TITLE: Some question ideal of variety
QUESTION [5 upvotes]: For an affine variety $X=V(x^{2}+y^{2}-1, x-1)$, I found the ideal of $X$,  $I(X)=\langle x-1,y\rangle$. But I don't know $I(X)=\langle x^{2}+y^{2}-1, x-1\rangle$.

REPLY [6 votes]: It's false that $I(X)=\langle x^2+y^2-1,x-1\rangle$. The ideal $\langle x^2+y^2-1,x-1\rangle$ is equal to $\langle x-1,y^2\rangle$, because $x^2-1=(x+1)(x-1)$ can be removed, but $\langle x-1,y^2\rangle$ is not equal to $\langle x-1,y\rangle$. 
This is fine, though - have another look at the Nullstellensatz:
$$I(V(J))=\text{rad}(J)$$
where $\text{rad}(J)$ is the radical of the ideal $J$. In general, it is only true that $\text{rad}(J)\supseteq J$.
Do you see why the radical of $\langle x-1,y^2\rangle$ is equal to $\langle x-1,y\rangle$?
Hint: it may help to first prove that
$$\text{rad}(J_1+J_2)=\text{rad}(\text{rad}(J_1)+\text{rad}(J_2))$$
for any ideals $J_1$ and $J_2$.