Ideal of polynomials vanishing on algebraic set containing a polynomial with roots only in the set.

Question

Lets say I have an algebraically closed field $\mathbb{K}$. I also have an algebraic set $Y \subset A^n(\mathbb{K})$.

Is it true that the ideal of polynomials vanishing on $Y$ denoted as $I(Y)$ contains a polynomial with roots only in $Y$?

Answer 1

If this were true, such a polynomial would have to vanish precisely on Y, since any element of $I(Y)$ vanishes on $Y$, and by assumption this polynomial does additionally not vanish outside of $Y$. So your question is equivalent to "is any algebraic set defined by a single polynomial?".

Spoiler alert: the answer is no. Algebraic sets which are the zero locus of a single polynomial are called hypersurfaces, and there are varieties which are not hypersurfaces. In fact, if we start out in $n$ dimensions, then the zero locus of a single non-constant polynomial will have dimension $n-1$, and the zero locus of $k$ non-constant polynomials will have dimension at least $n - k$. Proving this is overkill for now, but we can use this intuition to get a counterexample. For example the following:

Consider the case where $n > 1$, for simplicity say $n = 2$. Let's have $Y$ consisting of just one point: say $Y = \{(0,0)\}$, and let $f \in I(Y) \subset \mathbb K[X,Y]$. Suppose it does not vanish outside of $Y$. Since $f(0,0) = 0$, its constant term is zero.

Now let $0\neq z_1 \in \mathbb K$ and consider the polynomial $f(z_1,Y) \in \mathbb K[Y]$. We may assume this is non-constant, since otherwise $f$ would already vanish on the infinitely many points $(0,y)$ for $y \in \mathbb K$. Since $\mathbb K$ is algebraically closed, $f(z_1,Y)$ has a root $z_2$. Hence $f(z_1,z_2) = 0$, but $(z_1,z_2) \neq (0,0)$, a contradiction.