abstract algebra?

Question

Let K be ae field K(x) be the field of rational functions in x with coefficients from K.Let t in k(x) be the rational function P(x)/Q(x) with relatively prime polynomials P(x),Q(x) in K[x] with Q(x) is non zero.
1> Show the polynomial P(y)-tQ(y) in variable y and coefficients in K(t) is irreducible over K(t) and x as a root.

2> Show the degree of P(y)-tQ(y) as a polynomial in y with coefficients in K(t) is the maximum of degrees of P(y) and Q(y)

Answer 1

> All of the following can NOT be true of a single polynomial:
>
> A. It is irreducible.
> B. It has a root.
> C. It has a degree greater than 1.

Consider x^2 + 1 as a polynomial with coefficients in the reals and roots in the complex numbers, or x^ - 2 with coefficients in the rationals and roots in the real numbers.

The problem sets up three fields F <= G <= H with

a) F is the field K

c) H is the field K(x) of rational functions in x with coeefificents in K.

b) G is the field K(t), the smallest subfield of K(x) containing K and the specified element t = P(x) / Q(x) in K(x) - K.

The polynomial P(y) - t Q(y) is in the ring of polynomials G[y] = K(t)[y].

It makes perfect sense to ask if the polynomial is irreducible over G = K(t) but not over H = K(x).

Answer 2: Let the degree of P(y) be n and the degree of Q(y) be m, with d = max(n, m). The coefficients of P(y) are all in K. The coefficients of Q(y) are all in K. Therefore all of the nonzero coeffcicients of t Q(y) are in K(t) - K. t is non-zero so the degree of t Q(y) is still m = degree Q(y). Then degree P(y) - t Q(y) must be d = max(n, m) because there can be no cancellation of the highest degree term, given that all of P's nonzero coefficients are in K and all of t Q(y) 's nonzero coefficients are in K(t) - K.

Answer 1 second part: Just evaluate P(y) - t Q(y) at x [y is the indeterminate in the ring of polynomials K(t)[y], and x is an element of the larger field K(x) containing K(t)]. The result is P(x) - t Q(x) = P(x) - (P(x) / Q(x)) Q(x) = P(x) - P(x) = 0 [because t = P(x) / Q(x) ].

Answer 1 first part: I'm not quite sure off hand how to show this. Suppose P(y) - t Q(y) factors as a(y) b(y) in K(t)[y]. The coefficients of a(y) and b(y) cannot all be in K, as then the coefficients of their product would all be in K, but quite obviously P(y) - t Q(y) has at least one nonzero coefficient in K(t) - K. That may lead to something. Likewise, we know x is a root so a(x) b(x) = 0, and thus at least one of a(x) = 0 or b(x) = 0. Assume without loss of generality that a(x) = 0. If you could show the coefficients of a(y) are restricted to K + t K (as opposed to all of K(t)) then that would give you a reduced form of t = R(x) / S(x) contradicting that P(x) and Q(x) are relatively prime.

Answer 2

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Answer 3

No matter how many times you ask this, it remains wrong.

All of the following can NOT be true of a single polynomial:

A. It is irreducible.
B. It has a root.
C. It has a degree greater than 1.