Irreducible modules over a commutative ring are precisely its maximal ideal quotients

Let R be a commutative ring with 1 and let M be a unital left R module. Prove that M is irreducible if and only if M \cong_R R/I for some maximal ideal I \subseteq R. (Where \cong_R means “isomorphic as an R-module.”)

Suppose M is irreducible, fix m \in M nonzero, and define \varphi_m : R \rightarrow M by \varphi_m(r) = r \cdot m. Note that for all x,y \in R and r \in R, \varphi_m(x + r \cdot y) = (x+r \cdot y) \cdot m = x \cdot m + r \cdot (y \cdot m) = \varphi_m(x) + r \cdot \varphi_m(y). So \varphi_m is an R-module homomorphism. Now recall that since m \neq 0 and M is irreducible, then by this previous exercise, M = Rm. In particular, if b \in M, then there exists a \in R such that b = a \cdot m = \varphi_m(a). Thus \varphi_m is surjective.

FInally, we claim that \mathsf{ker}\ \varphi_m is a maximal ideal. To that end, let x + \mathsf{ker}\ \varphi_m be nonzero. In particular, x \cdot m \neq 0. Since M is irreducible, we have M = R(x \cdot m). In particular, m = y \cdot (x \cdot m) = (yx) \cdot m for some y \in R. Then 1 \cdot m - (yx) \cdot m = 0, so that 1 - yx \in \mathsf{ker}\ \varphi_m. That is, (y + \mathsf{ker}\ \varphi_m)(x + \mathsf{ker}\ \varphi_m) = 1 + \mathsf{ker}\ \varphi_m. So every nonzero element of R/\mathsf{ker}\ \varphi_m has a left inverse. Since R/\mathsf{ker}\ \varphi_m is commutative, R/\mathsf{ker}\ \varphi_m is a field, so that \mathsf{ker}\ \varphi_m is a maximal ideal of R.

By the First Isomorphism Theorem, we have M \cong_R R/\mathsf{ker}\ \varphi_m, where \mathsf{ker}\ \varphi_m is a maximal ideal.

Conversely, suppose I \subseteq R is a maximal ideal. Now R/I is a field. Let x + I be nonzero. Then there exists y \in R such that (y+I)(x+I) = (1+I). Now let r+I \in R/I. Note that r+I = (ry+I)(x+I), so that R/I = R(x+I). That is, R/I is generated (as an R-module) by any nonzero element. By this previous exercise, R/I is an irreducible R-module.

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  • RyanK  On April 20, 2011 at 12:08 pm

    So how would one, relate this to showing that a polynomial ring R[x] has a maximal ideal M such that R[x]/M is isomorphic to Complex numbers?

  • RyanK  On April 20, 2011 at 12:20 pm

    This is what I got so far consider M = (x^2 + 1)R[x]. (x^2 + 1) is irreducible over R. And M cannot be properly contained in another ideal of R[x] since [R[x]:M] = 2, the order of the irreducible polynomial (x^2 + 1) . So M is a maximal ideal. Therefore, R[x]/M is a field of dimension 2 over R. The only such field is C. Therefore, R[x].M is isomorphic to C.

    • nbloomf  On April 20, 2011 at 8:36 pm

      That might work. Your intuition is correct- the ideal you want is (x^2+1). However, I’d prove it in a slightly different way.

      Define a mapping \varphi : \mathbb{R}[x] \rightarrow \mathbb{C} that fixes the coefficients and sends x to i. Then show that this map (1) is a homomorphism of rings, (2) is surjective, and (3) has kernel (x^2+1). Using the first isomorphism theorem, we then get \mathbb{R}[x]/(x^2+1) \cong \mathbb{C}.

      In fact, I’m pretty sure that problem is around here somewhere…

  • Mathew  On July 31, 2012 at 2:56 pm

    to the person who made these solutions. You r work helped me a lot to understand many stuff. I would like to say thank you very very much.
    can I get these solutions as a pdf so that I can print and read them .
    thank you again

    • nbloomf  On July 31, 2012 at 6:34 pm

      I have no plans to make a PDF version anytime soon- sorry!

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