[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).

Solution:

- (a) Note that r is in ker f if and only if f(r) = ([r]
_{I},[r]_{J}) is 0; i.e., iff ([r]_{I},[r]_{J}) = ([0]_{I},[0]_{J}) iff r is in I and r is in J, hence iff r is in Int(I,J). - (b) Clearly, IJ is contained in both I and J , hence in Int(I,J). Conversely, Int(I,J) = (Int(I,J))(1) = (Int(I,J))(I+J) = ((Int(I,J))I + (Int(I,J))J, but Int(I,J) is in J so ((Int(I,J))I is in JI + IJ. And Int(I,J) is in I so ((Int(I,J))J is in IJ. Hence ((Int(I,J))I + (Int(I,J))J is in IJ + IJ = IJ.
- (c) First of all, g is well-defined: if ([a]
_{I},[b]_{J}) = ([a']_{I},[b']_{J}), then a - a' is in I and b - b' is in J. Thus a+b - (a'+b') = (a-a') + (b-b') is in I+J, so [a+b]_{I+J}= [a'+b']_{I+J}, hence g(([a]_{I},[b]_{J})) = g(([a']_{I},[b']_{J})). Next, g is an additive homomorphism: g(([a]_{I},[b]_{J})+([c]_{I},[d]_{J})) =g(([a+c]_{I},[b+d]_{J})) = [a+c-(b+d)]_{I+J}= [a-b]_{I+J}+ [c-d]_{I+J}= g(([a]_{I},[b]_{J})) + g(([c]_{I},[d]_{J})). Moreover, g is surjective: given any element [a]_{I+J}of R/(I+J), note that g(([a]_{I},[0]_{J}) = [a]_{I+J}, so g is surjective. Finally, Im f = ker g: first we see that Im f is contained in ker g, since anything in Im f is of the form ([r]_{I},[r]_{J}) for some r in R, but g(([r]_{I},[r]_{J})) = [r-r]_{I+J}= [0]_{I+J}. Conversely, if [a]_{I},[b]_{J}) is in ker g, then a-b is in I+J, so a-b = i+j for some i in I and some j in J. Thus a-i = b+j, so f(a-i) = ([a-i]_{I},[a-i]_{J}) = ([a]_{I},[b+j]_{J}) = ([a]_{I},[b]_{J}). - (d) By the First Isomorphism Theorem we know that R/ker f is isomorphic to Im f, but ker f = Int(f,g) so R/Int(I,J) -> R/I x R/J is a composition of the two injective maps R/Int(I,J) -> Im f -> R/I x R/J , so it is injective itself. Now, R/Int(I,J) -> R/I x R/J is surjective if and only if Im f = R/I x R/J iff ker g = R/I x R/J iff (R/I x R/J)/(ker g) = 0 iff R/(I+J) = 0 iff I+J = R iff I+J = (1). [Note: the claim "(R/I x R/J)/(ker g) = 0 iff R/(I+J) = 0" follows since by (c) and the First Isomorphism Theorem we have an isomorphism from (R/I x R/J)/(ker g) to R/(I+J).]

- (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) First solve a12 + b5 = 1. Then we can take x = 7b5 + 3a12. Since -2*12 + 5*5 = 1, we know that x = 7(5*5) + 3(-2*12) = 103 is a solution. Every other solution differs from this one by a multiple of 12*5 = 60. Thus 43 is the least positive solution.
- (b) We now know every solution of 7 mod 12 and 3 mod 5 is 40+r60, for some r. Thus we must find an r such that 40+r60 is congruent to 4 mod 19. I.e., we must solve now solve x = 43 mod 60 and x = 4 mod 19 simultaneously. Working as before, -6*60 + 19*19 = 1, so x = 4(-6*60) + 43(19*19) = 14083. Any other solution differs from this one by a multiple of 5*12*19 = 1140, so the least positive solution is 403 (since 14083 - 403 is divisible by 1140, but clearly this won't work for any psitive integer smaller than 403).

[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 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?

- (a) It's enough to show that {[1]} is a basis in k[x]/(f). But since 1 is not in (f), we know [1] is not 0 in k[x]/(f), so {[1]} is a linearly independent set. To see that it spans, let g be any element of k[x]. By the division algorithm we know g = fq+r for some polynomials q and r where either r is 0 or deg r < deg f. Since deg f = 1, this means r is a constant, so g - fq = r*1, hence [g] = r[1], so [1] spans. Thus {[1]} is a basis.
- (b) We'll prove it directly, using the same method as in (a).
We claim that S = {[1], [x], ..., [x
^{d}]} is a basis, where deg f = d+1. To see that S is a linearly independent set, assume that a_{0}[1] + a_{1}[x] + ... + a_{d}[x^{d}] = [0] for some constants a_{i}. This just means that the polynomial g = a_{0}+ a_{1}x + ... + a_{d}x^{d}is in (f); i.e., that f divides g. But unless all of the a's are 0, deg g < deg f so f can't divide g. To see that S spans, use the division algorithm again to show that every element [h] in k[x]/(f) is [r] for some r of degree at most d: h = fq+r for some q and r, where either r = 0 or deg r < deg f = d+1. - (c) Here we want to show that {[x
^{i}y^{j}] : 0 <= i <= n-1, 0 <= j <= m-1} is a basis for k[x,y]/(y^{m}, x^{n}). But certainly every polynomial in k[x,y] is a sum of monomials, and every monomial except thos in S are in (y^{m}, x^{n}). Thus [g] for every polynomial g in k[x,y] is equal to a linear combination of elements of S. Thus S spans. If S is not linearly independent, then some linear combination h of the monomials {x^{i}y^{j}: 0 <= i <= n-1, 0 <= j <= m-1} is in (y^{m}, x^{n}). But every term of any polynomial in (y^{m}, x^{n}) is divisible by either y^{m}or x^{n}. Since no nonzero term of h is divisible by either, we see h is not in (y^{m}, x^{n}) unless h is the 0 polynomial. Thus S is linearly independent, hence a basis. Now count the lements of S to show that the dimension is mn. - (d) Since f is a polynomial in x, we see (x, y - f) = (x,y), so
dim
_{k}k[x,y]/(x, y - f) = dim_{k}k[x,y]/(x, y) = 1 by (c) (or use dim_{k}k[x,y]/(x, y) = dim_{k}k[x]/(x) = 1, by (a)). - (e) Since h | f and h | g, we see that (f,g) is contained in (h).
But this means that k[x,y]/(f,g) maps onto k[x,y]/(h). In fact, by one of the
isomorphism theorems, (k[x,y]/(f,g))((f,g)/(h)) is isomorphic to k[x,y]/(h).
In any case, dim
_{k}k[x,y]/(f,g) >= dim_{k}k[x,y]/(h). So it is enough to check that dim_{k}k[x,y]/(h) is infinite. To see this, note that since deg h > 0, one of the variables, x or y (or both, but that doesn't matter), must appear in h. WLOG, assume x appears in h. Then S = {[1], [y], [y^{2}], ... } is a linearly independent set, which means that k[x,y]/(h) is infinte dimensional. For if S were not linearly independent, then some nonzero polynomial g which is a polynomial in y only is in (h), hence is divisible by a polynomial with an x in it. But if x appears in h, then it appears in hf for any nonzero f. Thus h can't divide g. - (f) What yuo probably found was that dim
_{k}k[x,y]/(f,g) = (deg f)(deg g) for every randomly chosen f and g. This isn't always true, since (d) gives a counterexample. What is true is that dim_{k}k[x,y]/(f,g) <= (deg f)(deg g), and equality holds except in special cases (which we'll talk about more later).

Note: To gather some data for [3](f), use

- 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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