4.3 Factor Rings of Polynomials over a Field

1.
a. Here A = {f img f(0) = 0} so A =img x img by Theorem 6 §4.1.
c. Put h = x(x − 1). Clearly h img Aso imgh imgA. If f img A then f(0) = 0 = f(1) so x and x − 1 divide f by Theorem 6 §4.1, say f = xq = (x − 1)p by Theorem 10, §4.2. But x and x − 1 are relatively prime in F[x] (indeed 1 = 1 · x − 1(x − 1)) so

img

Hence fimg img h img, proving that A = h .
2.
a.
c.
e.

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