If $X =V(x^{2}+y^{2}-1, x-1)$, what is $I(V)$? Here $V$ is affine variety and $I(V)$ its ideal

Question

As we can remove $x^2-1=(x+1)(x-1)$ then we have $I(V)=(x-1,y^2)$.

Could someone check whether it correct?

Answer 1

Clearly the second polynomial implies $x = 1$, sub this into the first polynomial so that $y = 0$. The only point in the variety is $(1,0) \in k^2$.

The ideal of polynomials vanishing on this point is $\langle x-1, y \rangle$. You can prove that $$I(\{ \, (1,0) \, \} ) = \langle x-1, y \rangle $$ by using the definition of $I$,

$$I(\{ \, (1,0) \, \} ) = \{ \, f(x,y) \in k[x,y] \, | \, f(1,0) = 0 \, \},$$

and showing containment in both directions. One direction is trivial, the other direction is a generalization of the fundamental theorem of algebra.