Why is $J/J^{2}$ is a vector space over $R/J$

Question

Let $R$ be a commutative local ring with $1$ and let $J$ be the Jacobson radical.

I have it written in my notes that $J/J^{2}$ is a vector space over the field $R/J$ that is an $R/J$ module.

Why is this? I don't understand why this is the case?

Answer 1

If $R$ is any ring and $I$ an ideal, then any $R$-module $M$ which is annihilated by $I$ has a unique $(R/I)$-module structure such that, if we let $R$ act on $M$ via the surjection $R\to R/I$, we recover the original $R$-module structure. Explicitly, one sets $(r+I)m=rm$ for $r\in R$ and $m\in M$. The fact that this multiplication is well-defined follows from the assumption that $I$ annihilates $M$.

In particular, since multiplication by $J$ on $J$ lands in $J^2$, $J/J^2$ is an $R$-module killed by $J$, and hence is naturally an $R/J$-module. Alternatively, $J/J^2=J\otimes_R(R/J)$, and so can be viewed as the extension of scalars of the $R$-module $J$ to $R/J$.