[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]

- (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).

[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.]

- (a) Let f = ax+b with a not 0. Show that dim
_{k}k[x]/(f) = 1. - (b) Let f be a polynomial in k[x] of degree n. Prove that
dim
_{k}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 dim
_{k}k[x,y]/(y^{m}, x^{n}) = mn. - (d) Given that f is a polynomial only in x,
determine dim
_{k}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 dim
_{k}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
dim
_{k}k[x,y]/(f,g) to be?

- 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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