How to conclude that this is a semisimple algebra

Question

Let $k$ be a field, $R$ a $k$-algebra and $J$ an ideal of $J$. I have proved that $$R/J\cong k \oplus \ldots \oplus k$$ which means that $R/J$ is semisimple as a $k$-module. But I want to conclude something more: I want to say that $R/J$ is semisimple as a $k$-algebra. I know about a result that says that an algebra $A$ is semisimple iff the $A$-module $A$ is semisimple. But I think I cannot use this here since I know that $R/J$ is semisimple as a $k$ module, not as a $R/J$ module. Or could I see the field $k$ as an $R/J$-module and therefore conclude that $R/J$ is a semisimple as an $R/J$-module and therefore as an algebra?

Edit: This was actually taken from Example 6 in page 57 of Lam's A first course in noncommutative rings. There, from the expression $R/J\cong k \oplus \ldots \oplus k$, the author concludes that $R/J$ is semisimple (he doesn't specify if its semisimple as an algebra or as a module) and therefore $\mathfrak{J}(R/J)=0$. We know that $A$ is a semisimple algebra iff $\mathfrak{J}(A)=(0)$. I assume this is the result that the author is using to conclude that $\mathfrak{J}(R/J)=0$, but I do not understand how can he be applying it if we don't know whether $R/J$ is a semisimple algebra.

Answer 1

You cannot conclude something like that.

Suppose $k$ is a field (as it often is in a lot of contexts). Then every $k$-module whatsoever is semisimple, even if $R/J$ is not a semisimple algebra.

Concretely, $R=\mathbb R[x]$ and $J=(x^2)$, you have that (as $\mathbb R$ modules) $R/J\cong \mathbb R\times \mathbb R$ is a semisimple $\mathbb R$ module, but $R/J$ is not a semisimple algebra because it has nonzero nilpotent elements and is commutative.

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After update with reference

This was actually taken from Example 6 in page 57 of Lam's A first course in noncommutative rings. There, from the expression $R/J\cong k \oplus \ldots \oplus k$, the author concludes that $R/J$ is semisimple (he doesn't specify if its semisimple as an algebra or as a module)

He says $R/J$ is isomorphic to the subset of diagonal matrices of $M_n(k)$ meaning that the ring $R/J$ is isomorphic to the ring $k^n$. Since $k^n$ is obviously a semisimple ring (in the sense of the Wedderburn-Artin theorem) then so is $R/J$. One can then say that $R/J$ is a semisimple ring, and that $R/J$ is a semisimple $R/J$ module. There isn't really any need to talk about $k$ algebras here.

A quotient may very well be isomorphic to some copies of $k$ without being a $k$ algebra. For example, $\mathbb Z/(3)$ is the field of $3$ elements but $\mathbb Z$ is not an algebra over any field.

Semisimplicity is just a statement about a module over a ring, whether or not that ring is an algebra. When the ring is an algebra, its modules do come equipped with a special action from $k$, but it doesn't modify what the modules are.