Is this proof of a irreducibility criterion in an integral domain correct?

Question

This is an exercise from Grillet's "Abstract Algebra" (page $145$, proposition $10.10$).

Let $R$ be an integral domain, let $I$ be an ideal of $R$, and let $\pi\colon R\to R/I$ be a canonical projection. If $f(x) = a_0 + ... + x^n \in R[x]$ and $\pi(a_0) + ... + \pi(1)x^n = (a_0 + I) + ... + (1 + I)x^n$ is irreducible in $(R/I)[x]$, then $f(x)$ is irreducible in $R[x]$.

My supposed proof:

Let $f(x) = g(x)h(x)$ where $g(x) = b_0 + ... + b_rx^r$ and $h(x) = c_0 + ... + c_sx^s$. Then $b_rc_s = 1$. Also, $(a_0 + I) + ... + (1 + I)x^n = (b_0c_0 + I) + ... + (b_rc_s + I)x^n = ((b_0 + I) + ... + (b_r + I)x^r)((c_0 + I) + ... + (c_s + I)x^s)$, hence at least one of $((b_0 + I) + ... + (b_r + I)x^r)$ and $((c_0 + I) + ... + (c_s + I)x^s)$, say, $g(x)$ is a unit. Since units of $(R/I)[x]$ are precisely units of $R/I$, we in particular have either $b_r \in I$ or $r = 0$. In the first case, as $I$ is an ideal, we have $1 = b_rc_s \in I$ and hence $I = R$. The polynomial in question is trivially irreducible in $R/R \cong \{0\}$ since every element of $\{0\}$ is trivially a unit. In the second case, $g(x) = b_r$ is a unit of $R$, hence a unit of $R[x]$.

However, the proof seems quite trivial. I wonder if I'm missing something.

Answer 1

Yes, you are missing some things (ie. the proof is not correct). Some comments:

• The goal is to assume that $f$ is irreducible in $(R/I)[x]$, then show that $f$ reducible in $R[x]$ results in a contradiction, thus proving that $f$ had to be irreducible in $R[x]$.

  In the case where $R=I$, you have not shown this contradiction - you have argued that $f$ is irreducible in $(R/I)[x]$ (actually, $0$ isn't irreducible - we usually exclude $0$ in the definition of an irreducible polynomial). Further, even if $f$ were irreducible in $(R/R)[x]$, this could never imply $f$ irreducible in $R[x]$, for arbitrary $f$ (otherwise every $f$ would be irreducible). So this should tell us that something has gone wrong.

  As pointed out by Tobias Kildetoft in the comments, altogether we have simply that the theorem is vacuously true in the case $(R/R)[x] = {0}$, because there are no irreducible polynomials in $(R/R)[x]$ (there could never be a counterexample where $f$ is irreducible in $(R/R)[x]$, but reducible in $R[x]$).

• The second case is also not done correctly: you have shown that $g(x)$ is a unit in $(R/I)[x]$. But this does not mean it was a unit in $R[x]$ (it is easy to construct a counterexample: $3$ is a unit in $\mathbb{Z}_5[x]$, but $3$ is not a unit in $\mathbb{Z}[x]$). However, I think you have already done enough to prove the second case - could you complete the proof?

Answer 2

The idea is good. There are a few things to fix, though.

Suppose $f(x)=g(x)h(x)$. You correctly note that both $g(x)$ and $h(x)$ have invertible leading coefficient.

Now a polynomial in $(R/I)[x]$, with invertible leading coefficient, is invertible if and only if it is a unit in $R/I$.

It is not generally true that units in $S[x]$ are the units in $S$, if the ring $S$ is not a domain. Consider, for instance, $1+2x$ and $1-2x$ in $(\mathbb{Z}/4\mathbb{Z})[x]$: since $(1+2x)(1-2x)=1$, those polynomials are units. However if a polynomial has invertible leading coefficients, then it is a unit if and only if it is an invertible constant. Indeed, if a polynomial has invertible leading coefficient, the the degree of the product with another polynomial is the sum of the degrees (easy proof).

If $I=R$, then $\pi(f)$ is not irreducible, because an irreducible element is, by definition, nonzero. So we can dismiss this case.

Now either $\pi(g)$ or $\pi(h)$ is a unit, so a constant, meaning that either $r=0$ or $s=0$. Hence either $g(x)$ or $h(x)$ is a unit in $R$ (because they have invertible leading coefficient).

The assumption that $f(x)$ is monic is essential: $2+2x$ is reducible in $\mathbb{Z}[x]$, but its image in $(\mathbb{Z}/3\mathbb{Z})[x]$ is $-1-x$, which is irreducible.