Proving that $V(R^*)=V(R)-1$

Question

Let $R$ be a noetherian local ring with Jacobson radical $J$.

Define $V(R)=\dim J/J^2$ where $J/J^2$ is considered as a vector space over $R/J$.

Now fix $x\in J-J^2$. If we then let $R^*=R/xR$ then I am trying to show that:

$V(R^*)=V(R)-1$

My attempt at the proof/understanding of a given proof is below. I am happy for an answer to be just a proof of the above or telling me what is wrong with the below:

Let us first not that $R^*$ is also a Noetherian local ring with Jacobson radical $J^*=J/xR$.

Now define $f:R\rightarrow R^*$ be the natural projection map and pick $y_1^*,\ldots, y_k^*$ to be a minimal generating set for $J^*$.

Then let us choose $y_1,\ldots ,y_k$ in $J$ such that $f(y_i)=y_i^*$. Then we are going to claim that $x,y_1,\ldots, y_k$ is a minimal generating set for $J$ and we will be done.

Note: this is the same as showing that it's image in $J/J^2$ is a basis of $J/J^2$.

So let us suppose $xr+y_1r_1+\cdots +r_ky_k\in J^2$,

that is, the sum is equal to zero, so we now want to show that that $r,r_i$'s are all zero, that is, in $J$- the zero's of $R/J$ which is the field we are considering $J/J^2$ to be over.

Then we have that $y_1^*r_1^*+\cdots +y_k^*r_k^*\in (J^*)^2$ and since we assumed that $y_1^*,\ldots, y_k^*$ we a minimal generating set then we know that the $r_i^*$ are all zero's, that is in $J^*$.

This then gives that we have $r_i\in J$ and so we have $y_1r_1+\cdots +y_kr_k\in J^2$ and so $xr\in J^2$

Now we have that $x\not\in J^2$ and so we have that $r\in J$ and we are done.

Is what I have above correct?

Answer 1

I think this is essentially right. I think you meant to say that $x \in J - J^2$ [set subtraction] and I would want to see a little more justification for why $r \in J$ at the end, but it's good.

I do think it looks little more complicated than it could be. I would try to present the argument as: to get the cotangent space of $R/(x)$ I take that of $R$, which is $J/J^2$, and then I quotient by the span of the nonzero vector $\bar{x}$ [the image of $x$]. Just to emphasize that it's linear algebra and nothing more.