Basic properties of quotients of a polynomial ring

Let R be a commutative ring with 1 \neq 0. Let f(x) \in R[x] be a monic polynomial of degree n \geq 1 (that is, the leading coefficient of f(x) is 1), and let bars denote the natural projection R[x] \rightarrow R[x]/(f(x)).

  1. Show that every element of R[x]/(f(x)) is of the form \overline{p(x)} for some polynomial p(x) \in R[x] of degree less than n.
  2. Prove that if p(x),q(x) \in R[x] are distinct and have degree less than n, then \overline{p(x)} \neq \overline{q(x)}.
  3. Suppose f(x) = a(x)b(x) for some a(x),b(x) \in R[x] of degree less than n. Prove that \overline{p(x)} is a zero divisor in R[x]/(f(x)).
  4. If f(x) = x^n-a for some nilpotent element a \in R, prove that \overline{x} is nilpotent in R[x]/(f(x)).
  5. Let p be a prime, let R = \mathbb{Z}/(p), and let f(x) = x^p - a for some a \in \mathbb{Z}/(p). Prove that \overline{x-a} is nilpotent in R[x]/(f(x)).

We begin with a lemma.

Lemma: Let R be a commutative ring with 1 \neq 0. If f(x) \in R[x] is monic, then for all g(x) \in R[x], \mathsf{deg}(fg) = \mathsf{deg}(f) + \mathsf{deg}(g). Proof: The inequality \mathsf{deg}(fg) \leq \mathsf{deg}(f) + \mathsf{deg}(g) always holds. Write f(x) = x^n + f^\prime(x) and g(x) = b_mx^m + g^\prime(x), where f^\prime and g^\prime have degree less than f and g, respectively. Then (fg)(x) = b_mx^{n+m} + x^ng^\prime(x) + x^mf^\prime(x) + (g^\prime f^\prime)(x); since b_m \neq 0, the degree of fg is the sum of the degrees of f and g. \square

  1. Let \overline{g(x)} \in R[x]/(f(x)), and write f(x) = x^n - f^\prime(x), where the degree of f^\prime is less than n. We proceed by induction on the degree of g.

    For the base case, if the degree of g is less than n, then we may use g(x) itself to represent \overline{g(x)}. For the inductive step, suppose that for some m \geq n-1, if h(x) is a polynomial of degree at most m then there exists p(x) of degree at most n-1 such that \overline{h(x)} = \overline{p(x)}. Let g(x) have degree m+1. We may write g(x) = b_{m+1}x^{m+1} + g^\prime(x), where the degree of g^\prime is at most m. Note that since \overline{f(x)} = 0, \overline{x^n} = \overline{f^\prime}. Now \overline{g(x)} = \overline{b_{m+1} x^{m+1} + g^\prime(x)} = \overline{b_{m+1}x^{m-n+1}}\overline{x^n} + \overline{g^\prime(x)} = \overline{b_{m+1}x^{m-n+1}}\overline{f^\prime(x)} + \overline{g^\prime(x)} = \overline{b_{m+1}x^{m-n+1}f^\prime(x) + g^\prime(x)}. Note that the degree of f^\prime is at most n, so that the degree of x^{m-n+1}f^\prime(x) is at most m. Thus by the induction hypothesis, we have \overline{g(x)} = \overline{p(x)} for some polynomial p(x) of degree less than n.

  2. Suppose p(x),q(x) \in R[x] have degree less than n and that p(x) \neq q(x). Suppose \overline{p(x)} = \overline{q(x)}. Then p(x) - q(x) = f(x)g(x) for some g(x) \in R[x]. Note, however, that p(x)-q(x) has degree less than n, while f(x)g(x) has degree at least n. Thus we have a contradiction.
  3. Suppose f(x) = a(x)b(x), where both a and b have degree less than n. Then \overline{a(x)}\overline{b(x)} = \overline{f(x)} = 0 in R[x]/(f(x)), but by the previous part, neither \overline{a(x)} nor \overline{b(x)} is zero. Thus a(x) is a zero divisor in R[x]/(f(x))..
  4. Suppose a^m = 0 in R. Now \overline{x}^nm = \overline{(x^n)^m} = \overline{a^m} = 0, so that \overline{x} is nilpotent in R[x]/(f(x)).
  5. Using Fermat’s Little Theorem, we have \overline{x-a}^p = \overline{(x-a)^p} = \overline{x^p-a^p} = \overline{a-a} = 0. Thus \overline{x-a} is nilpotent in R[x]/(f(x)).
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