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Let $\mathbb{A}_{2}$ denote the 2-affice space over $\mathbb{C}$. Now let $Y=V(y-x^{2})$, here $V$ denotes the zero-locus. Now by definition the coordinate ring is the quotient ring $k[x,y]/I(Y)$ where $I(Y)$ is the ideal generated by all polynomials vanishing on $Y$.

Now the author computes $k[x,y]/I(Y)$, then it computes $k[x,y]/$, so my question is: why is $I(V(y-x^{2}))=$?

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By the Nullstellensatz, $I(V(J))=\sqrt{J}$ for any ideal $J$. Because $y-x^2$ is an irreducible polynomial in $k[x,y]$, we have that $J=(y-x^2)$ is a prime ideal, and hence a radical ideal (i.e. $J=\sqrt{J}$). Thus $I(V(y-x^2))=(y-x^2)$.

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You don't need the Nullstellensatz. A polynomial $p(x, y) \in k[x, y]$ vanishes on the set of all points $(t, t^2)$ if and only if $p(t, t^2)$ is the zero polynomial when $k$ is infinite, and one can verify directly that this is true if and only if $p(x, y)$ is divisible by $y - x^2$ by replacing all occurrences of $y$ with $x^2$. In particular you do not need $k$ algebraically closed.