Due Friday, March 8, 2002

[1] Let I and J be ideals in a ring R. Let f : R -> R/I x R/J be the homomorphism f(r) = ([r]I,[r]J), where, for example, [r]I means the equivalence class r + I in R/I. Denote the intersection of I with J by Int(I,J).
• (a) Show that ker f = Int(I,J).
• (b) If I+J = (1), show that IJ = Int(I,J). [Hint: write a+b = 1 for some a in I and b in J, and note for any c in Int(I,J) that c = ca+cb is in IJ.]
• (c) Show that there is a surjective homomorphism g : R/I x R/J -> R/(I+J) of additive groups given by g([a]I,[b]J) = [a-b]I+J, and that ker g = Im f.
• (d) Conclude by the First Isomorphism Theorem that f induces an injective homomorphism R/Int(I,J) -> R/I x R/J, and that this is surjective if and only if I+J = 1 (in which case we get an isomorphism R/IJ -> R/I x R/J).
Note: What we have actually done here is to show that there is a short exact sequence 0 -> R/Int(I,J) -> R/I x R/J -> R/(I+J) -> 0 of additive groups. [A sequence of maps is exact if the image of each map is the kernel of the next one; it is called short if it has three terms in the middle with zeros on the end.]

[2]
• (a) Find a positive integer x such that x is congruent to 7 mod 12 and to 3 mod 5. (I.e., with R = Z, I = (12) and J = (5), find an x such that in Problem 1 we have f(x) = ([7]12,[3]5).)
• (b) Find a positive integer x such that x is congruent to 7 mod 12, to 3 mod 5 and to 4 mod 19. [Hint: One way to do this is to successively do two congruences at a time; i.e., first do R/(ab) -> R/a x R/b, and then do R/(abc) -> R/(ab) x R/c, which together give the isomorphism R/(abc) -> R/a x R/b x R/c.]
[3] In this problem you may assume that k is algebraically closed, if you like.
• (a) Let f = ax+b with a not 0. Show that dimk k[x]/(f) = 1.
• (b) Let f be a polynomial in k[x] of degree n. Prove that dimk k[x]/(f) = n. [You can prove (b) directly, which also proves part (a). Alternatively, you may assume that k is algebraically closed; then you can prove (b) using (a) and Problem 1.]
• (c) Show that dimk k[x,y]/(ym, xn) = mn.
• (d) Given that f is a polynomial only in x, determine dimk k[x,y]/(x, y - f).
• (e) Let f and g be polynomials in k[x,y]. Assume h divides both f and g, where deg(h) > 0. Show that dimk k[x,y]/(f,g) is infinite.
• (f) What happens in general? If f and g are randomly chosen polynomials of degrees m and n, what do you expect dimk k[x,y]/(f,g) to be?
Note: To gather some data for [3](f), use Macaulay2. Here are some commands that may help. First make sure you've defined your ring; say, R = ZZ/3[x,y] or R = ZZ/31991[x,y,z].
• random(R^4,R^5) : This gives a 4 x 5 matrix of randomly chosen constants.
• random(R^{2,3,6,7},R^5) : This gives a 4 x 5 matrix of homogeneous polynomials in R, where the entries in the first row have degree 2, those in the second row have degree 3, those in the third row have degree 6 and those in the fourth row have degree 7.
• J = ideal(random(R^{2,3,6,7},R^5)) : This makes an ideal named J generated by the entries of the 4 x 5 matrix.
• S = R/J : This makes a quotient ring named S, and makes S the "current" ring. (You may not want to use this much, because it resets your current ring to S.)
• degree(R/J) : IF R/J is a finite dimensional k-vector space, this gives the k-dimension of R/J. Thus degree(k[x,y]/ideal(x,y)) = 1. But "degree" applies to almost anything, and means different things depending on what you plug into it; for example, degree(k[x,y]) = 1 and degree(x^2) = {2}. So be careful.
• Let R = k[x,y,z] for some field k. Let J = ideal(random(R^{2,3},R^1)). Then J is an ideal generated by randomly chosen HOMOGENEOUS polynomials f and g, where deg(f) = 2 and deg(g) = 3. How can you randomly choose nonhomogeneous polynomials? Here's how to get the same effect: just throw z - 1 in. For example, k[x,y,z]/(J+ideal(z-1)) is the same thing as k[x,y]/(f(x,y,1),g(x,y,1)), but f(x,y,1) is really a randomly chosen nonhomogeneous polynomial of degree 2 and g(x,y,1) is a randomly chosen nonhomogeneous polynomial of degree 3.

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