A. The nice one is a + bt + (t^2) |-> the matrix with a's on the diagonal and b in the upper corner. In fact for any nonzero n you can safely rescale b by n before putting it up there.

Let t denote the coset t + (t^2) hereafter.

Let I be an ideal of Z[t]/(t^2). Consider its image under the unique homomorphism Phi from Z[t]/(t^2) -> Z, given by t|->0. This image is an ideal in Z, hence of the form (n) for some n. So I is contained in Phi^{-1}( (n) ) = (n,t), and must contain some element n+at.

Let's look also at I intersect (t), a subgroup of (t) = Zt. We know the subgroups of Z; they look like all multiples of some number b. Note that nt is in I, so b must be a divisor of n. (There is a tricky special case of n=0; note that any b divides 0.)

Now I claim that I = (n + at, bt). If i is in I, then Phi(i) is a multiple k of n, so i - k(n + at) is in I intersect (t), hence is a multiple of bt. QED. So every ideal is of the form (n + at, bt) where b is a divisor of n.

If we don't want to overcount (for those who care), we can first insist that n,a,b are all nonnegative. Then if n=0, take a=0. If n is not zero, then b is automatically not zero, and we can insist that a is less than b.

2. Let M be a module over a commutative ring R, and r an element of R. Let rM denote the subset of M consisting of all {rm : m in M}. (One might call this a "principal submodule".)

i. Plainly rM is contained in M. Since r is a unit, rs=1 for some s. So for any m in M, m = 1m = (rs)m = r(sm) in rM.

ii. If R=M=Z and r=2, then rM is just the even integers.

iii. If R=Z, M=a finite abelian group of odd order, and r=2, then rM = M since every element of M is twice something.

3. Let M be a left module over R, and m an element of M. Let I be {r in R : rm = 0}. Show that I is a right ideal, the "annihilator".

A. For all r in R, i in I, (ri)m = r(im) = r0 = 0, so ri in I. Hmm, maybe this is a left ideal. I am never sure about these things.

13.1#4. Let S denote the square root of 2. Consider the map phi: a+bS |-> a-bS from Q[S] to itself. This plainly preserves addition and negation, so we only have to check multiplication.
phi( (a+bS) (c+dS) ) = phi(ac + 2bd + (ad+bc)S) = ac+2bd - (ad+bc)S = (a-bS)(c-dS) yup

13.2#2. We already did F_2[x] / (x^2+x-1) last week. For the other, we mainly want to show that there's an element of multiplicative order 8. If we square x, we get x^2 = 1-x. If we square that, we get x^4 = 1-2x+x^2 = 1-2x+1-x = -1. If we square that, we get x^8 = (-1)^2 = 1. So the order of x divides 8, but isn't 1,2,4 so it is indeed 8. (I'm not going to bother writing down the full multiplication table.)

Incidentally, in the field F_2[x]/cubic, the nonzero elements correspond in a natural way to the vertices in the Fano plane (remember way back when?). Multiplying by an element other than 0,1 gives a permutation of the vertices that takes edges to edges. So, some new insight on that old problem.

13.2#4. Q[2+sqrt 3] = Q[sqrt 3] which is degree 2 over Q. Let a = the positive cube root of 2. Then Q is in Q[1+a+a^2] is in Q[a], which is degree 3 over Q. Since
3 = [Q[a]:Q] = [Q[a] : Q[1+a+a^2]] * [Q[1+a+a^2] : Q],
one of the two numbers on the RHS is 1 and one is 3. Q[1+a+a^2] is not Q, therefore that number isn't 1, hence it's 3.

There are other ways to do this problem, this just looked like the most fun.

13.2#14. Ignore the oddness condition, at first, and just look at F[a] over F[a^2]. I claim that the set {1,a} spans F[a] as a vector space over F[a^2]. Proof: given any polynomial in p(a), we can write it as even(a) 1 + odd(a)/a a where even(a),odd(a) are the even and odd parts of the polynomial p.

That shows that the dimension of F[a] is at _most_ 2. (It might not have been a basis, just a spanning set.) So it's 1 or 2.

Now use the pair of containments F[a] contains F[a^2] contains F. Since [F[a]:F] is odd, so is [F[a]:F[a^2]], so it's not 2 -- it's 1.

13.2#19a. This is just commutativity and distributivity.
b. Pick a basis {b_1...b_n} of K over F; we'll use it to F-linearly isomorph K with F^n. Define a map phi: K->{nxn matrices over F} by
ijth entry of phi(k) = coefficient of b_i when expanding k b_j in the basis
It's easily seen to be a ring homomorphism, so must be 1:1 since it's coming from a field, hence by the First Iso Thm of rings the image is a subring isomorphic to K.