A power series ring over $\mathbb C$ 
I have two questions around the ring of formal power series $R=\mathbb C[[x^2,x^3]]$. What is the global dimension of $R$? Is it a local regular ring?

The global dimension of a ring is the supremum of the projective dimensions of all modules over the ring. I guess that $R$ is semisimple, so all $R$-modules are of projective dimension zero whence the global dimension of $R$ is zero.
Thanks for any suggestion!
 A: It seems that $R$ has infinite global dimension.  Consider the algebra $S=\mathbb{C}[[x]]$ and its maximal ideal $M=(x)$.  Since $R$ is a subalgebra of $S$, it acts on $M$, so $M$ is an $R$-module.  Let us prove that $M$ has infinite projective dimension.
The projective cover of $M$ is $p:R\oplus R\stackrel{(x,x^2)}{\to} M$ (see below for a proof of this fact).  The kernel of $p$ is the ideal generated by $(x^3, -x^2)$; this ideal, viewed as an $R$-module, is isomorphic to $M$ via the map $M\to ((x^3, -x^2)): x\mapsto (x^3, -x^2)$.  Thus a minimal projective resolution of $M$ has the form
$$ \ldots \to R\oplus R \to R\oplus R \to M \to 0,$$
so $M$ has infinite projective dimension.

Proof that $p$ is a projective cover: it suffices to show that $\ker p$ is a superfluous submodule of $R\oplus R$; in other words, we need to prove that if $N$ is another submodule of $R\oplus R$ such that $\ker p + N = R\oplus R$, then $N= R\oplus R$.  If $N$ is such a submodule, then $(1,0)$ and $(0,1)$ are in $\ker p + N$.  Thus there are lements $a,b$ of $R$ and $(n_1, n_2), (m_1, m_2)$ of $N$ such that 
$$ (ax^3, -ax^2) + (n_1, n_2) = (1,0)$$
and
$$(bx^3, -bx^2) + (m_1, m_2) = (0,1).$$
This implies that $n_1$ and $m_2$, viewed as elements of $R$, have non-zero  constant terms; thus they are invertible in $R$.  From there, it is possible to write $(1,0)$ and $(0,1)$ as linear combinations of $(n_1, n_2)$ and $(m_1, m_2)$, so $N$ contains a set of generators of $R\oplus R$.  Thus $N=R\oplus R$.
