# Show a homomorphism of local rings of two varieties

Shafarevich, Basic Algebraic Geometry 1, II.1.1

"Prove that the local ring $\mathcal O_x$ of the curve $xy=0$ at $(0,0)$ is isomorphic to the subring $\mathcal O \subset \mathcal O_1 \oplus \mathcal O_2$, where $\mathcal O_1$ and $\mathcal O_2$ are copies of the local ring of $0$ in $\mathbb A^1$ consisting of functions $f_1,f_2$ with $f_1\in \mathcal O_1$ and $f_2\in\mathcal O_2$ such that $f_1(0)=f_2(0)$."

I know that $\mathcal O_1=\mathcal O_2=\mathbb C[t]_{(t)}$ and that $\mathcal O=(k[x,y]/(x,y))_{(x,y)}=\{\frac{p(x)+q(y)}{r(x)+t(y)}:r(0)+t(0)\neq 0\}$.

I can't figure out the isomorphism, though. I was thinking $\frac{p+q}{r+t}\to (\frac{p}{r},\frac{q}{t})$. I think I've shown this is a homomorphism, but I can't figure out what the kernel has to do with the hint in the problem. Thanks!

Let $f \in {k[x,y]/(x,y)}_{(x,y)}=\mathscr O$ (the local ring at the origin of the curve $xy=0$.
Then define a map $\mathscr O \to \mathscr O_1 \times \mathscr O_2$ by $f(x,y) \mapsto (f(x,0),f(0,y))$. This is a well-defined ring homomorphism.
The kernel consists of all $f$ such that $f(x,0)=f(0,y)=0$. The first condition implies that $f \equiv 0 \pmod y$, and the second condition implies that $f \equiv 0 \pmod x$. Thus $f \in (y) \cap (x)=(xy)$, meaning that $f$ is already zero in the quotient. Hence the map is injective.
That the map lands in $\mathscr O$ is obvious, and that it is surjective onto $\mathscr O$ is also easy to see, since the condition is for $f_1,f_2$ to have the same constant term.