude conception of freedom, let alone the more refined idea of competencies. The failure of world governments to eradicate smallpox in 1900 even though it was possible to do so would kill 300-500 million people. Private health care is so inefficient that it requires consumers to spend more money out of pocket (in the US, its effect is equivalent to an across-the-board tax hike of 8%). And the only thing robber baron capitalism did was make people like communism.

In response to Dana’s final assertion, all I can say is, liberals don’t care if you smoke or eat junk food or do illegal drugs – at least, the civil libertarians don’t. All they care about is that you don’t kill people via second-hand smoke, or eat junk food because nothing else is available, or do illegal drugs because a drug dealer laced them with addictive substances.

Civil liberties can’t exist without some government enforcement. The freedom of the Internet is only possible thanks to net neutrality. Privacy was nonexistent in Gilded Age company towns. Non-enforcement of equal rights laws leads to nothing but wanton discrimination. Maximal freedom requires government interventions, unless you distort the meaning of the word “freedom” so much you might as well make up a new word.

The Party that Never Stops Sleeping

October 31, 2006

Stentor links to an old post on saying that the Democratic Party does have ideas, only the Republicans shoot them down.

And, with just weeks to go before the August 7 summer adjournment, an analysis of the 203 roll call votes taken in the Senate through July 13, reveals a Republican-dominated body that, far from practicing what they preach and extending a hand of cooperation, went out of their way to scuttle almost every amendment and bill sponsored by Democratic senators.


Well, of course, there was S.Amdt. 4322, Ted Kennedy’s (D-MA) umpteenth attempt at raising the minimum wage, that went down to defeat once again, along with the Massachusetts Senator’s S.Amdt. 3028 to restore Bush-administration cuts for vocational education and increase the maximum Pell Grant.

Hillary Clinton (D-NY) had two major pieces of legislation related to the Federal Emergency Management Agency (FEMA) scuttled. S.Amdt. 4563 would have established FEMA as an independent agency and eliminated the bonehead move by Team Bush in which they made it subordinate to the Department of Homeland Security. Clinton’s S.Amdt. 2716 pushed for a Congressional commission to scrutinize the federal screw-up in Hurricane Katrina response — and every single Republican on the floor voted against it.

The only thing I’ll say about the first two bills mentioned is that if the Democrats can’t get a minimum wage increase that 83% of Americans support and 68% say should be a top priority, they have the political competence of an amoeba; and that in a country where making public colleges free would cost $20-30 billion per year, Pell grants are the worst kind of middle of the road solution.

Instead, I’d like to focus on the next two bills, both of which are managerial in character. They go well with the Democrats’ latest billing of themselves as the competent party, compared with the incompetent Republicans who screwed up Iraq and Katrina. The Republicans’ failures are a good way of getting people disillusioned with the Republicans, but a horrible way of getting people behind the Democratic Party.

For a start, “We’ll make the trains run on time” is a really bad election strategy. Nobody supports people who merely make the trains run on time, without a more basal reason. The Nazis and the Italian fascists gave people jobs and promised national glory, so the people were amenable to say things like “he made the trains run on time” and “we no longer have the freedom to starve.”

In fact Mussolini didn’t make the trains run on time, and on the eve of World War Two Germany still had a lower quality of life than the US and Britain even though Germany was alone among the three in being out of the Depression. This alone is enough to imply that perceptions of managerial competence follow the people’s established biases rather than shape them. In an American context, Katrina provided a Kuhnian surprise, which shifted many Americans from a pro-Bush to an anti-Bush bias.

But people who already consider you more competent won’t decide to vote against you simply because you emphasize real issues. The Democratic Party is to the right of the majority of American voters on so many issues – prescription drugs, single-payer health care, Iraq, and gay rights to name four – that creating a positive campaign around them should be both easy and popular.

Even if the Democrats do win based on managerial competence, they’ll have a weak mandate at best. They’ll probably be able to increase the minimum wage, but that’s it; on other issues, the resistance from conservative Democrats will be too stiff. And thanks to the technocracy-based campaign, they won’t be able to say, “53% of the people voted for us because of single-payer health care, immigration, and civil unions.”

Computing Class Groups

October 30, 2006

Recall from the previous post (now with errata) that for every number field K, there exists a constant m depending on K such that every ideal of K is equivalent in the class group H to an ideal of norm <= m. In addition, every ideal of norm <= m is generated by prime ideals of norm <= m; hence, it’s enough to look at the behavior of prime integers less than m under decomposition into ideals.

If K = Q(SQRT(-19)), then O(K) = Z[0.5+0.5SQRT(-19)], with integral basis {1, 0.5+0.5SQRT(-19)}. With the identity conjugate, m1 = |1| + |0.5+0.5SQRT(-19)| = 1 + SQRT(5). Similarly, m2 = 1 + SQRT(5), so m = 6 + 2SQRT(5) ~= 10.42. So it’s enough to look at the behavior of 2, 3, 5, and 7.

The element 2 is prime in O(K). Let s = 0.5+0.5SQRT(-19), and note that s^2 = -4.5 + 0.5SQRT(-19) = s – 5. If 2 is not prime, it divides (a + bs)(c + ds) = ac – 5bd + (ad + bc + bd)s, but not a + bs or c + ds. If b is even, then a is odd, so as ac – 5bd is even, c is even; and as ad + bc + bd is even, d is even, which is a contradiction. Hence, b is odd. Similarly, a, c, and d must be odd. But in that case, ad + bc + bd is odd, which is a contradiction. So the ideal (2) is prime.

The element 3 is similarly prime. As with 2, the element 3 can’t divide a, b, c, or d. Modulo 3, we have equations ac – 5bd = 0, ad + bc + bd = 0. -5 = 1 mod 3, so the first equation becomes ac + bd = 0. If a = 1, then c = –bd, and we get dbbd + bd = 0 –> b^2 – b – 1 = 0 which is impossible for any of the two possible values of b. If a = -1, then c = bd, and we get –d + bbd + bd = 0 –> b^2 + b – 1 = 0 which is again impossible.

Note that N(s) = 5, so 5 = s*w2(s) = s*(1-s). The ideals (s) and (1-s) are prime since they have prime norm. Also, 7 = (1+s)(2-s), and N(1+s) = N(2-s) = 7, so the ideals they generate are prime. Therefore, all ideals of O(K) are equivalent to products of principal ideals, so that H is the trivial group, and K is a PID. In fact, it’s one of four PIDs known not to be Euclidean by any function, the other three being generated by SQRT(-43), SQRT(-67), and SQRT(-163).

If K = Q(SQRT(-5)), then m = 10.42 again, so we need to check 2, 3, 5, and 7 again. We have 5 = -SQRT(-5)^2. But the other elements don’t split so nicely. If r = SQRT(-5), then we have (2) = (2, 1+r)(2, 1+r), (3) = (3, 1+r)(3, 1-r), (7) = (7, 3+r)(7, 3-r). The ideal (2, 1+r) has a principal square, obviously. From way earlier, (3, 1+r) is equivalent to (2, 1+r). Thus, by division, so is (3, 1-r). Also, (7, 3+r)(2, 1+r) = (14, 7+7r, 6+2r, -2+4r) = (3+r), so by division so are (7, 3+r) and (7, 3-r). Hence H has just two elements, the class of principal ideals and the class of principal ideals times (2, 1+r). So H = Z/2Z. We say that h = 2, where h is the size of H.

In the next post, I’ll show how to use the fact that h = 2 in Q(SQRT(-5)) to partially recover unique factorization in the Diophantine equation x^2 + 5 = y^3.

What is “Grown at Home”?

October 30, 2006

About two hours ago, while on the subway, I saw an ad for an oil heater called Intelligent Warmth that praised its product in several different ways. One of them, “more and more of our energy is grown at home” (rough quote), was conspicuous if only for its nationalist assumption of what “home” means.

The energy in question is not grown in New York. It’s probably biofuel grown in the American corn belt, heavily subsidized by the government because its energy return on energy investment ratio is lower than 1. For some reason, New Yorkers are supposed to believe that sending their tax money to Iowans who are already grossly overrepresented in politics is great, but sending a fraction of that money to impoverished Brazilians is evil.

The most frustrating argument for reducing dependence on Saudi oil is “energy independence is a national security issue.” Together with “food self-sufficiency is a national security issue,” it’s about the most irrationally nationalistic argument in politics that you don’t need to be a religious nut to believe.

The havoc that globalization has wreaked on some countries can obscure the fact that overall, autarky is a lot more conducive to war than free trade. Democracies generally don’t fight one another. Countries that freely trade with one another, regardless of their political system, fight one another even less. Warmongering leaders seek autarky for a reason.

I know that being self-sufficient is good for the USA as a country. But what’s good for the national dick size isn’t necessarily good for the people or for the world. Subsidies to inefficient industries are bad for everyone but the select few privileged people who get them. The consequences to the world at large are even more disastrous; the threat of an oil embargo is one of the reasons the US can’t quite bomb every non-nuclear third-world country.

Besides, the idea that faraway Americans are more important than faraway non-Americans makes about as much sense as the average scene in Alice in Wonderland. Americans bitch about foreign aid, which amounts to 0.11% of the USA’s GDP. The bitching tends to be strongest in states that routinely receive much more than that from the federal government, courtesy of the high-income state taxpayer. But nationalism dictates that New Jersey has a moral obligation to subsidize Montana while the country as a whole is considered charitable when it sends the third world peanuts.

Sunday Miscellany

October 29, 2006

First, to my readers in the US, Canada, and Mexico: switch back your clocks if you haven’t done so already (I presume everyone in the EU and Russia has already switched).

Second, you can celebrate the extra hour of sleep you got tonight by reading the 52nd Carnival of the Godless, posted on Skeptic’s Rant.

The Ideal Class Group is Finite

October 29, 2006

In this post I’m going to prove that every number field K has only finitely many ideal classes. Next post, I’m going to compute some explicit class groups and show how this whole thing is applicable to Diophantine equations where we must use rings of integers that don’t have unique factorizations.

First, if X = {x1, x2, …, x(n)} is an integral basis of O(K), and the set of ring homomorphisms from K to C is {w1, w2, …, w(n)}, then let x(i, j) be w(i)(x(j)), that is that ith conjugate of x(j). Each x(i, j) has an absolute value inherited from C, which may or may not be equal to N(x(j)). Now, define m(i) to be the sum of |x(i, j)| over all j, and m to be product of m(i) over all i. Then m is a positive real number depending on K and X.

Second, every ideal I of O(K) has some element a such that |N(a)| <= m*N(I). To see why, define k to be the integer such that k^n <= N(I) < k^(n+1) (see errata below), where n is the degree of K over Q. Now, consider the set S of elements of the form a1x1 + a2x2 + … + a(n)x(n), where every a(i) is an integer between 0 and k inclusive. S has k^(n+1) elements, so the natural function from S to O(K)/I can’t be injective.

Now, we can find two distinct elements in S with the same image in O(K)/I, defined by, say, b(j) and c(j). Then bc = (b1-c1)x1 + … + (b(n)-c(n))x(n) is in I, and for each j, –k <= b(j)-c(j) <= k. |N(bc)| is the product of the absolute values of the conjugates of (b1-c1)x1 + … + (b(n)-c(n))x(n); each absolute value is at most the absolute value of the sum of that conjugate of each (b(j)-c(j))x(j) over all j, which is at most kx(j). The |N(bc)| is at most the product of the sum of kx(i, j), which is just (k^n)*m <= N(I)*m.

The bound here, m, is really bad. Using some theorems on lattices in space, it’s possible to get a far lower bound; for instance, if K = Q(SQRT(-5)), the lowest m is derived from the integral basis {1, SQRT(-5)}, for which m = 10.47 to two decimal places. The theorem on lattices gives m = 1.42.

The third step is showing that every ideal I is equivalent in the class group to some ideal of norm less than m. To see why that holds, we use fractional ideals. If I^(-1) is the inverse of I, then we can find some c such that cI^(-1) is an integral ideal of O(K). Then there exists an a in I with |N(a)| <= m*N(cI^(-1)). Since (a) is contained in cI^(-1), it is divisible by it, so their quotient, (a/c)I, is an integral ideal. But now N((a/c)I) = |N(a)|/N(cI^(-1)) <= m.

Recall that the definition of the class group is H = F/P, where F is the multiplicative group of fractional ideals and P is the multiplicative group of principal fractional ideals. Two ideals I and J have the same representative in H if we can find some element b in the field such that I = bJ. But that shows that (a/c)I is equivalent to I in H.

Finally, there are only finitely many ideals of norm at most m. But this means that every ideal is equivalent in H to an ideal drawn from a finite set. In particular, every element of H has a representative in a finite set. So H is finite, and we’re done.

Errata: the bound on Q(SQRT(-5)) is not 1.42 by the theorem on lattices. The bound is never less than 2. The actual bound is ((4/pi)^s)*(n!/n^n)*SQRT(|d(K)|), where n is the degree of K and n = r + 2s, where r is the number of real conjugates of K and s is the number of complex conjugate pairs of conjugates of K. For K = SQRT(-5), n = 2, r = 0, and s = 1, so the bound is (4/pi)*(1/2)*SQRT(20) = 2.85. The estimate of 1.42 was obtained by erroneously letting d(K) = 5.

Also, thanks to commenter Bob: k^(n+1) is a mistake. The set S has (k+1)^n elements. There are k+1 choices for each a(i), and n different i‘s. We can’t define k to satisfy k^n <= N(I) < k^(n+1), because we’re not guaranteed to have such a k, let alone a unique one. What we need is k^n <= N(I) < (k+1)^n.

The Religion of Peace

October 29, 2006

There’s a website called The Religion of Peace, which documents how Islam is not really a religion of peace, despite what too many apologists say. The basic concept of the site is sound: highlight violence caused in the name of Islam, and show that mainstream Muslim groups don’t do enough to admonish Islamists. I’ve written a novel that does the same to Christianity.

But what is not sound is the formation of an anti-Muslim nationalism, to use Orwell’s term. The people on The Religion of Peace display the exact same characteristics that anti-Russian Trotskyists did back when they were more than a tiny fringe: they downplay moderation, portray fundamentalism as an exclusively Muslim problem, and pretend Muslims don’t suffer from anything.

Exhibit 1: the site links to a news story, “Relaxed sheik gets rock star treatment.” The sheikh in question said uncovered women were meat for cats. But the rock star treatment was only within the sheikh’s own congregation. Most Muslim organizations in Australia immediately disowned him; even his own mosque suspended him from preaching for three months.

Exhibit 2: the site not only believes the most tenuous attacks on the latest Lancet study of Iraqi civilian casualties, but even has a counter, “Civilian lives saved by military intervention.” How am I supposed to take seriously a website that ignores serious epidemiology in order to support an invasion that only strengthened Islamism?

Exhibit 3: the site features a counter, “Islamic Terrorists Have Carried Out More Than x Deadly Terror Attacks Since 9/11″ (right now, x = 6264). But as both looking at the list of attacks and reading what liberal Muslims say will confirm, most of these attacks are on other Muslims. The purpose underlying the counter isn’t to highlight that the Middle East has lots of violence, but to scare Westerners into hating all Muslims, even though the number of Muslim terrorist attacks on first-world countries is scant.

It’s not surprising that in unstable regions of the world, there will be endemic terrorist attacks. In Sri Lanka, a country with 20 million people, the war between the Buddhist Sinhala government and the Hindu Tamil Tigers has killed more than 50,000 people in a little over 20 years. The LTTE doesn’t support Hindutva, but then again many Muslim terrorists aren’t Islamists, either, and many more are Islamists but have primarily nationalist rather than religious aspirations.

The website reminds me of some libertarian anti-communist rhetoric. Not content with merely showing that communism has caused mass murder, they proceed to support McCarthyism, contend that every mass murder in the last 100 years can be traced to Marxist influence, and downplay the evils of fascism and religious fundamentalism.

Look, Islamism is bad. So is bombing random countries and occupying them incompetently. There’s something totalitarian about turning opposition to one totalitarian movement into rationalization of warmongering so long as the victims are on the other side.

Gay Marriage Politics

October 28, 2006

Lindsay contributed to the Washington Post’s latest question to bloggers, “Should homosexuals be allowed to marry?”. Lindsay’s own response is a fairly good summary of the civil rights case for gay marriage, but I’d rather focus on another response, by Carl Senna. Carl Senna talks mostly about the political angle,

But as far as the politics is concerned, I believe that it has been a major political error for homosexuals to have forced the social decision to recognize homosexual marriages on the legislature. In hindsight, an organized religious authority set up for homosexuals would have been the least fractious politically for homosexuals to defend their right to marriage. A religion that preached homosexuality for homosexuals would have been able to participate in inter-faith discussions.

Actually, forcing civil rights on legislatures has been the politically smartest move in the last 20 years not only for the American gay rights movement but for the entire American left. By forcing the issue out in the open, gay marriage advocates managed to squeeze into mainstream liberalism in a time of a political shift to the right. Public support for civil unions in the US has soared in the last 6 years. In 2000, it was a fringe liberal issue; in 2002, it was a mainstream liberal one; by 2004, it was supported by the majority of Americans.

The anti-SSM propositions that have passed in many states didn’t change much. All they did was proclaim, “Our state doesn’t recognize gay marriage.” In the one state where it would’ve changed anything, Massachusetts, no such proposition has passed. Bush won in 2004 because of terrorism, not gay marriage. Even now, it looks like the Republicans are not going to be able to get votes out of the New Jersey decision.

The alternative proposal in the response, a gay church, just wouldn’t work. Interfaith coalitions don’t include everyone; they only include the largest churches, and typically the ones that acquiesce to the status quo the most. An interfaith coalition can work when different homophobic religions join together and proclaim that homosexuality is evil. It can’t work when a gay church wants to join in.

Besides, the greatest support for gay rights has come from secularists. The sort of people who support Michael Newdow almost invariably support SSM, which they see as another civil right the religious establishment suppresses. Even more moderate secularists, who are just passionate about evolution in schools, tend to view SSM bans as an extension of religious fundamentalism. Working via the church will seriously undermine that support base for no reason.

Welcome all Kossacks

October 28, 2006

And thanks to DarkSyde for the plug. The plugging post is worth reading, if only because DarkSyde explains cosmic inflation in a remarkably clear, remarkably awe-inspiring way. After explaining the mechanism, he analogizes,

And that’s why today,at very large scales, the universe looks like someone took a stick of dynamite, put it in a can of paint (Paint starstuff/galaxies), set it off, and took a high speed photo of the resulting explosion showing streamers, sheets, drops, and fliaments of paint flying apart.

Go read the rest. Ultimately it’s a summary of a much deeper but less comprehensible post by Sean Carroll about inflationary models in cosmology.
Now I need to make sure I go to sleep at normal hours instead of just before sunrise, so that among other things I don’t post a welcome ten hours after the fact.

Norms, Part 2

October 28, 2006

If I and J are coprime ideals of a Dedekind domain R, that is I + J = R, then by unique factorization into prime ideals, I and J have no prime ideal factors in common. The intersection of I and J is just IJ, since it’s defined by the maximum powers of the ideal factors, and IJ is defined by the sum of powers. More importantly, R/IJ and (R/I)*(R/J) have the same structure, where R1*R2 is defined like Z^2 = Z*Z.

To see why, it’s enough to construct an invertible ring homomorphism from one to the other. We can send a + IJ to (a + I, a + J); it’ll clearly be a ring homomorphism. The homomorphism is 1-to-1, since (a + I, a + J) = (b + I, b + J) implies that ab is in I and J, so that it’s in IJ and a + IJ = b + IJ. It’s onto, since given a + I and b + J, we can find c with c + I = a + I and c + J = b + J.

The last statement is true because otherwise, a + I and b + J have empty intersection. That would mean that I and (ba) + J have empty intersection. But that’s a contradiction: R = I + J implies that (ba) = r1 + r2 where r1 is in I and r2 is in J, so I and (ba) + J intersect in r1 = (ba) – r2.

In particular, rings with the same structure have the same number of elements. So if R is a ring of integers of a number field, we have N(IJ) = N(I)N(J).

To show it holds even when I and J are not coprime, it’s enough to show it holds when I is prime and J is a power of I, I^n. Then it’ll hold for all ideals by decomposing them into primes. In fact, it’s enough to show N(I^n) = N(I)N(I^(n-1)). But it’s true that R/I^(n-1) has the same structure as (R/I^n)/(I^(n-1)/I^n); just think about it for a little while, and remember that I^(n-1) is a bigger ideal than I^n since it contains more elements.

Now, that sameness means that N(I^(n-1)) times the number of elements of I^(n-1)/I^n is equal to N(I^n). So it’s enough to show that I^(n-1)/I^n has N(I) elements. To see that, fix an element c in I^(n-1) that isn’t in I^n, and consider the function that sends a + I to ac + I^n. This is not a ring homomorphism – it’s only a homomorphism of the additive groups – but it’s good enough.

This function is 1-to-1, since if ac + I^n = bc + I^n, then (ab)c is in I^n. The set of all elements ab of R that satisfy that condition is an ideal containing I and not equal to R, since 1 is not in it by choice of c. I is prime, so it’s a maximal ideal, which means that ideal is just I; then a + I = b + I.

This function is also onto. Otherwise, there’s some b + I^n, where b is in I^(n-1), that doesn’t intersect (c) + I^n. But the ideal factorization of c includes I to the power of n-1 by the choice of c, so (c) + I^n = I^(n-1), and we can write b in I^(n-1) as cr + d where r is in R and d is in I^n. Then b + I^n and (c) + I^n both contain cr + d.

This shows that for all ideals I and J, N(I)N(J) = N(IJ). This is useful for, among other things, showing that there are only finitely many ideals of a given norm.

That is true since if P is a prime ideal, then the set of all rational integers in P is a prime ideal of Z, generated by p. Since p is in I, N(p) = p^n (since all conjugates of the field K fix Q) is divisible by N(P), so that N(P) = p^k for some integer k. In particular, P contains its norm, p^k. So if I = P1P2…P(r) then I contains N(P1)N(P2)…N(P(r)) = N(I).

So every ideal of norm m contains m. Writing out (m) as P1P2…P(r), we get that every ideal of norm m is some combination of the P(i)’s, since in divides (m). There are only finitely many combinations; hence, only finitely many ideals of norm m. More generally, there are only finitely many ideals of norm at most m.

Male Versus Female Circumcision

October 28, 2006

When I was 9 days old, my parents paid a moyel to hack off part of my penis. Judaism mandates 8 days, but they lied about my age to make the ceremony not fall on the Sabbath. Of course, I don’t remember anything about it, save what they’ve told me, since 8 day old babies are still in or barely out of the gray area between sentient and non-sentient.

Now that Jill’s writing about the trial of a man in Georgia who cut off his daughter’s clitoris with scissors when she was 2, I feel I should explain why the two practices are entirely different.

First, as commenters both on Feministe and Majikthise have explained, male circumcision only affects the foreskin, while female circumcision affects the entire clitoris. The equivalent of female circumcision would be cutting off the entire penis head.

Second, what the commenters haven’t pointed out is the age difference. I don’t have any psychological trauma from my experience more than 18 years ago. Why would I? I don’t have any trauma from my birth, either, even though it was at least as violent. Obviously, if it had been done to me when I was 2, let alone 12, I’d have remembered it vividly.
Third, circumcision was never intended to sexually restrict men – and indeed it doesn’t (circumcision makes you and your partner enjoy sex less, apparently, but they don’t destroy sexual function). Female genital mutilation is and does, especially when combined with sewing the victim’s vagina shut.

There’s legitimate men’s rights activism around issues like circumcision, the emotional stunting of boys in education, conscription, male victims of domestic violence, and ignorance of male victims of sexual assault. The problem is that men’s rights activists never concentrate on them; instead, they concentrate on making it look as if gender inequality doesn’t exist, and as if cutting off a 12-year-old woman’s clit to deprive her of sexual pleasure is not any worse than cutting off a newborn’s foreskin.

The Black Vote

October 28, 2006

High-up Democrats are complaining that many blacks will not vote because of disillusionment with the process. Usually it’s something I’d attribute entirely to the Democrats’ conservatism on welfare reform, education funding, and health care; if you can no longer take a group’s vote for granted while screwing it when in power, it’s your fault, not the group’s.

But in this case, it’s not about social welfare, but about education. Says the New York Times,

Democrats’ worries are backed up by a Pew Research Center report that found that blacks were twice as likely now than they were in 2004 to say they had little or no confidence in the voting system, rising to 29 percent from 15 percent.

And more than three times as many blacks as whites — 29 percent versus 8 percent — say they do not believe that their vote will be accurately tallied.

Voting experts say the disillusionment is the cumulative effect of election problems in 2000 and 2004, and a reaction to new identification and voter registration laws.

Fatalism among the lower classes isn’t a particularly shocking thing; to a great degree, electoral politics is about civic activists finding ways to mobilize the fatalist masses to vote for them (and reneging on their promises after winning). Even doubts about the voting system aren’t an out of the ordinary thing in a country with a low-income ethnic subculture. Judging by political commercials, they are or at least used to be pretty serious among Russian immigrants in Israel, where voting is done by paper ballots.

What is unique to the US is a voting system that causes people to be so fatalistic they’re expressing doubts about the validity of their vote. One person cited in the article says, “I realized that maybe the poll tax isn’t gone after all.” I’ve written before about how some of the peculiarities of American bureaucracy contribute to a difficult election process.

First, there are no mandatory ID cards. This is a good thing, but one of the side effects of that is the proliferation of fakable IDs. Passports could put an end to this, but most Americans don’t have passports – when the world ends at your country’s borders, you don’t need a document useful only to travel abroad – and the government doesn’t even take the basic step of making sure everyone has some government-issued photo ID that it’s not trivial to forge.

Second, the standard method of ID verification in this country, the social security number, is too porous. It’s workable when you ask for information over the phone, but in person, it’s vastly inferior to a federally-issued card that has your name, photo, place of residence, and social security number in it.

And third, electronic voting reminds me of a joke I once got by email. “American astronauts found that their pens couldn’t work in zero gravity; NASA then contracted devising a pen that could work in zero gravity to a corporation that spent ten million dollars on developing a pen that worked underwater, in zero gravity, and in a vacuum. The Russians just wrote in pencil.” India manages to have elections with hardware-based voting machines, and Canada uses paper ballots. I’m going to venture a guess that either country spends vastly less per capita on holding elections than the US.

Although the Democrats aren’t doing anything to make sure everyone’s vote is counted, except after the fact, the Republicans bear the brunt of the blame. It’s not the Democratic Party that’s circulating pamphlets in majority-minority areas telling people they can’t vote if they got parking tickets. It’s not the Democratic Party that sends operatives to challenge the legitimacy of black voters and then calls people who complain about voter suppression paranoid.

No, the Democratic Party’s problem is in failing to solve the problem. It can’t be that hard to tell the Republican majority, “Okay, let’s mandate photo IDs for voting, as long as we can ensure that every eligible voter can get a free government ID by election day.” Enough moderate Republicans will accept the deal for it to pass; if they don’t, that’s where comparisons to totalitarian countries’ ban on emigration come in (every US citizen who can afford a passport can get one, but the restriction on traveling without a passport comes from the government).

Friday Night Links

October 27, 2006

Liza of CultureKitchen writes about misogyny and the Sandinistas, in light of the total abortion ban its Congress passed yesterday, which imposes a 30-year prison sentence on women who abort. Liza explains how the ban is related to the betrayal of socialist principles by Daniel Ortega and his supporters, and talks about the intimate connection between feminism and Liberation Theology. Finally, she concludes,

Abortion would be completely contrary to Liberation Theology. There is no question about it. But for liberation theologists to push for a complete ban on abortion? That’s completely unheard of. I find it hard to believe that Liberation theologists would go on a full attack of women’s sovereignty. I believe the case is more about the corruption of the revolutionary movement, a corruption that is symbolized by Daniel Ortega’s 20 years of sexual domination and raping of his stepdaughter, Zoilamérica Narváez.

The French police seems determined to make sure that the riots of last year happen again this year. I didn’t mention the trigger of the original riots in my original article about them (sequel promised if the riots recur): two Arab-French youths were electrocuted while running from the police. France has enough ghettoization and police racism that putting more police officers in Arab areas will just make things worse.

As a consequence, the National Syndicate of Police Officers (SNOP) has demanded that reinforcements be deployed in the departement of Seine-Saint-Denis, just north of Paris, because ‘the delinquents in certain housing estates are preparing to violently ‘celebrate’ (last year’s) events.’

So great is the fear about a renewal of violence, that Interior Nicolas Sarkozy said this week he will draw up a bill that would make it a crime – rather than a misdemeanor, as it is currently – to attack police officers, gendarmes or fire-fighters and will propose a law to treat juvenile repeat offenders as adults in court.

The degree of repression countries will go to to avoid cracking down on discrimination is astounding. When Syria or Jordan slaughters Palestinians it’s understandable – neither has been famous for its democratic governance – but why France keeps trying the failed law-and-order solution is beyond me.

DarkSyde interviews Karen Wehrstein, a highly talented science and science fiction illustrator who’s responsible to, among other things, the cover of Kosmos, and many of DarkSyde’s Science Friday pictures. Besides being a great drawer and photoshop artist, she apparently has world creation ideas that make most popular science fiction look cretinous.

Gordo attacks coded racism in Republican ads and the Republican response to it.

When a Republican gets called out for being insensitive or for using coded racism, he turns around and says that his critics are trying to use race to divide the electorate. It goes like this:

Democrat: What’s your position on affirmative action?

Republican: I’m against it. If you want my opponent’s position, you’ll have to ask one of the many white women who work on his staff?

Democrat: Are you trying to say I’m too friendly with white women?

Republican: I don’t want to hug THAT tar baby.

Citizen: Tar baby? Isn’t that a bit insensitive?
Republican: I think YOU’RE being OVERsensitive. But I understand that you people are hot-blooded.
Citizen: What? That’s a bigoted stereotype! You take that back!
Republican: Take it back? Do you think I’m an Indian giver?
Citizen: Apologize!
Republican: See, this is what happens. Every chance you get, you people start mau-mauingg on the race issue, trying to divide us instead of uniting us.

The Republican base may hate the French, but its ethnic policies are remarkably similar to France’s: sweep the issue under the carpet, pretend everyone’s an equal citizen and no racism exists, and bash everyone who says that the emperor has no clothes for trying to divide the country.


October 27, 2006

Although the main result of playing with discriminants in number fields is the existence of integral bases, there are a few other important consequences. For one, proving the finiteness of the class group is based on some juggling around with norms of ideals.

There are three ways of defining norms of ideals, which turn out to be equivalent. The first is by trying to extend the definition of the norm of an element by multiplication; but for that we need to first show that the class group is finite. The second is by letting the norm of an ideal I of O(K) be the number of elements of O(K)/I. The third is by playing with integral bases more.

The proof that every ring of integers of a number field has an integral basis relies on two ideas: first, that O(K) has n elements that are linearly independent over Q (where n is as usual the degree of K), and second, that if such a set is not an integral basis then we can find another set of smaller discriminant. Part two works for any nonzero ideal of O(K) – just do a global search and replace and write I instead of O(K).

For part one, the key fact is that every ideal contains a rational integer, say m. If X = {x1, x2, …, x(n)} is a linearly independent set of O(K), then so is mX = {mx1, mx2, …, mx(n)}, since m is in Q. Also, since the x(i)’s are in O(K), every mx(i) is in I.

Now, given the prominence of the smallest (or highest if it’s negative) discriminant in the proof, it makes sense to give it a special name – the discriminant of K, denoted d(K). The proof shows that every set of discriminant d(K) is an integral basis. In fact, the converse holds: every integral basis has discriminant d(K).

To see why, let A be the matrix corresponding to some integral basis X of discriminant d(K). Let Y = {y1… y(n)} be another integral basis, and write y(j) as b(1j)x1 + b(2j)x2 + … + b(nj)x(n). Let B be the matrix of the elements b(ij). B has integer elements, since X is an integral basis.

We can multiply matrices. The multiplication is weird at first, because it’s based on the original definition of matrices as linear functions on vector spaces. If AB = C, then c(ij) = a(i1)b(1j) + a(i2)b(2j) + … + a(in)b(nj). In general, AB != BA; for example, if A = [1, 0; 1, 1], B = [2, 0; 0, 1], then AB = [2, 0; 2, 1], but BA = [2, 0; 0, 1]. But in fact, |AB| = |BA| = |A|*|B|.

The reason you should care about this is that the matrix corresponding to Y is just AB, by the definition of the b(ij)’s as coefficients of y(j)’s. The conjugates don’t give us any problems since for every rational number, k*w(x) = w(kx). So disc(Y) = |AB|^2 = (|A|^2)(|B|^2) = disc(X)(|B|^2).

Since B has integer entries, |B| is an integer. So |B|^2 is a positive integer, and all discriminants of integral bases have the same sign. Further, disc(Y) is clearly divisible by disc(X). But by the same argument, disc(X) is divisible by disc(Y), so that disc(X) = disc(Y). It’s justified to define d(K) to be disc(X) for any integral basis, then.

By the same argument, we get that disc(I) is the same regardless of which integral basis for I we choose. We can even recover |B|^2 via disc(I)/d(K), and then take a square root. There are two possible square roots, corresponding to two different values of |B|; we can just take the positive one, and call it N(I).

If c is an element of O(K), then to find the norm of (c), let’s look for a good integral basis of (c). If X is an integral basis of O(K), then cX is an integral basis of (c). All elements of cX are clearly in (c), and we can write every a  in O(K) as a sum of x(i)’s times integers, which lets us write ca as a sum of the cx(i)’s times integers. If A is the matrix of X, then the matrix of cX has a first row equal to this of X times c = w1(c), a second row equal to this of X times w2(c), and so on. So the matrix has determinant |A|*w1(c)*w2(c)*…*w(n)(c) = |A|*N(c). So the norm of (c) is the absolute value of the norm of c.

Finally, N(I) is equal to the size of O(K)/I. Defining integral bases X of O(K) and Y of I, and writing the matrix of Y as B as before, we can add and switch rows or columns, which won’t change the absolute value of |B|. Further, it won’t change I: playing with rows will change X to another integral basis but do nothing to Y, and playing with columns will change Y to another integral basis of I.

There’s a procedure for performing these operations in a way that will change B to a diagonal matrix D, that is one with nonzero values only on the diagonal running from top-left to bottom-right. Then |D| is just the product of the diagonal entries, d11*d22*…*d(nn). But regarding O(K) just as a group and D as a subgroup, we get that O(K)/D has the same structure as Z/d11Z * Z/d22Z * … * Z/d(nn)Z, which has |d11*d22*…*d(nn)| elements.

Feminists for Whose Life?

October 26, 2006

Via Pandagon: Patricia Heaton of Feminists For Life apparently came out publicly against stem cell research. FFL strongly supports the right of a zygote to live, provided it doesn’t spontaneously abort the way 80% of zygotes do. It’s less enthusiastic about the rights of people like those in this ad:

Maybe far right conservatives want to give embryos and people the same rights, because they think people have the same cognitive ability as a blastula.

There are 6.4 Billion People, not 300 Million

October 26, 2006

PZ links to a site that offers American student bloggers scholarships. The basic idea – recognizing bloggers who are college students – is good, but for some reason, the rules require every contestant to be a US citizen. Not just a resident – a citizen. I couldn’t apply even without that restriction since I’m no longer in college, but it still showcases to me how Americans tend to treat non-US citizens like dirt.

A good example of a person who’d never be able to get this scholarship even if she did make it to college is Adama Bah. Bah is an 18-year-old girl who lives in New York is about to deported to Guinea, where she lived until she was 2, because the US government decided she was a potential suicide bomber despite having no evidence to support that belief.

[Link] Adama Bah’s schoolmates were jubilant when she returned to 10th grade at Heritage High School in Manhattan in May 2005 after six weeks in a distant juvenile detention center. Her release put to rest the federal government’s unexplained assertion that Adama, a popular 16-year-old who wore jeans under her Islamic garb, was a potential suicide bomber.

But a year and a half later, with many of her friends planning proms and applying to college, Ms. Bah, now 18, was still wearing an electronic ankle bracelet and tethered to a 10 p.m. government curfew, restrictions that were conditions of her release.

The ability of people not to care about the civil liberties of those who they consider different would astonish me if I weren’t already a cynic. The political meaning of “different” underlies Niemöller’s poem, “First they came…”; the racial meaning is why segregationists torpedoed Roosevelt’s race-neutral universal health care plan; the national meaning is why many American libertarians are literally up in arms about violations of American citizens’ rights but cheer when the US government shreds civil liberties abroad.

Her income fell far short of needs. And though a few community agencies tried to help with diapers for the youngest and trips to a food pantry, she said, the financial crisis deepened. In the end, it was an Islamic political activist in Maryland who came through, taking three of Ms. Bah’s siblings into his home for the summer, and paying $500 a month toward household expenses so she could attend summer school and re-enroll in Heritage this fall.

Incidentally, if the US government is looking for a self-centered reason not to treat non-Americans like dirt, here it is. If people find the government oppressive or even uncaring, they’ll turn to extremists like the activist described above, a former Black Panther who mainstream Muslim groups shy away from.

The Algebraic Closure of the Rational Numbers

October 25, 2006

A while ago, I asserted that the field A is not a number field, since it has infinite dimension over Q. This is because the set {SQRT(2), SQRT(3), SQRT(5), SQRT(7), …} is linearly independent over Q. A commenter named Esther asked me for a proof, so here it is.

Suppose on the contrary that the set is linearly dependent. That is, that there exists a finite subset, say {SQRT(2), SQRT(3), …, SQRT(p)} such that a2*SQRT(2) + a3*SQRT(3) + … + a(p)*SQRT(p) = 0 and the a(p)’s are rational numbers not all zero.

There must be a minimal p that satisfies that condition, that is a p such that {SQRT(2), SQRT(3), …, SQRT(p)} is linearly dependent, but {SQRT(2), SQRT(3), …, SQRT(p‘)} is linearly independent, where p‘ is the largest prime less than p. We can write SQRT(p) as a rational combination of {SQRT(2), SQRT(3), …, SQRT(p‘)}, i.e. a2*SQRT(2) + … + a(p‘)*SQRT(p‘).

Now, let us look at the conjugates of the elements we have. SQRT(p) has minimal polynomial x^2 – p, so it has exactly two conjugates: itself, and -SQRT(p). The conjugates of a2*SQRT(2) + … + a(p‘)*SQRT(p‘) are of the form (+/-)a2*SQRT(2) (+/-) … (+/-) a(p‘)*SQRT(p‘).

Given any two primes between 2 and p‘, say q and r, we can look at a conjugate of a2*SQRT(2) + … + a(p‘)*SQRT(p‘) defined by SQRT(q) –> SQRT(q), SQRT(r) –> -SQRT(r), if it exists. If this conjugate is equal to SQRT(p), then by subtraction, we get that a linear equation on {SQRT(2), SQRT(3), …, SQRT(p‘)} whose SQRT(r) coefficient is 2a(r) is 0. By linear independence, then, a(r) = 0. Similarly, if this conjugate is equal to -SQRT(p), then a(q) = 0.

In other words, given any two different primes, at least one must have a zero coefficient. This means that at most one prime can have a nonzero coefficient, say q. Then a(q)*SQRT(q) = SQRT(p), i.e. (a(q))^2 * q = p in Q, which is false since p/q is never a perfect square for distinct primes p, q.

The final step, showing that the conjugate we want exists, requires working with Q(SQRT(q), SQRT(r)). It has at most four homomorphisms, defined by the images of SQRT(q) and SQRT(r). Showing it has exactly four, i.e. that it has degree 4, will force one of them to be what we want. I’ll do it in another post; here I’ll just show it has degree at least 3. This is enough because one homomorphism either sends SQRT(q) to SQRT(q) and SQRT(r) to -SQRT(r), or SQRT(q) to -SQRT(q) and SQRT(r) to SQRT(r); either will be sufficient for the proof.

But suppose that b1 + b2*SQRT(q) + b3*SQRT(r) = 0, where b(i) is in Q. Then b1 + b2*SQRT(q) = -b3*SQRT(r). Squaring both sides, we get b1^2 + 2b1b2*SQRT(q) + q*b2^2 = r*b3^2. Only SQRT(q) is irrational, so b1b2 = 0. If b1 = 0, then b2 and b3 are 0 since by assumption, SQRT(q) and SQRT(r) are contained in a linearly independent set. If b2 = 0, then b1 + b3*SQRT(r) = 0, so straightforwardly b1 = b3 = 0. Then b1 = b2 = b3 = 0, which means the set {1, SQRT(q), SQRT(r)} is linearly independent, and Q(SQRT(q), SQRT(r)) has degree at least 3.

Counting Arguments

October 25, 2006

In discrete mathematics, people often use counting arguments, which are based on deducing things from sizes of sets.

For example, suppose we have a function from a set with n elements to another set with n elements. This function is 1-to-1 iff it’s onto (1-to-1, or injective, means that f(x) != f(y) when x != y; onto, or surjective, means that every element in the range can be written as f(x) for some x; invertible, or bijective, means 1-to-1 and onto).

To see why, let’s say that the function f from X to Y is not onto. There is some y in Y that isn’t of the form f(x) for some x in X. So the image of f has at most n-1 elements. But X has n elements, so the function can’t be 1-to-1. This is called the pigeonhole principle: if there are more pigeons than pigeonholes, at least two pigeons must share a pigeonhole. Similarly, if the function is not 1-to-1, we must have f(x1) = f(x2) for some distinct x1 and x2 in X. But then the image has at most n-1 elements, whereas Y has n.

In algebra, it’s relevant because it implies that in a finite ring, every element that isn’t a zero divisor is a unit. If a satisfies the condition that ab = 0 implies a = 0, then the function from R to itself that sends x to ax is 1-to-1. This is because ax = ay implies a(xy) = 0, so xy = 0, i.e. x = y. Then the function is onto, and we can find some b with ab = 1. In particular, if R is a finite integral domain, it’s a field. I should add that this requires no additional assumptions; the existence of a implies that R has 1, and even if R is not commutative, the condition implies that ab = ba = 1 for some b.

Going Overboard

October 25, 2006

Living in the West, one is hard-pressed to find extreme pro-Palestinians. There exist nuts like Noam Chomsky, and anti-Semites like Pat Buchanan, but they’re outside the political mainstream. In this climate, anyone who thinks killing innocent civilians is unconditionally wrong and the majority of Israelis and Palestinians would have no trouble accepting a peace agreement that didn’t shaft either side.

However, the blogosphere is hardly representative of the mainstream. It’s not left or right of mainstream, but just has a higher standard deviation from the mean; hence nuts who think either side is composed of genocidal maniacs. Now that Majikthise is being trolled by a commenter who sounds like a Little Green Football regular in reverse, I’d like to make some clarifications about the issue.

Palestine is under occupation. It doesn’t matter who the land originally belonged to; what matters is that it has 3 million people without citizenship and with very limited autonomy. The occupation is illegal and, more importantly, immoral. The same applies to settlements, many of which were created with the explicit purpose of splintering Palestinian land.

Whatever historical injustice may have been done to the Palestinians in 1948-9 is a red herring. It delegitimizes Israel to the same extent that post-WW2 ethnic cleansing means that Germany has a right to annex Posen and East Prussia. And it says nothing about Israel’s demographic future, considering that only 10% of refugees would go to Israel if given the choice.

But the most idiotic thing I’ve seen is the assertion that Israel is committing genocide, and anyone who says otherwise is a Holocaust denier. I’ve seen the word “holocaust” used for four acts: the Jewish holocaust, in which Hitler killed 6 million Jews out of a population of 11 million in Europe over 4 years; the Ukrainian holocaust, in which Stalin engineered a famine that killed 8 million Ukrainians out of a population of 30 million over 9 months; the Cambodian holocaust, in which Pol Pot murdered 1.5-2 million Cambodians out of a population of 8 million over 4 years; and the Armenian holocaust, in which the Ottoman Empire killed 1 million Armenians out of a population of 2 million over 3 years.

Since the beginning of the second Intifada, a little over 4,000 Palestinians have been killed, out of a population of 3 million. It’s bad, but calling it a Holocaust is going overboard by three orders of magnitude. A school shooting that kills 5 people is closer in orders of magnitude to that than the Jewish holocaust.

I still maintain that the best solution to the conflict is to round up the extremists in Israel who believe in ethnically cleansing all Palestinians into Jordan and the extremists in Palestine who want an Islamic state in the entire Cisjordan, and throw them into a desert island where they can kill each other over historical grudges all they like without involving 10 million civilians in the process.