A generalization of Eisenstein’s Criterion

Prove the following generalization of Eisenstein’s Criterion:

Let R be a unique factorization domain with field of fractions F and let P \subseteq R be a prime ideal. If p(x) = \sum c_i x^i \in R[x] is a polynomial such that c_n \notin P, c_i \in P for 0 \leq i < n, and c_0 \notin P^2, then p(x) is irreducible in F[x].

[Thanks to Marc van Leeuwen for suggesting this proof.]

We begin with a lemma.

Lemma: Let D be an integral domain. If q,s,t \in D[x] are polynomials such that q = st and q = cx^k has only one nonzero term, then both s and t also have only one nonzero term. More specifically, s = ax^n and t = bx^m where k = n+m and c = ab. Proof: Let a_1x^{n_1} and a_2x^{n_2} be the (nonzero) terms of highest and lowest degree in s, and likewise let b_1x^{m_1} and b_2x^{m_2} be the nonzero terms of highest and lowest degree in t. Now the term of highest degree in q is cx^k = a_1b_1x^{n_1+m_1} and the term of lowest degree is cx^k = a_2b_2x^{n_2+m_2}. Thus we have n_1+m_1 = n_2+m_2, so that (n_1-n_2) + (m_1-m_2) = 0. Note that n_1-n_2 and m_1-m_2 are nonnegative, so that n_1 - n_2 = m_1 - m_2 = 0. That is, n_1 = n_2 and m_1 = m_2, so that s and t each have only one nonzero term. \square

Now suppose p(x) is reducible in R[x]; say p(x) = s(x)t(x), where s and t are nonunits. Reducing the coefficients mod P, we have \overline{p}(x) \equiv \overline{c_n}x^n \equiv \overline{s}(x) \overline{t}(x) in (R/P)[x]. Because P is prime, R/P is an integral domain, and thus by the lemma both \overline{s}(x) and \overline{t}(x) have only one nonzero term. Suppose both \overline{s}(x) or \overline{t}(x) have positive degree. Then in particular, s_0 and t_0 (the constant terms of s and t) are both in P, so that p_0 = s_0t_0 \in P^2, a contradiction. Thus without loss of generality, \overline{s}(x) has degree 0, so that s(x) has degree 0. That is, p(x) = s_0t(x) where s_0 \in R and t(x) is irreducible over R.

Suppose now that p(x) is reducible in F[x]. Say p(x) = a(x)b(x). Clearing denominators, we have dp(x) = a^\prime(x)b^\prime(x), where a^\prime and b^\prime are in R[x]. Then ds_0t(x) = a^\prime(x)b^\prime(x); considering degrees, without loss of generality, a^\prime is constant. But then a(x) is constant, and thus a unit in F[x]. Thus we have a contradiction, and so p(x) must be irreducible in F[x].

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  • Marc van Leeuwen  On March 25, 2011 at 6:26 am

    The proof appears to establish a stronger result, namely that p(x) is irreducible in R[x], because reducibility in R[x] is all that is deduced from reducibility in F[x]. However, this stronger result is not true because of possible non-trivial divisors of degree 0 in R[x]; this points to the omission of mentioning that a(x) and b(x) are non-constant. The proof is fairly unclearly stated, so it is hard to pinpoint where the tacit assumption of non-constance is actually used; however what it appears to prove by induction is that as long as $c_i$ is in $P$, $b_i$ is as well, so the conclusion that $b_k$ is in $P$ depends on the assumption that $c_k$ is in $P$, and so implicitly on $k<n$ (since $c_n$ is NOT in $P$), and this depends on $a(x)$ non-constant.

    Also the proof is confused, using equality instead of congruence mod $P$ for the step concerning $b_1$.

    Here's a cleaner version of the proof. The essential lemma is: if $D$ is an integral domain, and $P=ST$ in $D[x]$ where $P$ has exactly one non-zero term, then $S$ and $T$ each have exactly one non-zero term as well (as there can be no cancellation in either the product of their leading terms or the product of their lowest-degree terms). Now supposing $p(x)$ decomposes in $R[x]$ as a product of non-constant polynomials, this is true for its image in $(R/P)[x]$ as well, but then by the lemma the factors in $(R/P)[x]$ both have zero constant term, contradicting $c_0\notin P^2$. This allow to conclude that $p(x)$ is irreducible in $F[x]$, and it is irreducible in $R[x]$ as well unless it has a non-trivial constant divisor (it is non-primitive so to speak).

    • nbloomf  On March 25, 2011 at 4:33 pm

      Thank you for pointing that out! Thanks also for suggesting a fix. I sincerely appreciate it.

  • Marc van Leeuwen  On March 25, 2011 at 8:00 am

    One more remark: the hypothesis that $R$ be a UFD is only used to apply Gauss’s lemma. So the conclusion “if $p(x)$ satisfies the given conditions relative to a prime ideal $P$ and has no non-trivial constant divisors, then it is irreducible in $R[x]$” is valid independently of this hypothesis.

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