If $X$ is a affine variety and $Y$ is a quasiaffine variety of $X$, then $I(X)=I(Y)$?

Question

Let $X$ be an algebraic affine variety, that is, a closed irreducible subset of $\mathcal{A}^n$, and let $Y$ be an open subset of $X$. Does it happen that $I(X)=I(Y)$? I think it happen because if $f=0$ on a open dense subset of $X$ then it must be 0 in the whole $X$, but I'm not sure.

Answer 1

$X$ affine variety means $X = V(I)$ for some prime ideal $I = I(X)$ of $k[t_1,\ldots,t_n]$.

$I(X) = I(Y)$ is the definition of $Y$ is dense.

$Y$ is open in $X$ means $C=X \setminus Y$ is closed that is $C = V(J)$ (for some ideal of $k[t_1,\ldots,t_n]$ containing $I(X)$)

Let $Z= V(I(Y))$ be the closure of $Y$ (closure in $\Bbb{A^n}$ or $X$ it gives the same), then $X = Z \cup C$ where $Z,C$ are closed.

That none of $Z,C$ is the whole of $X$ means $X$ is not irreducible (and $I(X)$ is not a prime ideal since by definition of closed we can find $f,g$ vanishing on one but not the other so $fg = 0,f \ne 0,g\ne 0$ in $k[t_1,\ldots,t_n]/I(X)$).

We know it isn't the case, thus either $Z= X$ and $I(X)=I(Z) = I(Y)$, or $C= X$ and $Y = \emptyset$.