I have met the statement below many times, either here on this site or while reading through books, but I am for my life is unable to prove why it is correct. Here is the statement:
If $R$(commutative ring with $1$) is semisimple, then the intersection of all maximal ideals is $0.$ Also, why the fields associated with each maximal ideal has empty intersections with each other?
Could someone provide an explanation of this to me, please?
In generality by the Artin-Wedderburn Theorm, a ring with unity $R$ is semisimple if and only if there is an isomorphism $$ R \cong \prod_{k= 0}^{m} \operatorname{Mat}_{n_k}(D_k) $$ where $m \in \mathbb{N}$ and each ring $\operatorname{Mat}_{n_k}(D_k)$ is the ring of $n_k \times n_k$ matrices over a division ring $D_k$. Note that if $R$ is commutative, each $n_k = 1$ and each division ring $D_k$ is a field. It can be shown that the Jacobson radical is a proper two-sided ideal of $R$ with $$ J(R) = \bigcap_{\substack{\text{Maximal right ideals}\\\mathfrak{m} \triangleleft R}}\mathfrak{m}. $$ In particular, the Jacobson radical of $R$ coincides with the intersection of all maximal right ideals of $R$; note that in the case $R$ is commutative, this is just the radical of $R$ as all right ideals are two-sided for free. We are then done once we show the following result:
Theorem For a ring $R$ with unity, if $R$ is semisimple then $R$ is Artinian and $J(R) = 0$.
Proof Assume that $R$ is semisimple and note that by the Artin-Wedderburn Theorem we can assume that $R$ is a finite product of matrix rings over division rings. In fact, by a routine induction we can even assume that $R = \operatorname{Mat}_m(D) \times \operatorname{Mat}_n(E)$ for division rings $m$ and $n$. At any rate, it is routine to check that for any rings with unity $A, B$ $$ J(A \times B) = J(A) \times J(B). $$ Using this we find that $$ J(R) = J(\operatorname{Mat}_m(D) \times \operatorname{Mat}_n(E)) = J(\operatorname{Mat}_m(D)) \times J(\operatorname{Mat}_n(E)) = 0 \times 0 = 0 $$ because each ring $\operatorname{Mat}_{n_k}(D_k)$ is a simple ring and $J(\operatorname{Mat}_{n_k}(D_k))$ is a two-sided ideal of $\operatorname{Mat}_{n_k}(D_k)$ not equal to $\operatorname{Mat}_{n_k}(D_k)$. Finally, $R$ is Artinian because it is a finite product of simple rings. $\square$
This explains why the intersection of all maximal ideals of a finite product of fields is $0$. Essentially, the maximal ideals of a commutative unital semisimple ring look like $$ \mathfrak{m}_i = \left(\prod_{j=0}^{i-1} 0\right) \times D_{n_i} \times \left(\prod_{j = i+1}^{m} 0\right) $$ i.e., $\mathfrak{m}$ is $0$ in all terms of the product except one. Intersecting two distinct such ideals gives you $0$ because the only nonzero terms do not live in common pieces of your product.