Prove the following generalization of Eisenstein’s Criterion:

Let be a unique factorization domain with field of fractions and let be a prime ideal. If is a polynomial such that , for , and , then is irreducible in .

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

We begin with a lemma.

Lemma: Let be an integral domain. If are polynomials such that and has only one nonzero term, then both and also have only one nonzero term. More specifically, and where and . Proof: Let and be the (nonzero) terms of highest and lowest degree in , and likewise let and be the nonzero terms of highest and lowest degree in . Now the term of highest degree in is and the term of lowest degree is . Thus we have , so that . Note that and are nonnegative, so that . That is, and , so that and each have only one nonzero term.

Now suppose is reducible in ; say , where and are nonunits. Reducing the coefficients mod , we have in . Because is prime, is an integral domain, and thus by the lemma both and have only one nonzero term. Suppose both or have positive degree. Then in particular, and (the constant terms of and ) are both in , so that , a contradiction. Thus without loss of generality, has degree 0, so that has degree 0. That is, where and is irreducible over .

Suppose now that is reducible in . Say . Clearing denominators, we have , where and are in . Then ; considering degrees, without loss of generality, is constant. But then is constant, and thus a unit in . Thus we have a contradiction, and so must be irreducible in .

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## Comments

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).

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

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.