I have been looking at some examples of affine schemes to help learn some of the key concepts, most recently "the union of a parabola with the $x$-axis," i.e the spectrum of $A=k[x,y]/\big(y(y-x^2)\big)$. Before looking at the intersection I've started with a more boring point $(1,0)$, corresponding to the local ring $B=A_{(x-1,y)}$, i.e inverting all polynomials outside the ideal $(x-1,y)$. Since this corresponds to just a point on a 1D curve, I would guess that there's a ring isomorphism $B\cong k[t]_{(t-1)}$, which I'm pretty sure comes from showing that $y=0$ in $B$, but I'm having trouble proving this concretely.
I think the result follows from the abstract properties of local rings: since $y-x^2+1$ is in the maximal ideal $(x-1,y)\subseteq B$, we know that $y-x^2$ is invertible which means $y=0$. But I haven't been able to find an explicit inverse for $y-x^2$, or a more concrete proof that $y=0$ (e.g by manipulating the relation $y(y-x^2)=0$,) which would help a lot with getting a feel for why local rings work the way they do. Does anyone know if one can find an explicit inverse for $y-x^2$ or some 'concrete' proof that $y=0$? I'd appreciate any answer, or any other help in sharpening my ring theory skills for future study of schemes. Thanks!
The element $y - x^2 \in A$ is not in $(x-1, y)$, hence you get an inverse $\frac{1}{y - x^2} \in B$. On the other hand, $y (y-x^2) = 0$, so that $$0 = \frac{0}{y - x^2} = \frac{y(y-x^2)}{y-x^2} = y.$$
Quite generally, if you have zero divisors $ab = 0$, and you invert $a$, then $b$ becomes zero: $$0 = \frac{0}{a} = \frac{ab}{a} = b.$$
The most extreme case of this is when you have a ring $A$ and you localize with respect to a multiplicative system $S$, which contains $0$ itself. Then $S^{-1}A = 0$.