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Over an algebraically closed field, every regular function on a nonempty irreducible projective variety is constant. Regard the function as a morphism to ${\mathbb{P}}^1$ avoiding infinity. Its graph is closed; the closedness of projection from projective space makes its image closed in ${\mathbb{P}}^1$. A proper closed subset there is finite, and an irreducible finite image is a singleton. Irreducibility is the usual variety convention; on a disconnected projective algebraic set different components may have different constants.

Work in the classical setting of varieties over an algebraically closed field $k$. A prevariety is separated when its diagonal morphism ${\Delta }_X:X\to X{\times }_{k}X$ is a closed immersion; for prevarieties the diagonal is already an immersion, so this is equivalent to its image being closed. A complete variety has the following closed-projection property: for every variety $Y$, projection $X{\times }_{k}Y\to Y$ maps closed subsets to closed subsets. This is universal closedness; since a variety is separated and of finite type, it is equivalent to the structural map being proper.

First consider projective space. In two sets of homogeneous coordinates its diagonal is defined byThese equations hold exactly when the two nonzero coordinate vectors are proportional. They are homogeneous in each coordinate set, so they define a closed subset of ${\mathbb{P}}^n\times {\mathbb{P}}^n$. On a common affine chart, the diagonal map is the usual closed immersion defined by equality of the affine coordinates. Thus projective space is separated. If $X\hookrightarrow {\mathbb{P}}^n$ is a projective embedding, its diagonal is the intersection of that closed diagonal with $X\times X$. Consequently every projective variety is separated.

Here is an algebraic proof of completeness, rather than an appeal to projective compactness in an analytic topology. It suffices to prove closedness of projection from projective space over an affine chart $Y$, with coordinate ring $A=k[Y]$. A closed subset $Z\subseteq {\mathbb{P}}^n\times Y$ is cut out by a homogeneous ideal $I\subseteq A[T_0,\dots ,T_n]$. Put $B=A[T_0,\dots ,T_n]/I$ and write $B_d$ for its degree-$d$ part. Each $B_d$ is a finitely generated $A$-module, because it is a quotient of the free module on the degree-$d$ monomials.

Suppose the fibre over $y\in Y$ is empty. The specialized homogeneous equations then have no nonzero common zero. By the Hilbert Nullstellensatz, some power of each $T_i$ belongs to the specialized ideal. If those powers are $T_i^{r_i}$, then every monomial of degree $d=1+{\sum }_i(r_i-1)$ belongs to that ideal. HenceThe Nakayama lemma gives $(B_d{)}_{{\mathfrak{m}}_y}=0$. Since $B_d$ is finitely generated, there is $a\in A\setminus {\mathfrak{m}}_y$ with $(B_d{)}_a=0$: choose generators and multiply finitely many elements annihilating their localizations. On the principal open subset $D(a)$, all graded pieces of degree at least $d$ vanish too, since the quotient is generated in degree one. Every fibre over $D(a)$ is therefore empty: a projective point would have a nonzero coordinate whose degree-$d$ power could not vanish. Thus the complement of the projection of $Z$ is open, proving the required closed-projection property.

An affine open cover of an arbitrary $Y$ establishes the same result there. Finally, a closed subset of $X\times Y$ is closed in ${\mathbb{P}}^n\times Y$ when $X$ is a projective variety, so its image is closed by the preceding argument. ThereforeOver a non-algebraically-closed field, the corresponding statement concerns schemes or geometric points. Projection of rational points alone need not be closed, so it must not replace the completeness definition.

By part (a), work first with an affine base and its homogeneous equations. Put $S=k[y_0,\dots ,y_n]$, write $S_N$ for its finite-dimensional degree-$N$ vector space, and use $S_j=0$ for $j<0$. At each base point define the graded multiplication map for a projective fiberIts image is exactly $(I_x{)}_N$, since taking the degree-$N$ part of any expression in the homogeneous ideal discards all other degrees. Moreoverbecause degree-$N$ monomials generate ${\mathfrak{m}}^N$.

Choose monomial bases for the domain summands and codomain. Every entry of the resulting matrix is a coefficient of one of the $f_i$, hence a regular function of $x$. If $h_N=\dim S_N$, failure of surjectivity is the vanishing of all $h_N$-by-$h_N$ matrix minors. It is therefore a closed rank-defect locus $C_N$. If the matrix has fewer than $h_N$ columns, the failure is automatic and $C_N=X$, also closed. Constants and $N=0$ are covered by the same construction.

Part (c) gives $\pi (Y)={\bigcap }_{N\ge 0}{C}_N$, an intersection of closed subsets, hence closed. Apply part (a) to pass back to arbitrary $X$. Thus projection from $X\times {\mathbb{P}}^n$ sends every closed subset to a closed subset. This proves the closedness of projection from projective space without assuming the properness result being proved.

The field convention matters: a rational-point projection need not be Zariski closed. Over $\mathbb{R}$, the closed equation $y_0^2=t{y}_1^2$ in ${\mathbb{A}}^1\times {\mathbb{P}}^1$ has real-point image $[0,\infty )$. Indeed $y_1=0$ would force $y_0=0$, so a solution exists precisely when $t$ is a square in $\mathbb{R}$. This infinite proper subset of the affine line is not Zariski closed. The valid arbitrary-field theorem uses scheme or geometric images, for which the matrix argument remains valid at every residue field.