### Projective morphisms

This post is about the morphisms between projective varieties. There are some aspects of such morphisms that I’m troubled about. The development will closely follow that in Karen Smith’s “Invitation to Algebraic Geometry”.

First, say we have a morphism $\phi:\Bbb{P}^1\to\Bbb{P}^2$ such that $[s,t]\to[s^2,st,t^2]$. We will try and analyze this map.

This map has to be homogeneous: in that each coordinate has to be homogeneous, and the homogeneity has to be of the same degree. This is the only way that such a map between projective varieties can be well-defined.

Now let us talk about the mappings from affine charts. Essentially, the affine charts cover the projective space, and hence every projective variety that lives in that space. When we talk about a particular affine chart, we can reduce the number of variables by $1$. Because the value of one variable is always $1$: hence it can be neglected. However, is the image also an affine chart? That depends. In this case of $[s,t]\to [s^2,st,t^2]$, the image of an affine chart will be an affine chart. This is because $s=1\implies s^2=1$. Similarly, $t=1\implies t^2=1$.

We’ve covered all the possible points in the domain by picking out the affine charts. Hence, we have fully described the map.

A map $f$ between projective varieties is a projective morphism if for each $p\in V$, where $V$ is the domain, there exists an neighbourhood $N(p)$ such that $f|_{N(p)}$ is a homogeneous polynomial map. Is an affine chart an open set? Yes. If it is the $z$th affine chart it is the complement of the algebraic set $z=0$ in $\Bbb{P}^n$.

Let us now consider a different map: consider $V(xz-y^2)\in\Bbb{P}^2$. Let us call this curve $C$. Now consider the map $C\to \Bbb{P}^1$, defined as $[x,y,z]\to [x,y]$ if $x\neq 0$ and $[x,y,z]\to [y,z]$ if $z\neq 0$. What does this mean?

First of all, why is the option $y\neq 0$ not included? If $y\neq 0$, both $x,z\neq 0$ is implied. Hence, this case is a subcase of the two cases considered earlier. Secondly, what does it mean to map to a projective space of a lower dimension? The curve is one-dimensional. Is that the reason why we can embed it in $\Bbb{P}^1$? Probably. Note that we haven’t yet proven that this mapping is an embedding. However, this will indeed turn out to be the case.

Is this map consistent? In other words, are the two maps the same in the intersection of the open sets $x\neq 0$ and $y\neq 0$? Let us see. $[x,y]\to [xz,yz]\to[y^2,yz]\to [y,z]$. Hence, when $x,z\neq 0$, this map is consistent.

Why do we have to have such a broken up map? Why not one consistent map? First of all, mapping from affine charts seems like a systematic way to map. You can always ensure that at least one coordinate is non-zero; both in the domain and range. That is really all there is to it. Sometimes on restricting to affine charts, you write affine maps, like in the precious example. In other cases, including this one, you write a projective map. Defining the various projective maps, whether they change with affine charts or not, is of paramount importance. The affine map part is just an observation which may or may not be made.

How to map a curve $f(x_0,x_1,x_2,\dots,x_n)$ to $\Bbb{P}^1$ in general? This seems to be a very difficult question. [This](http://web.stanford.edu/~tonyfeng/Brill_Noether1.pdf) suggests that every smooth projective curve can be embedded in $\Bbb{P}^3$. That seems to be the best we can do.

For completeness, I would like to mention that the two maps given above are inverse to each other, although this is unrelated to the motivation for this article.

### Projective Closures of Affine Varieties

This blog post is going to be on the projective closures of affine varieties. This too is something that has confused me for some time. I did manage to make sense of most of it eventually, but I want to still write it down for my peace of mind.

Say we have a polynomial $p(x_1,x_2,\dots,x_n)$ in affine space $\Bbb{A}^n$. We want to find the projective closure of that. How do we do it? We first homogenize it by writing it as a homogeneous polynomial $p'(x_1,x_2,\dots,x_n,z)$. Note that $z$ is the extra variable that has been added here. Then, after having sketched the zeroes of the polynomial in the $z$-th affine chart in projective space, we find the projective closure in the whole space by assuming $z=0$, and then determining what are the zeroes of the remainder of the polynomial.

Why does this work? The complement of the $z$-th affine chart is that space in projective space that the affine curve could never have reached in affine space. Remember that the whole affine curve, point by point, lies inside the affine chart. Hence, the points in the complement of the chart will contain all the “new” points of the curve. Heuristically speaking, the “new” points are all those points that the curve tends to, if it could go infinitely far, whatever that might mean.

Now how do you find the projective closure of a general algebraic set (and not one that is generated by a single polynomial)? In other words, say you have an affine algebraic set $X$. What changes do you need to make to $I(X)$ in order for $V(I(X))$ to be the projective closure of $X$? Turns out, homogenizing the generators of $I(X)$ is not the answer, but homogenizing *all* the elements of $I(X)$ is. Is is the explanation of this part that is the aim for writing this post.

Why does homogenizing the generators not work? This is because we may count extra points of projective closure. How? Take the polynomials $x+y+1$ and $x+y+2$ in $\Bbb{R}^2$. Their corresponding homogeneous polynomials are $x+y+z$ and $x+y+2z$ in $\Bbb{P}^3$. The variety corresponding to the ideal $I(x+y+1, x+y+2)$ is obviously $\emptyset$. Hence, the affine chart of $\Bbb{P}^2$ will also contain the empty set, the projective closure of which is also the empty set. However, the projective closures of both $x+y+z$ and $x+y+2z$ contain the point $x+y=0$. From this, we can see that a set of algebraic sets in affine space may all contain a set of points of projective closure that their intersection may not.

This problem is solved by homogenizing all the polynomials in $I(X)$, and then generating an ideal from it. The variety corresponding to this ideal is exactly the projective closure of $X$. Let this ideal be referred to as $\tilde{I}(X)$. Why does this work?
First of all, it is clear that the projective closure of $X$ lies inside $V(\tilde{I}(X))$. Now we need to prove the other inclusion. If we can produce a homogeneous polynomial such that the corresponding variety on the affine chart is the same as $X$, then we’ll be done. This we can do by just adding the homogenized versions of all the generators in $I(X)$, but also ensuring that the homogenized versions are of different degrees!!! The key insight behind this is that as we have a sum of homogeneous polynomials of different degrees, the corresponding variety will contain only those points that lie in the intersection of all zero sets of the individual polynomials. The affine chart will consequently only contain those affine points which lie in the zero sets of all those affine polynomials, which is precisely $X$. To complete the proof, we need to assume that each affine variety has a unique projective closure.

Such a polynomial (sum of homogenized versions, to different degrees, of all the generators of $I(X)$) does in fact exist in $\tilde{I}(X)$. Hence proved.