Exercise I.6.5 of Hartshorne

algebraic-geometrycommutative-algebra

Let X be a nonsingular projective curve. Suppose that X is a (locally closed) subvariety of a variety Y (Ex. 3.10). Show that X is in fact a closed subset of Y. See (II, Ex. 4.4) for generalization.

If we use the fact that regular morphism defined in Harris (or proper morphism) is closed, then this is just direct consequence of the fact. However, it seems that Hartshorne did not introduce any of such notions of proper map in Chapter I. Is there a way to prove this using materials of Chapter I?

What I also tried to do is to assume $\overline{X}$ in $Y$ and apply Theorem 6.8; however, since we may not assure that whether $\overline{X}$ is nonsingular or not, this doesn't work.

Best Answer

The key theorem for this exercise in the context of Hartshorne chapter I is theorem I.5.7A:

Theorem (Elimination Theory). Let $f_1,\cdots,f_r$ be homogeneous polynomials in $x_0,\cdots,x_n$, having indeterminate coefficients $a_{ij}$. Then there is a set $g_1,\cdots,g_t$ of polynomials in the $a_{ij}$, with integer coefficients, which are homogeneous in the coefficients of each $f_i$ separately, with the following property: for any field $k$, and for any set of special values of the $a_{ij}\in k$, a necessary and sufficient condition for the $f_i$ to have a common zero different from $(0,0,\cdots,0)$ is that the $a_{ij}$ are a common zero of the polynomials $g_j$.

This lets us prove that any map out of a projective variety is closed. $\renewcommand{\PP}{\mathbb{P}}$

To do this, we need to show that $\PP^n\times Y\to Y$ is a closed map for any variety $Y$. By covering $Y$ with affines, we may reduce to the case that $Y$ is affine. Now suppose $Z\subset \PP^n\times Y$ is closed: this means that it's cut out by a list of equations $f_1,\cdots,f_r$ which are homogeneous polynomials in the homogeneous coordinates on $\PP^n$ with coefficients taken from $A(Y)$. By theorem I.5.7A, we may find polynomials $g_1,\cdots,g_s$ in the coefficients of these $f_i$ which vanish iff all the $f_i$ simultaneously vanish. But this exactly means that the image of $Z\subset \PP^n\times Y$ under the projection $\PP^n\times Y\to Y$ is closed, or that the projection $\PP^n\times Y\to Y$ is closed.

Now consider a morphism $f:Z\to Y$ where $Z$ is projective and $i:Z\to \PP^n$ is the inclusion of $Z$ in to some projective space (guaranteed by the definition of projective). We can factor $f$ as $Z\to \PP^n\times Y\to Y$ where the first map is $z\mapsto (i(z),f(z))$. To show the image of this map is closed, cover $\PP^n\times Y$ by affine open subsets of the form $U_i\times Y$ for $U_i$ the standard affine opens. Then on each of these opens, the set of points $(i(z),f(z))$ is closed because it's cut out by the equations for $Z\cap U_i$ and then also the equalities $y_i=f_i(z)$ where $y_i$ are the coordinate functions on $Y$ and $f_i$ are the components of the map $f$. So the image of $Z$ in $\PP^n\times Y$ is closed and therefore the image of $Z$ in $Y$ is closed since the projection $\PP^n\times Y\to Y$ is closed.

To apply this to our problem, write $i:X\to Y$ for the inclusion. Then by the above, $i(X)$ is closed in $Y$.

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