How to calculate the embedding dimension of this ring Define $$R=\frac{k[x,y,z]_{(x,y,z)}}{(x^2z,y^2z)}.$$
How can I prove that $\operatorname{edim}R-\dim R\leq1$ (where edim means minimal number of generators of the maximal ideal) ?
It seems to me that $\operatorname{edim}R=3$ but how can I prove it?
 A: Let $M = (x,y,z)$ and $I = (x^2 z, y^2 z)$ be ideals of $k[x,y,z]$. Let $\mathcal{M}=(x,y,z)=M/I$ be a maximal ideal of $k[x,y,z]/I$. Let $\mathfrak{m}$ be the maximal ideal of $R$. Note that $R = k[x,y,z]_M / I k[x,y,z]_M  = (k[x,y,z]/I)_\mathcal{M}$.
It is clear that $\mathfrak{m}/\mathfrak{m}^2 = \mathcal{M}/\mathcal{M}^2 = (M/I) / (M/I)^2 = M/ (M^2 + I)$. But $I \subseteq M^2$, then $M/(M^2 +I) = M/M^2$ has $\{x,y,z\}$ as a basis over $k$. Then
$$
\mathrm{emb.dim} R = \dim_k \mathfrak{m}/\mathfrak{m}^2 = \dim_k M/(M^2+I) = 3.
$$
A: When $A$ is a local ring with maximal ideal $\mathfrak m$ and $\mathfrak a\subseteq\mathfrak m^2$, then edim $A/\mathfrak a=\operatorname{edim}A$. (This follows easily if you note that $(\mathfrak m/\mathfrak a)^2=(\mathfrak m^2+\mathfrak a)/\mathfrak a=\mathfrak m^2/\mathfrak a$.)
In your case $R=A/\mathfrak a$, with $A=k[x,y,z]_{(x,y,z)}$ and $\mathfrak a=(x^2z,y^2z)$. Therefore $\operatorname{edim}R=3$, $\dim R=2$ (since $(x^2z,y^2z)\subset (z)$, and $(z)$ is a prime ideal of coheight $2$), and you have $\operatorname{edim}R-\dim R=1$.
