Contact Me

You can contact me at alon_levy12@hotmail.com or alon_levy1@yahoo.com. I check both several times a day, so feel free to email either. I check even the spam folder only slightly less regularly than the inbox, so if your email gets junked, I’ll still probably see it within 24 hours.

Any online stalker worth his weight in stone can find both my university and my department email addresses, but please don’t mail anything to them, because a) I use them less often, and b) if your email gets marked as spam, it’ll never see the light of day.

40 Responses to Contact Me

  1. Justin says:

    I just recently stumbled upon your blog, and I’d just like to say that I enjoy it very much.

    Keep up the good work,

    Justin Gilmore

  2. nida says:

    i just had a question
    that if we have a reducible polynomial, from R=R[x] , polynomials over real numbers and we have a quotient ring R/f(x)R with this particular reducible f(x), how can we show that this quotient ring would not be a field since f(x) is reducible, can we come up with a contradiction somehow that an inverse of f(x) would not exist?
    thanks

  3. Alon Levy says:

    In the quotient ring, f(x) = 0, so it’s trivial that f has no inverse. But to find a nonzero number without an inverse, write f(x) as g(x)h(x); you can do that since f is reducible. Now g(x) and h(x) are nonzero, but g(x)h(x) = 0, so that they’re zero divisors. No zero divisor can have an inverse, because if there exists p(x) with g(x)p(x) = 1, then it means that h(x) = 1*h(x) = g(x)h(x)p(x) = 0, a contradiction.

  4. nida says:

    thanks, and i had another question.
    how can i show that the set of real numbers is a subfield of R/f(x)R
    i know that the real numbers itself is a field, and that R/f(x)R is closed under addition and multiplication, so is it enough to show that if multiplicative inverse and 0 exist then it would be a subfield?

  5. nida says:

    and from the previous proof, why are we saying that g(x)h(x)=0? it has to be f(x) which can be nonzero right?

  6. Alon Levy says:

    Since we’re modding out by f(x), it means that f(x) = 0. Think of it this way: if we want to construct the complex numbers, i.e. C, then we’ll mod out by x^2 + 1. Then x = i, and indeed i^2 + 1 = 0.

    You’re assuming f(x) is reducible, so we can express it as g(x)h(x).

    Finally, the real numbers arise as a subfield of R[x]/f(x) by taking the subring of constants. The sum, product, additive inverses, and reciprocals of constants are all constants, regardless of what we’re modding out by.

  7. nida says:

    but why are we modding it out by f(x),if we had have the field R/f(x) we could have done mod f(x) but our field is R/f(x)R, so we shud mod by f(x)R, but then again thats a set, so we cant do that, but i still dont get the reason to do mod f(x)?

  8. Alon Levy says:

    Oh, I’m just abusing notation. Modding out by the ideal f(x)R[x] can be written as “modding out by f(x),” just like modding out by the ideal (2) in Z is often written as “modding out by 2.”

  9. sehr says:

    can we do this using a specific example, say f(x) = x^2 + 1

  10. Alon Levy says:

    Well, not if you want a reducible f… but try f(x) = x^4 + 1, which for a while I was sure was irreducible, go figure. f breaks down as (x^2 + SQRT(2)x + 1)(x^2 – SQRT(2)x + 1). Then, if we let j be the image of x in the quotient ring, we get that j^2 + SQRT(2)j + 1 is not a unit, or else we eventually get 1 = 0.

    If a quartic is too offputting, let f(x) = x^2 – 1…

  11. sehr says:

    hey
    if R is a ring and r is an element of R then rR is always a subring?
    how can we prove this?

  12. Alon Levy says:

    rR is an ideal; it’s a subring if you allow rings not to have 1. To prove that, note that rR is closed under the ideal operations because ra + rb = r(a + b), ra – rb = r(a – b), bra = rab.

  13. tanya says:

    how can we show that for a field F, F-> F[x]/(x), can be a surjective mapping?

  14. Alon Levy says:

    In F[x]/(x), every f(x) = a(n)x^n + … + a1x + a0 is equivalent to a0. So given any f(x) in F[x]/(x), a0 in F maps to it.

  15. tanya says:

    prove that if n divides p – 1 then the congruence x^n – 1 congruent to 0 mod p, has exactly n solutions

  16. nida says:

    can be solve the congruences x^2 = 1 mod 7
    x^2 = 1 mod 13, for x. using chinese remainder theorem.
    and are these sol unique?.
    and hence is it true for all prime numbers.

  17. Alon Levy says:

    Nida, just use x = (+/-)1 mod 7 and mod 13.

    Tanya, give me a day or two; this is relevant to my research, so once I get around to explaining it, I’ll prove that x^n = 1 mod p has n solutions iff n divides p-1.

  18. Tanya:

    Since Z_p is a field, x^n – 1 has at most n solutions.

    x^n-1 = 0 iff x^n = 1, so by looking at the multiplicative group U(Z_p), we’re looking for elements of order dividing n. Since U(Z_p) is cyclic, we reach the needed result (if G is cyclic of order m, and k divides m, look at the set of all g^{m/k}.)

  19. nida says:

    If f(x) is an element on F[x], polynomial ring, and f(x) has n disticnt roots,
    >in its splitting field E, then galois(E/F) is isomorphic to a subgroup
    >of the symmetric group Sn.

    as far as i know is that :
    The elemnts of Gal(E/F) are the F-automorphisms of E.Any
    h in Gal(E/F) sends roots of f(x) into roots of f(x)
    but where do go from here?

  20. Mary Wang says:

    Could you help me with this very standard problem in Abs Alg:
    Find the Glaois Group of x^8 – 1 over Q.

  21. Alon Levy says:

    Okay… x^8 – 1 factors as (x^4 + 1)(x^2 + 1)(x + 1)(x – 1). The last two factors can be ignored. The second factor splits as (x + i)(x – i) in Q[i], where the first factor splits as (x^2 + i)(x^2 – i). Since -1 is a square in Q[i], it means that if we have a square root of i, we have a square root of -i. So the splitting field has degree 2 over Q[i], hence degree 4 over Q[i]. To see that the Galois group is V rather than Z/4Z, it’s enough to find 3 quadratic subfields (or even 2). Q[i] is obvious. Less obvious is Q[SQRT(2)], since SQRT(i) = (1 + i)/SQRT(2) and SQRT(-i) = (1 – i)/SQRT(2), so adding them we get SQRT(2). Likewise, subtracting them we get SQRT(-2), hence Q[SQRT(-2)].

  22. J.A. says:

    The next time you though indirect- & unconsciously threaten me, God’s attorney’s just like that welcome+ to take your motivated opinions about ‘it’ up with me, so that I can more easily find out, what we motivatedly owe each other & e.g. ourselves, is there anything, I don’t know, if I can explain to myself, it’s a false attitude & This Buddhism, I ‘hopefully’ seem like no member of, what do the challenging matter to ME? Who keeps quiet, agrees with me!
    Greet’s from Yours, faithfully,
    J.A., to be continued.

  23. tc says:

    I am supervising Galois Theory and found your site…hopefully my students don’t. Do you always answer questions or do you often give hints? I would prefer that if my students stumbled upon your site…they would not have found someone to do their work but rather another resource to help them learn the material! But in the end, it is your site and your call :-)

    Cheers,
    –tc

  24. DOUGLAS FIELD says:

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    Most Americans are well aware that there are plenty of innocent poorer US citizens in our prison systems nationwide,but very few of us have ever heard about our US Congressional representatives who are mostly lawyers themselves, denying our Middle Class and Working Poor Americans proper legal counsel for their federal appeals. When any poorer American is charged with a Federal crime all legal costs and appeals are paid for,but when one is charged in the various 50 states and run out of higher court state appeal opportunities and need to appeal to the Federal courts,it appears that these poorer and mostly uneducated prison inmates are left to their own best lawyering skills to attempt to write a federal appeal that could sway a US Federal judge to grant them a new retrial.
    Our US Congressional Representatives sleep soundly every night knowing there are reported to be an estimated 100,000 innocent Americans (some residing for decades even on death row) in our US Prisons who have been denied proper legal counsel to help them attempt to exonerate themselves with their Federal Appeals?
    *USA WHERE FREE SPEECH IS RESPECTED AND PROTECTED

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  39. Laniaz says:

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  40. José says:

    Could you help me with this very easy problem but with which I am having troube: prove that the single transcental extention of the field of the rational numbers is not normal? Q(t):Q is not normal?
    Thank you for the help.

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