# Must the projective closure of a closed subset of affine space have points at infinity?

let $k$ be an algebraic closed field. For $I\subseteq k[x_1,\ldots,x_n]$, we donote $I^*$ to be the ideal generated by the set $\{F^*|F\in I\}$, here $F^*=x_{n+1}^{deg F}F(x_1/x_{n+1},\ldots,x_n/x_{n+1})\in k[x_1,\ldots,x_{n+1}]$.

Now, $V=Z(\mathfrak{p})$ is an irreducible algebraic set of $\mathbb{A}^n(k)$. $V^*=Z(\mathfrak{p}^*)$ is a irreducible algebraic set of $\mathbb{P}^n$, called the projective closure of $V$. It is the smallest algebraic set in $\mathbb{P}^n$ containing $V$.

My question is: When is $V^*$ really bigger than $V$ ? That is to say $V^*$ has infinities.

Trivial case: $V$ is a point, then $V^*$ is a point too, thus $V^*$ do not increase points.

So I want to know if this will be the only case that $V^*$ equals $V$?

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The projective closure $\overline{S}$ of a subset $S$ of affine $n$-space over an algebraically closed field will contain no points "at infinity" iff $\overline{S}$ is an affine variety. But the only subvarieties of projective space which are both affine and projective are the finite sets.
(One can see this for instance by observing that any regular function on a projective variety must be constant on all of the finitely many connected components of the variety, so the coordinate ring of a projective variety is a finite-dimensional $k$-algebra. When $k = \mathbb{C}$ this boils down to the usual open mapping theorem of complex analysis.)
Thus the answer is that $\overline{S}$ is strictly larger than $S$ iff $S$ is infinite.
Added: I missed the word "irreducible" when I read the question the first time. Since a finite algebraic set is irreducible iff it has exactly one point, yes, assuming irreducibility the only possibility is that $S$ consists of a single point.