## Over an integral domain, every maximal ideal is irreducible

Let $R$ be an integral domain. Prove that every maximal ideal in $R$ is irreducible.

We begin with a lemma. This lemma is actually true (with a minor modification) over a much larger class of rings. However, I am not entirely comfortable with the proof of the general version. I am, however, comfortable with this more specific version. I referred to a discussion at MathForum.org for a proof of the lemma.

Lemma: Let $R$ be an integral domain and let $I \subseteq R$ be an ideal which is finitely generated as a $\mathbb{Z}$-module. If $I^2 = I$, then either $I = 0$ or $I = R$. Proof: Suppose $I \neq 0$. Let $A = \{\alpha_i\}_{i=1}^n$ be a generating set for $A$ over $\mathbb{Z}$; in particular, some $\alpha_i$ is nonzero. Since $I^2 = I$, there exist $u_{i,j} \in A$ such that $\alpha_i = \sum_j u_{i,j} \alpha_j$ for each $i$. In particular, $0 = \sum_j (u_{i,j} - \delta_{i,j})\alpha_j$ for each $i$, where $\delta_{i,j}$ is the Kronecker delta. That is, $[\alpha_1\ \cdots\ \alpha_n]^\mathsf{T}$ is a nontrivial solution to the matrix equation $Mx = 0$, where $M = [u_{i,j} - \delta_{i,j}]$. Thinking of $R$ as embedded in its field $F$ of fractions, we have $\mathsf{det}(M) = 0$. On the other hand, by the Leibniz expansion of $\mathsf{det}(M)$, we have $\mathsf{det}(M) = \beta - 1$, where $\beta \in I$ (using the fact that $I$ is an ideal). In particular, $-1 \in I$, so that $I = R$. $\square$

Now let $M \subseteq R$ be nonzero and maximal. If $M = AB$, then $M \subseteq A$ and $M \subseteq B$, so that $A,B \in \{M, R\}$. If $A = B = M$, then $M^2 = M$. By the lemma, either $M = 0$ (a contradiction) or $M = R$ (also a contradiction). So either $A$ or $B$ is $R$. Since $R$ is a unit in the semigroup of ideals of $R$ under ideal products, $M$ is irreducible.