# Segre Embedding And Heights

# Product

The Segre embedding allows us to define the product of projective varieties reasonably, and we will discuss it right now. To begin with we consider the product of $\mathbb{P}^m$ an $\mathbb{P}^n$. In this section the ground field is an arbitrary algebraically closed field.

Definition 1.TheSegre embeddingis defined as follows:Clearly, $N=(m+1)(n+1)-1=mn+m+n$. The image on the right hand side has $X_iY_j$ ordered lexicographically.

First of all we make sure that this function is well-defined, otherwise our work will be useless.

Proposition 1.The Segre embedding is a well-defined injective map.

*Proof.* Assume $X_i’=\lambda X_i$ and $Y_j’=\mu Y_j$ for some $\lambda,\mu \ne 0$, then

Next suppose that $[X_iY_j]=[X_i’Y_j’]$. Without loss of generality we can assume that $X_0 \ne 0$ and $X_0’ \ne 0$ so that we can put them to $1$. Then by looking at first $n+1$ elements we can identify $[Y_0:\dots:Y_n]$ with $[Y_0’:\dots:Y_n’]$. Then using $X_1Y_i=\lambda X_1’Y_i’$ we can immediately identify $X_1$ and $X_1’$. Likewise, other components are identified. $\square$

Next we study the image further using linear algebra

## Image of the Segre Embedding And Rank of A Matrix

We can write elements in $\mathbb{P}^N$ as a $(m+1) \times (n+1)$ matrix, which can make things easier:

Therefore the image of $\iota$ is given by $Z_{ij}=X_iY_j$. Through an elementary observation, we see the matrix

has rank $1$. The question is, is the converse true? For this reason we study the set

Note $Z_{ij}Z_{kl}-Z_{kj}Z_{il}$ is the determinant of all $2\times2$ submatrices the matrix $[Z_{ij}]$. This $Z$ contains all $[Z_{ij}]$ with rank $1$. To show the converse, we consider the standard affine cover. Let $U_{kl}=Z(Z_{kl})$ and put $V_{kl}=\mathbb{P}^N \setminus U_{kl}$. Then $\{V_{kl}\}$ is the standard affine cover of $\mathbb{P}^N$ as we all know. Then likewise we use the affine open subset $U_k \subset \mathbb{P}^m$ and $U_l’ \subset \mathbb{P}^n$, to obtain

This is indeed the inverse map of $\iota$ on $U_k \times U_l’$. Hence the converse is true.

Therefore, the image of the Segre embedding is a projective variety. As a classic example, the image of $\mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^3$ is determined by the polynomial $xy-zw=0$.

# Height

In this section we offer a way to understand the Segre embedding in number fields. To begin with, we need some definition.

Height is computed by absolute values on a field so we first normalise all absolute values on $\mathbb{Q}$. Recall that two absolute values $|\cdot|_1$ and $|\cdot|_2$ are equivalent if there exists some $\lambda>0$ such that $|\cdot|_1=|\cdot|_2^\lambda$. The question is, which $\lambda$ should we pick.

The ordinary absolute value $|\cdot|_\infty$ has nothing to worry about. But for $p$-adic absolute values $|\cdot|_p$, instead of using other equivalent ones, we have a restriction that $|p|_p=\frac{1}{p}$. All these absolute values will be denoted by

Likewise we define $M_K$, where throughout $K$ will always be a number field. $M_K$ consists of the ordinary absolute value and extensions of $p$-adic ones defined as follows:

In particular, $M_K$ satisfies the product formula:

for all $x \in K^\times$. This restriction allows us to work fine with projective spaces, as we will see later.

Definition 2.The (absolute logarithmic)heightof $x\in \mathbb{P}^n_{\overline{\mathbb{Q}}}$, with coordinates $(x_1,\dots,x_m) \in K$, is defined by

Actually, the height function can show the “algebraic complication” of $x$, and is well-defined in many senses.

Proposition 2.The height $h(x)$ is independent of the choice of $K$.

*Sketch of the proof.* Let $L$ be another number field containing $x_0,\dots,x_m$, then we can assume that $K \subset L$, then $L/K$ is a finite separable extension. But in this case,

which implies that

Therefore

gives what we want. $\square$

Proposition 3.$h(x)$ is well-defined on $\mathbb{P}^n_{\overline{\mathbb{Q}}}$.

*Proof.* It remains to show that $h(x)$ is independent of the choice of coordinates. For $\lambda \ne 0$, we see

Note $\sum_{v \in M_K}\log|\lambda|_v=0$ because o the product formula. $\square$

To highlight the ability of height to measure algebraic complication, let’s mention the following theorem of Kronecker.

Theorem 1 (Kronecker).The height of $\zeta\in\overline{\mathbb{Q}}^\times$ is $0$ if and only if $\zeta$ is a root of unity.

One direction is straightforward. To prove the converse, one may need some combinatorics, symmetric functions and Dirichlet’s pigeon-hole principle. See theorem 2.4 of this note for a proof.

## Segre Embedding And Height of Polynomials

Now let’s invite the Segre embedding into the party:

Using the fact that $\max_{i,j}|x_iy_j|_v=\max_i|x_i|_v \cdot \max_j|y_j|_v$, we see immediately that

The Segre embedding is immediately used after introducing the height of a polynomial.

Definition 3.For $f(t_1,\dots,t_n) \in K[t_1,\dots,t_n]$, we writeThen the

heightof $f$ is defined to bewhere

Likewise, it can show the complication of $f$ in some way, but we are not interested in it at this moment. Notice that products of multivariable polynomials in **different variables** can be understood as tensor products and therefore we have the following fact

Proposition 4.Let $f(t_1,\dots,t_n)$ and $g(s_1,\dots,s_m)$ be polynomials in different sets of variables, then

Note: if $f$ and $g$ share the same variable, the story is different. Say we have $f_1,\dots,f_m \in \overline{\mathbb{Q}}[X_1,\dots,X_n]$, and put $f=f_1 \dots f_n$, then

where $d$ is the sum of the partial degrees of $f$.

Segre Embedding And Heights