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

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.

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