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I want to show that:

$R$ is semi-simple iff $R$ is isomorphic (as a ring isomorphism)to a direct product of a finite number of fields.

Definition: $R$ is a semi-simple ring if it is a direct sum of simple ideals.

My attempt:

$\Leftarrow$

Assume that $R$ is isomorphic to a direct product of a finite number of fields, I want to show that $R$ is semi-simple. I was able to proof that any field is a semi-simple ring, then $R$ is isomorphic to a direct product of a finite number of semi-simple rings. But then how can I proof that the direct product of a finite number of semi-simple rings is semi-simple (I think that is very obvious and does not require any proof ... am I correct?)

$\Rightarrow$

Assume that $R$ is a semisimple ring and we want to show that $R$ is (ring) isomorphic to a direct product of a finite number of fields.

Since $R$ is a semisimple ring, then $R$ is the direct sum of a finite number of simple modules $i.e., R = \bigoplus_{i=1}^m M_i$(I know that direct sum is the same as direct product if we have a finite number). Now we want to show that $M_i$ is a field for every $i.$ So it suffices to show that every non-zero element in $M_i$ has an inverse but I do not know how to show this, any help in this will be greatly appreciated!

Also, I do not know how to show the ring isomorphism, could anyone help me in this, please?

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  • $\begingroup$ What about the finitely many part? $\endgroup$ Apr 8, 2021 at 22:58
  • $\begingroup$ Please give the definition of semi simple $\endgroup$
    – Matt
    Apr 11, 2021 at 1:13
  • $\begingroup$ @Matt I added it. $\endgroup$
    – user889267
    Apr 11, 2021 at 18:40
  • $\begingroup$ IF R is semisimple, than intersection of all maximal ideals is 0. Therefore, for every maximal ideals m_{1},m_{2}, R/m_{1} intersected with R\m_{2} is 0. And when we divide ring by maximal ideal, we obtain field. It is more or less correct? $\endgroup$
    – robin3210
    Apr 12, 2021 at 6:37
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    $\begingroup$ A simple commutative ring is a field. Done. $\endgroup$ Apr 13, 2021 at 0:42

1 Answer 1

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Suppose $R$ is a ring such that it can be written as $R=k_1\times \cdots \times k_n$ for some fields $k_i$. The ideals of any ring $A_1\times A_2$ are direct sums of the form $I_1\oplus I_2$, where $I_i$ is an ideal of $A_i$. If we think of $A_1\times A_2$ as a module over itself, then its submodules are precisely the ideals, in particular $A_1\times A_2 = A_1\oplus A_2$ as a module over itself. Applying this to $R$, we have the module $R=k_1\oplus\cdots\oplus k_n$. Since a field is a simple module over itself, the module $R$ is a finite direct sum of simple modules, and therefore a semisimple module over itself. This means it is semisimple.

Conversely, suppose $R$ is semisimple and of the form $R=L_1\oplus\cdots\oplus L_n$. Then we can write the unit element $1$ as a finite sum $1=e_1 + \cdots + e_n$, where $e_i\in L_i$. The important thing to notice here is that $e_ie_j = 0$ if $i\neq j$ ($L_iL_j \subseteq L_i\cap L_j = 0$), and $e_i^2=e_i$ ($e_i = 1e_i = (e_1 + \cdots + e_n)e_i = e_1e_i + \cdots + e_ne_i$, where all but $e_i^2$ are zero by the previous observation). We can therefore write $L_i = \langle e_i\rangle = \{ re_i\ |\ r\in R\}$ (since, again, we have the direct sum decomposition, $re_i=0$ whenever $r\not\in L_i$. Furthermore, since we know that each $L_i$ is an ideal, it is additively closed and contains $0$, but we also have $(re_i)(se_i) = (rs)e_i^2 = (rs)e_i$ for any $r, s\in R$, so it is also multiplicatively closed, and $e_i$ is its unit element ($e_i(re_i) = re_i^2 = re_i$ for any $re_i \in L_i$). Therefore, $L_i$ is a subring of $R$.

We may now start quotienting out the $L_i$. First, some notation. Let $R_j = R/(L_1 \oplus \cdots \oplus L_{j-1})$, the quotient with the subrings from $1$ to $j-1$ quotiented out, and $R^m_j = R_j/(L_m\oplus\cdots\oplus L_n)$, the subrings starting from $m$ quotiented out. First, $L_n = R_n$, and it is a simple module both over $R$ and itself, so it is a simple commutative ring (a field). Next, $L_{n-1} = R_{n-1}/R_j^n$, which is (by the same arguments) a field. This holds for each $L_i = R_i/R_i^{i+1}$, so we conclude that $R$ is indeed a finite direct product of fields.

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  • $\begingroup$ Oh, and to be added: the rings in the direct product are themselves unital, but their units are not the sames that of the product ring (if many rings in the product)! That is, $1_R \neq e_i = 1_{L_i}$ $\endgroup$
    – user914108
    Apr 13, 2021 at 14:09
  • $\begingroup$ but where is the ring isomorphism? $\endgroup$
    – user889267
    Apr 13, 2021 at 17:01
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    $\begingroup$ There's no need for one explicitly if it can be shown that semisimplicity implies being a product of fields and vice versa. $\endgroup$
    – user914108
    Apr 13, 2021 at 17:04

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