Thank you for spreading out and making sure we as a class finished Step 3 of Field of Fractions. (Although a few more selected the easier ones than I would've liked.)
For all of this being explicit about where you use properties of D would be great. Working within the Field of Fractions as described (equivalence classes of ordered pairs) is required. Please also work with the equivalence as defined, rather than as we're familiar with. So, please do not "multiply by b/b" or similar.
21.6 For associativity, I recommend starting at each end and computing until you reach a common result in the middle. This requires using the distributive properties of D and the associative property of addition. No equivalence is needed.
21.7 This is perhaps the most straightforward of the bunch, and probably why the multiplicative identity wasn't even asked to be proven. You only need to prove in one direction since step 1 was commutativity of addition. Be sure to use the definition of addition given in Lemma 21.3. No equivalence needed. This uses the 0 and 1 properties for D.
21.8 This does require equivalence, and uses properties of 0 and 1 in D. It also requires multiplicative commutativity of D and inverses in D.
21.9 For associativity, I recommend starting at each end and computing until you reach a common result in the middle. Notice that this hinges on associativity of D in the middle. Almost nothing else is used. No equivalence needed.
21.10 For commutativity, I recommend starting at each end and computing until you reach a common result in the middle. Notice that this hinges on commutativity of D in the middle. Almost nothing else is used. No equivalence needed.
21.11 This is surely one of the more involved ones. It does require equivalence. It's probably, again, easiest to compute both and show they reach the same in the middle.
For the next set, I hope everyone did look at my notes on our coursepage.
21.12 [t,t] is a good unit for t in T (notice [(1,1)] may not be in Q(R,T) if 1 is not in T), and letting i(t) = [tt,t], then [t,tt] is a good inverse. Once you have these ideas, the proof should be direct, as long as you aren't scared with many ts in a row.
21.13 If we let T = {a^n|n a natural number}, then T is multiplicatively closed and does not contain zero divisors (if a^2 is a zero divisor, a^2b=0, then a(ab)=0 and a is zero divisor). So, we can apply the construction in 21.12 to R,T. This does not change the characteristic, but does use (and also preserve) commutativity. 19.30 does not use commutativity.
21.14 As I talked about in the Canvas question period, RxT = {(0,1),(1,1),(2,1),(3,1),(0,3),(1,3),(2,3),(3,3)}. It turns out because of multiplication mod 4 that (0,1) ~ (0,3), (1,1) ~ (3,3), (2,1) ~ (2,3), (3,1) ~ (1,3) (note: the last two do require Z_4 multiplication). There are no other equivalent pairs, so the set of elements in Q(R,T) ={[(0,1)],[(1,1)],[(2,1)],[(3,1)]}, there are four.
21.15 The subring of R is {a/(2^n)|a an integer, n a natural}. The isomorphism is [(a,b)] ->a/b. You need to prove this is an isomorphism, including well-definedness (i.e. if (a,b) ~ (c,d) then a/b = c/d in R).
21.16 The subring of R is {3a/(6^n)|a an integer, n a natural}. The isomorphism is [(a,b)] ->a/b. You need to prove this is an isomorphism, including well-definedness (i.e. if (a,b) ~ (c,d) then a/b = c/d in R).
21.17 The first problem here is in transitivity of the equivalence relation (the last part of step 1). Using the notation at the bottom of p. 191, we have asd = brd, but if d, in T, is a zero divisor, and as makes asd = 0, then we get 0 = 0 and may not conclude as = br. With respect to the set T = {1, 2, 4}, we want to set this up to cause problems, so, we see using d = 2 and a = 3 and s = 1 will make asd = 0. So, in this context we are trying to prove: if (3,2) ~ (0,2) (this is true mod 6, but does produce 0 = 0), and (0,2) ~ (0,1) (this is true and unsurprising), but (3,2) is not ~ (0,1) (see 3 ≠ 0). So, the equivalence relation is not transitive. This is why Fraleigh says "this is a crucial step in the argument".
22.24 We're proving "if D has no zero divisors, then D[x] has no zero divisors" (at least that's the key step, all other properties inherit nicely). Notice the "no" in both clauses. This is much easier with contrapositive "If D[x] has zero divisors, then D has zero divisors." Suppose a(x), b(x) not zero in D[x] are such that a(x)b(x) = 0. If they are constants, we are done. In any case, look at the leading coefficients (which must also not be zero), a_n b_k multiply together to be zero and we have zero divisors in D.
22.25 b. The units in Z[x] are {1,-1}, c. The units in Z_7[x] are {1,2,3,4,5,6}, a. the units in D[x] are the units in D (this is why 2 is not a unit in Z[x]).
22.26 Nothing profound here, but good practice with generic polynomial operations. Be sure you're explicit about where you use properties of R. Also, as he and I both did for associativity, it is better to use different indices for each polynomial summation.
22.27 D is actually differentiation. It's straightforward to show D(p(x) + q(x)) = D(p(x)) + D(q(x)) by computing both sides. It is not a ring homomorphism, because you know about derivatives, D(x*x) = D(x^2) = 2x, but D(x)D(x) = 1*1=1. The kernel is the set of all polynomials with zero derivative. Those are the constant polynomials. The image of F[x] under D is all of F[x], the function is onto. An onto proof is required which will require identifying an antiderivative for each polynomial.
22.28 This is actually rather boring. I am glad most didn't do it.
22.29 All properties here come from the fact that \phi(r) and \psi(r) are in R.
22.30 a. If you add, multiply and negate polynomials, you get polynomials, b. following the hint, in Z_2, F^F has 4 elements, but there are infinitely many polynomials of different degrees. The catch is many different polynomials give the same functions.
22.31 a. 4, 27, b. For Z_2 -> Z_2 we have the constant zero function, the constant 1 function, the identity function, and the function that exchanges 0 and 1. Each of these functions when added to itself is the zero function, so since all elements have order 2, this group is addititively isomorphic to Z_2 x Z_2. Similarly for 9 elements the functions from Z_3 -> Z_3 can either be isomorphic to Z_27, Z_9 x Z_3, but only if one has order nine, which is impossible, so it is also isomorphic to Z_3 x Z_3 x Z_3. c. is fascinating - all functions are polynomials! This makes so much of mathematics much easier. Perhaps unsurprisingly no one got this far. I would be very happy to talk to anyone who wants to sort through the details here.
23.34 By Fermat's little theorem, we know that x^p = x mod p for any x in Z, i.e. in Z_p x^p =x for any x in Z_p. Let x = (-a). So (-a)^p = (-a). Notice then that (-a)^p +a = -a +a = 0, and so (-a) is a zero of x^p+a, so there is a factor, namely x-(-a) = x+a, by the (root-)factor theorem.
23.35 f(a)=0. Evaluate sum(a_{n-i} x^i) at x = 1/a. We get sum(a_{n-i} a^-i). Now multiply it all by a^n, this produces sum(a_{n-i} a^n-i), which is exactly f(a), which is zero.
23.36 The remainder theorem is standard secondary material. f(x) = (x-a)q(x) + r(x) by the division algorithm. Note that the degree of r(x) < degree of (x-a) = 1, so r(x)=r, a constant. We now have f(x) = (x-a)q(x) + r. Notice, though, that if we evaluate at x=a we get f(a) = (a-a)q(a) + r= r, and hence f(a) = r. (and, I think I proved this for you in a video by accident; no regret.)
23.37 Our last problem. 37a is tedious, but not terribly interesting. You need to check homomorphism properties for both operations. Keep track of the fact that the overbar function is for polynomials, and the one without is for numbers. b. Prove the contrapositive, if you have a a factorisation in Q you have one in Z by Gauß's lemma, and you have one in Z_m by applying the function in a to it. c. is an example of this. Reducing mod 7 gives x^3+3x+1, which, if it had a root would either be 1 or -1, and neither works. If you claim it has no zeros in some Z_m, you need to demonstrate that.