"1. Let R = C[x_1..x_n], and R_d := the homogeneous polynomials of degree d.
For n=2, compute dim(R_d)."
R_d is spanned by the monomials x_1^a x_2^{d-a}, of which there are d+1.
"2. An ideal I in a polynomial ring is called graded if I is the sum over d of (I intersect R_d), i.e., if for every p in I, when we break p into its homogeneous components p_d each of those is also in I. Give an example of an ideal in C[z] that is not graded, and prove that it isn't."
Let I = < x+1 >. Since 1 is not a multiple of 1+x, this is not a graded ideal.
"3. If I,J are graded ideals, show I+J and (I intersect J) are graded too."
Anything in I+J is of the form i+j. When we break i,j into their homogeneous components, we get things in I and J, hence in I+J.
If p is in I intersect J, then p is in I and in J, so each p_d is in I and in J, so each p_d is in I intersect J.
"4. Let M_n(C) denote the noncommutative (!) ring of nxn complex matrices, where multiplication is matrix multiplication.
Fix T an element of M_n(C).
Define f : C[z] -> M_n(C), taking p(z) |-> p(T).
Let I = ker(f). This is principal (being an ideal of C[z]), i.e. I = < m(z) > for some m(z). What is this polynomial called in a linear algebra class?
Compute it for T = identity matrix."
This polynomial is called the minimal polynomial (and divides the characteristic polynomial, by the Cayley-Hamilton theorem). For the identity matrix it's
"5. Let p(z),q(z) be polynomials, and I = < p(z),q(z) >. Since I is principal, it's generated by some element r(z) = a(z)p(z) + b(z)q(z). What is this r(z) called? (some name involving p,q)"
The greatest common divisor of p & q. Pretty cool that it's also a linear combination of them.
[Do you see how to use the division algorithm to compute the a(z),b(z)?]
Let's do induction on deg(p)+deg(q). If deg(p) > deg(q), switch them without loss of generality. The division algorithm says q = mp + s, with deg(s) < deg(p) <= deg(q), so the pair (p,s) has lower total degree than (q,p). By induction, we can write gcd(p,s) = cp+ds for some polynomials c,d. Hence gcd(q,p) = gcd(p,s) = cp + ds = cp + d(q-mp) = dq + (c-md)p, which is what we wanted to compute.
"6. If p,q are two polynomials in C[z] such that p^2 = q^3, show that there exists another polynomial r such that p = r^3, q = r^2. (Hint: use the Fundamental Theorem of Algebra, that every polynomial in C[z] factors as a number times a product of (z-a_i)^{m_i}, where the m_i is the multiplicity of the root a_i.)"
If we factor, we find that the factor z-e shows up in p^2=q^3 the same number of times, M_e. Since M_e has to be a multiple of 2 (from the p^2) and 3 (from the q^3), it's got to be a multiple of 6. Hence p^2=q^3 is a 6th power of a polynomial, r, and this r does the job. Moreover, this r satisfies rq = p.
"7. Let R = C[x,y] / < x^2 - y^3 >. Use #6 to show that R is not isomorphic to C[z]."
Assume phi : R -> C[z] is an isomorphism, taking x |-> p(z) and y |-> q(z). Then by #6, there is some r(z) in C[z] with r^3 = p, r^2 = q. Since phi is a correspondence, there must be some f in R such that phi(f) = r. Since it is a ring isomorphism, this f must have fy = x. We want to show that no such f can exist.
This is a little confusing inside the quotient ring R, so pick F(x,y) an honest polynomial in the translate f = F +
Another way to say this is to look at the equation "fy=x" inside the quotient ring R / < y >
