0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Are you sure $I$ is not supposed to be a prime ideal? $\endgroup$ – egreg Apr 17 at 10:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.