= Solution
Work in the classical setting of <varieties> over an algebraically closed <field> $k$. A prevariety is separated when its <diagonal morphism> $\Delta_X:X\to X\times_kX$ 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 $X\times_kY\to Y$ 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 by
$$
X_iY_j-X_jY_i=0\qquad(0\le i,j\le n).
$$
These 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 \b[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> $I\subseteq A[T_0,\ldots,T_n]$. Put $B=A[T_0,\ldots,T_n]/I$ 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. Hence
$$
B_d\otimes_A k(y)=0.
$$
The <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. Therefore
$$
\boxed{\text{Every projective variety is separated and complete.}}
$$
Over 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.
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