In the book "Commutative Algebra with a View Toward Algebraic Geometry (Eisenbud, 1995), exercise 1.10 one has to find the ring associated to the union of the circle $C:(y+1)^2+x^2=1$ and the parabola $P:y=x^2$.
Beginning: as $I(C\cup P)=I(C)\cap I(P)$ then one has to find $(y-x^2)\cap(y^2+2y+x^2)$ as intersection of ideals in $k[x,y]$. As all these ideals are radicals the ring is then $k[x,y]/(y-x^2)\cap(y^2+2y+x^2)$. I think we have to simplify this intersection and my idea was $(y-x^2)\cap(y^2+2y+x^2)=((y-x^2)(y^2+2y+x^2))$.
We have $((y-x^2)(y^2+2y+x^2))\subseteq(y-x^2)\cap(y^2+2y+x^2)$ but since we don't have $(y-x^2)+(y^2+2y+x^2)=1$ (geometrically: intersection not empty) I can't conclude equality. The fact is that I don't know if it is an equality (condition $\mathfrak{a}+\mathfrak{b}=1$ is only a sufficient condition isn't?)
I then make the following reasoning with $I=(y-x^2)$ and $J=(y^2+2y+x^2)$: one has $Z(IJ)=Z(I)\cup Z(J)$ so with the Nullstellensatz $\sqrt{IJ}=I(Z(I)\cup Z(J))=IZ(I))\cap IZ(J)=\sqrt{I}\cap\sqrt{J}$ so $\sqrt{I}\sqrt{J}=\sqrt{I}\cap\sqrt{J}$ and so as my two ideals are radicals I get $IJ=I\cap J$. Problem: I find the relation $\sqrt{I}\sqrt{J}=\sqrt{I}\cap\sqrt{J}$ very suspicious because with it we would get for all prime ideals $\mathfrak{p},\mathfrak{q}$ the relation $\mathfrak{p q}=\mathfrak{p}\cap\mathfrak{q}$ without $\mathfrak{p}+\mathfrak{q}=1$.
An other idea (not so far) is to tell that $Z(IJ)=Z(I)\cup Z(J)$ so $Z(IJ)=P\cup C$ so with Nullstellensatz $\sqrt{IJ}=I(P\cup C)$ so $IJ=I(P\cup C)$. With that I get the solution but I don't finally know if $(y-x^2)\cap(y^2+2y+x^2)=((y-x^2)(y^2+2y+x^2))$.
Hint. $y-x^2$ and $y^2+2y+x^2$ are irreducible polynomials.