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.

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