Show that if $R$ is a commutative, semisimple ring with unity then $R$ is a direct sum of fields.

I have looked at the math.stackexchange post Commutative ring is semisimple iff it's isomorphic to a finite direct product of fields. but I have gone about trying to prove the $\Rightarrow$ direction in a different way.

Here is my attempt:

First we state a lemma.

Lemma: If $J$ is a maximal ideal of a commutative ring $R$ with unity, then the quotient ring $R/J$ is a field.

Since $R$ is semisimple, $R = \oplus_{i \in I} \space R_i$ where each $R_i$ is simple. Now let $J_i$ be a maximal ideal of $R_i$. Such is guaranteed to exist by Zorn's Lemma.

Now we show that $J_i$ is a submodule of $R_i$. Let $x, y \in R_i$ and $r \in R$. Then the sum $x + ry$ is an element of $J_i$ because $ry \in J_i$ by definition of ideal and ideals by definition are subgroups under addition. Hence, $x + ry \in J_i$ and $J_i$ is a submodule of $R_i$. But since $R_i$ is simple, this implies $J_i = \{0\}$.

Hence, $R_i/J_i = R_i$ and by the lemma, $R_i$ is a field. $\blacksquare$

Is this proof legitimate? I am concerned with the argument I make implying that $J_i$ is the trivial module. How about the case when $J_i = R_i$? Then the quotient is $0$ and I am unsure how to proceed.

  • $\begingroup$ You mean to say that $J_i$ is a maximal (proper) ideal of $R_i$ (that is what is guaranteed by Zorn) and so you would not have $J_i = R_i$. $\endgroup$
    – user357980
    Mar 3, 2018 at 3:05
  • $\begingroup$ What definition of semisimple are you using? $\endgroup$ Mar 3, 2018 at 3:14
  • $\begingroup$ @MarianoSuárez-Álvarez A ring R is semisimple, if it is semisimple when regarded as a module over itself. Then I use the definition of semisimple such that $R$ is a direct sum of simple modules. $\endgroup$ Mar 3, 2018 at 3:16
  • $\begingroup$ Well, then the direct summands are modules, not rings. $\endgroup$ Mar 3, 2018 at 3:20

1 Answer 1


This line of argument works, but there's one more step needed to finish it. You know that $R=\bigoplus R_i$ as an $R$-module, not as a ring. To say $R$ is a product of fields, you need to additionally show that the map $R\to \prod R_i$ is a ring-homomorphism. This may not actually be true unless you define the ring structure on $R_i$ carefully (as noted below the ring structure on them is not automatic).

Here's how you get the ring-isomorphism. The direct sum decomposition $R=\bigoplus R_i$ gives us a projection homomorphism $p_i:R\to R_i$ for each $i$ (these are $R$-module homomorphisms). Now choose the ring structure on $R_i$ such that each $p_i$ is a ring-homomorphism as well (in other words, give $R_i$ the ring structure of the quotient ring $R/\ker(p_i)$). Then our bijection $R\to \prod R_i$ is a ring-homomorphism, since the $i$th coordinate is just $p_i$ which is a ring-homomorphism.

Some other miscellaneous comments:

  • The $R_i$ are not, a priori, rings. They are just submodules of $R$ (i.e., ideals of $R$). But since they are simple modules, they are in particular cyclic, so they are isomorphic to modules of the form $R/I_i$ for some ideals $I_i$ in $R$. So you can consider them as rings, and any ideal in them would give an $R$-submodule, and so following your argument they must be fields. (The ring structure on $R_i$ depends on the choice of generator as an $R$-module, though! To get the correct one for which $R\to\prod R_i$ is a ring-isomorphism, you need to choose your generator to be the image of $1\in R$.)
  • There is no issue with having $J_i=R_i$, since a maximal ideal is by definition a proper ideal. To get the existence of a maximal ideal, though, you need to know that $R_i$ is not just a ring but a nonzero ring. This is true because $R_i$ is a simple $R$-module and in particular is a nonzero module, so it has more than one element.
  • $\begingroup$ Great response, thank you. Can you just say a few more words about simple modules $R_i$ being isomorphic to $R/I_i$? And how choice of generator of $R_i$ leads us to turn the map into a ring homomorphism? $\endgroup$ Mar 3, 2018 at 3:23
  • $\begingroup$ If $a\in R_i$ is a generator as an $R$-module, then there is a surjective homomorphism of $R$-modules $f:R\to R_i$ sending $r$ to $ra$. The kernel of $f$ is some ideal $I_i$, and so $f$ gives an isomorphism of $R$-modules $R_i\cong R/I_i$. $\endgroup$ Mar 3, 2018 at 3:36

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