A. Let z = a+bi+cj+dk. When does it commute with i?
zi = ai - b - ck + dj VS. iz = ai - b + ck - dj
zj = aj + bk - c - di VS. jz = aj - bk - c + di
so to commute with i, we need c=d=0, and to commute with j, we need b=c=0. Therefore the only central {z}s are in the reals. Conversely, any real commutes with any H, so the reals are the whole center.

The set {a+bi} is closed under addition, subtraction, and multiplication, and has 0, so it's a subring. Since i is in it it's not central (i doesn't commute with j).

7.1#13a. If n = a^k b, then (ab)^k = a^k b^k = n b^{k-1}, and is therefore a multiple of n.
b. If a + nZ is nilpotent in Z/nZ, then a^k is a multiple of n. Hence every prime dividing n divides a^k. Exactly the same primes occur in a as in a^k. So every prime dividing n divides a. If n=72, then a must be a multiple of 2*3.
c. Let f be a nonzero function from X to a ring S with no nilpotents. (For example, S could be a field.) Then since f is not zero, there exists an x such that f(x) is not zero. Hence f(x)^n is not zero for any n, since S had no nilpotents. This is the value of f^n at the point x, by definition. So f^n is not zero either.

7.1#15. First note that for all a, a+a = (a+a)^2 = a^2+a^2+a^2+a^2 = a+a+a+a. So 0 = a+a for all a.

Now look at a+b = (a+b)^2 = a^2 + ab + ba + b^2 = a + ab + ba + b. So 0 = ab+ba for all a,b, i.e. ab=-ba. But by the first part, -ba=+ba. So R is commutative.

7.4#7. The main thing to note is that R[x] / (x) is isomorphic to R. Then S/I is a domain <=> I is prime, S/I is a field <=> I is maximal.

#8. If (a)=(b), then a is in (b) and b is in (a). So a = bx, b = ay for some x,y. Thus a = ayx, b = bxy, so either a=b=0 or xy=1. The other direction is even easier.

#13a. Let R -phi-> S -> S/P. Then the kernel of R->S/P is phi^{-1}(P). If P is prime in S, then S/P is a domain, therefore so is the image of R in S/P. That image is isomorphic to R/phi^{-1}(P) by the First Iso Thm for rings. So phi^{-1}(P) is either all of R or a prime ideal.

b. Pretty much the same argument, except now S/M is a field, and by assuming phi is surjective we get R/phi^{-1}(P) is also a field (indeed, the same one). Hence phi^{-1}(P) is maximal.

If there were an example R -> S -> S/P, we could just replace the middle S by S/P and the ideal by 0 (the only maximal ideal in S/P). Okay, now we're thinking about R -> S where S is a field. Conversely, the only statement we care about is that the image of R in S is not a field, so we may as well assume R is injecting into S (since only the image matters). Okay, now we're looking at a subring R of a field S and asking for it not to be a field. Z into Q is a fairly obvious example.

#15a. Any polynomial in F_2[x] can be rewritten in the quotient to have no x^2 in it, using the relation.
b. Since 1+1=0, any y+y=y(1+1)=y*0=0. So we get a four-element group with no order 4 elements, hence the Klein.
c. x^2 = x+1, using the polynomial relation. Also, the multiplicative group has 3 elements, so of course it's Z/3.

#19. Let R be finite. Show every prime ideal is maximal.

A. If I is prime, then R/I is an integral domain. Let c be a nonzero element. Then left multiplication by c is 1:1 (since it's a domain) hence it's onto (since it's finite), so it hits 1. I.e. for some d, cd=1. Hence it's a field. So I is maximal.

#30. If i in I, then i^1 in I, hence i in rad I.
Say r+I is in the nilradical of R/I. This means (r+I)^n = 0+I for some n, i.e. r^n + I = I, i.e. r^n in I. So r in rad I. The reverse implication is exactly the same.