# Semi-simple rings and fields.

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?

• What about the finitely many part? Apr 8, 2021 at 22:58
• Please give the definition of semi simple
– Matt
Apr 11, 2021 at 1:13
– user889267
Apr 11, 2021 at 18:40
• 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? Apr 12, 2021 at 6:37
• A simple commutative ring is a field. Done. Apr 13, 2021 at 0:42

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.

• 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, $1_R \neq e_i = 1_{L_i}$
– user914108
Apr 13, 2021 at 14:09
• but where is the ring isomorphism?
– user889267
Apr 13, 2021 at 17:01
• There's no need for one explicitly if it can be shown that semisimplicity implies being a product of fields and vice versa.
– user914108
Apr 13, 2021 at 17:04