## 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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### Comments

• 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…

• nbloomf  On April 20, 2011 at 8:38 pm

Yup- here it is.

• Mathew  On July 31, 2012 at 2:56 pm

Hello,
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!