## 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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### Comments

• 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.