10
$\begingroup$

I am trying to prove the following:

Let $R$ be a commutative ring. Prove that $R$ is semisimple if and only if it is isomorphic to a direct product of a finite number of fields.

Suppose $R$ is a semisimple commutative ring. By Wedderburn, $R \cong \bigoplus_{i=1}^n M_{n_i}(D_i)$, where $D_i$ is a division ring for each $i$. Since $R$ is commutative, each $M_i:=M_{n_i}(D_i)$ is a simple commutative ring. I'll show that each element has an inverse: take $a \in M_i$ and consider $aM_i$, then $aM_i=M_i$ so there is $b \in M_i : ab=1$. With the same argument, we get that there is $c \in M_i$ with $bc=1$, but then $a=a.1=a(bc)=(ab)c=1.c=c$. It follows each $M_i$ is a field.

I don't know how to show the other implication, any help would be appreciated.

$\endgroup$
1
  • 6
    $\begingroup$ You can save yourself a little work in the above: Once you know that $M_i$ is a matrix ring, you know immediately that it is a 1x1 matrix ring, because otherwise it's easy to write down two matrices that don't commute. $\endgroup$
    – WillO
    Nov 7, 2014 at 17:38

2 Answers 2

Reset to default
5
$\begingroup$

A finite product of fields is clearly Artinian.

It's also then clearly Jacobson semisimple (since it has no nonzero nilpotent elements, and the radical is nilpotent.)

Since Jacobson radical zero and Artinian combine to make "semisimple," we're done.

$\endgroup$
2
$\begingroup$

One way is to show that every ideal of a finite direct product of fields is a direct summand. Can you see why?

$\endgroup$
5
  • $\begingroup$ Because then a submodule (ideal) of $R$, which is in correspondence with an ideal of the finite product, is a direct summand? $\endgroup$
    – user16924
    Nov 7, 2014 at 17:45
  • $\begingroup$ @user16924 The question is why an ideal of a direct product of fields is a direct summand. (Think about the form of ideals in finite direct products of rings.) $\endgroup$
    – user26857
    Nov 7, 2014 at 17:53
  • $\begingroup$ I kept thinking about this problem but I couldn't follow your suggestion, however, it occurred to me that if $R \cong K_1 \times ... \times K_n$, with each $K_i$ a field, then taking $M_{1}(K_i)$ (the one by one matrices for each field), then by Wedderburn theorem, we have $R$ is semisimple (the theorem states that $R$ is semisimple iff it is isomorphic to a finite product of matrices over division rings). Sorry I forgot to answer. $\endgroup$
    – user16924
    Nov 13, 2014 at 14:06
  • $\begingroup$ So, to answer your question, maybe I would need more details to follow your approach because up to now I couldn't figured out how to work on them. $\endgroup$
    – user16924
    Nov 13, 2014 at 14:08
  • 1
    $\begingroup$ @user16924 A characterization (equivalent definition) of semisimple modules/rings says that a module $M$ (a ring $R$) is semisimple iff every submodule of $M$ (ideal of $R$) is a direct summand of $M$ (of $R$). If $R=K_1\times\cdots\times K_n$ is a direct product of fields, then every ideal of $R$ is a direct product of ideals of the $K_i$'s, that is, of $(0)$ and $K_i$ (because $K_i$ is a field). Now if sum such an ideal $I$ with the ideal $J$ which has $(0)$ where $I$ has $K_j$, respectively $K_j$ where $I$ has $(0)$ you get $I+J=R$ and $I\cap J=0$, that is, $I$ is a direct summand of $R$. $\endgroup$
    – user26857
    Nov 13, 2014 at 14:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .