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 b-c = (b1-c1)x1 + … + (b(n)-c(n))x(n) is in I, and for each j, -k <= b(j)-c(j) <= k. |N(b-c)| 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(b-c)| 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.


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