Let,$R = \mathbb K[x_0,...,x_n]. A = \mathbb K[x_0,...,x_{n-1}]$, where $\mathbb K$ is an algebraically closed field. My question is the following :
Is it true that $R/(x_1,...x_n)^2 \otimes R/(x_n) = A /(x_1,...,x_{n-1})^2$ ?
I know that $R/(x_1,...x_n)^2 \otimes R/(x_n) = R/(x_1,...x_n)^2 + (x_n)$ as we have $R/I \otimes R/J \cong R/I+J$.
But then how can we show that $R/(x_1,...x_n)^2 + (x_n) \cong A /(x_1,...,x_{n-1})^2$ ?
If we define a map $\phi :A \to R/(x_1,...x_n)^2 + (x_n)$ in an obvious way then I can see the map is surjective and $(x_1,...,x_{n-1})^2 \subset Ker (\phi)$,But how to show the equality?( This is basically coming from an attempt to compute scheme theoretic intersection of two closed subschemes of a scheme)
Any help from anyone is welcome.