Suppose that $X$ is a reduced projective scheme over a field $k$ (algebraically closed, say), such that it has two irreducible components, both isomorphic to $\mathbb P^1$, which meet at an ordinary double point $P$ (i.e. $\widehat{\mathcal O}_{X,P} \cong k[[x,y]]/(xy)$). Why is $X$ uniquely determined up to isomorphism?
I thought one possible strategy is to remove one smooth point on each component and show that the remainder is isomorphic to $Y := \{(x,y) \in \mathbb A^2 : xy = 0\}$ (and then glue isomorphisms). I know this locus is affine (by EGAII, Corollary 6.7.3) but I don't know how to show it's isomorphic to $Y$.
Added: OK, suppose $Z = Spec(A)$ is an affine variety over $k$ with ordinary double point and two irreducible components isomorphic to $\mathbb A^1$ each. Then on coordinate rings we get surjections $A \to k[x]$, $A \to k[y]$, hence an injection $A \to k[x] \times_k k[y] \cong k[x,y]/(xy)$. Why is this an isomorphism?
(The assumption on the ordinary double point is clearly necessary, otherwise we could take e.g. $k[x+y,x^2] \subset k[x,y]/(xy)$, which is isomorphic to $k[s,t]/(t(s^2-t))$.)