# 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:\[ \begin{aligned} \iota:\mathbb{P}^m \times \mathbb{P}^n &\to \mathbb{P}^N \\ ([X_0:\cdots:X_m],[Y_0:\cdots:Y_n]) &\mapsto [X_0Y_0:X_0Y_1:\cdots:X_mY_n] \end{aligned} \]

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

\[ \begin{aligned} \iota([X_0':\dots:X_m'],[Y_0':\dots:Y_n'])&=[X_0'Y_0':\dots:X_m'Y_n'] \\ &=[\lambda\mu X_0Y_0:\dots:\lambda\mu X_mY_n] \\ &=\lambda\mu[X_0Y_0:\dots:X_mY_n] \\ &=[X_0Y_0:\dots:X_mY_n] \\ \end{aligned} \]

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:

\[ \begin{bmatrix} Z_{00} & \dots & Z_{0n} \\ \vdots & \ddots & \vdots \\ Z_{m0} & \cdots & Z_{mn} \end{bmatrix}. \]

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

\[ \begin{bmatrix} X_0Y_0 & \cdots & X_0Y_n \\ \vdots & \ddots & \vdots \\ X_mY_0 & \cdots & X_mY_n \end{bmatrix} \]

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

\[ Z=Z(\{Z_{ij}Z_{kl}-Z_{kj}Z_{il}:1 \le i,k \le m,1 \le j,l \le n\}). \]

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

\[ V_{kl} \cap Z\to U_k \times U_l', \quad [Z_{ij}] \mapsto ([Z_{0l}:\dots:Z_{ml}],[Z_{k0}:\dots:Z_{kn}]) \]

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

\[ M_\mathbb{Q}=\{|\cdot|_p:p\text{ is prime or}=\infty\}. \]

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:

\[ |x|_v=|N_{K_v/\mathbb{Q}_p}(x)|_p^{1/[K:\mathbb{Q}]},\quad \forall x\in K,v|p. \]

In particular, \(M_K\) satisfies the product formula:

\[ \prod_{v \in M_K}|x|_v=1 \text{ or } \sum_{v \in M_K}\log|x|_v=0 \]

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\[ h(x)=\sum_{v \in M_K}\max_j \log |x_j|_v. \]

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,

\[ \sum_{w|v}[L_w:K_v]=[L:K], \]

which implies that

\[ \sum_{w|v}\log |x|_w=\log |x|_v. \]

Therefore

\[ \sum_{w \in M_L}\max_j\log|x_j|_w=\sum_{v \in M_K}\sum_{w|v}\max_j\log|x_j|_w \]

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

\[ \begin{aligned} h(\lambda{x})&=\sum_{v \in M_K}\max_j\log|\lambda x_j|_v \\ &=\sum_{v \in M_K}\left(\log|\lambda|_v+\max_j\log|x_j|_v \right) \\ &=\sum_{v \in M_K}\log|\lambda|_v+\sum_{v \in M_K}\max_j\log|x_j|_v \\ &=\sum_{v \in M_K}\max_j\log|x_j|_v \\ &=h(x). \end{aligned} \]

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:

\[ \begin{aligned} \iota:\mathbb{P}^n_{\overline{\mathbb{Q}}} \times \mathbb{P}^m_{\overline{\mathbb{Q}}} &\to \mathbb{P}^N_{\overline{\mathbb{Q}}} \\ (x,y) &\mapsto x \otimes y \\ &:=(x_iy_j). \end{aligned} \]

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

\[ h(x \otimes y) = h(x) + h(y). \]

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 write\[ f(t_1,\dots,t_n)=\sum_{\mathbf{j}}a_{\mathbf{j}}\mathbf{t^j}. \]

Then the

heightof \(f\) is defined to be\[ h(f)=\sum_{v \in M_K}\log |f|_v \]

where

\[ |f|_v=\max_{\mathbf{j}}|a_{\mathbf{j}}|_v. \]

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\[ h(fg)=h(f)+h(g). \]

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

\[ \left|h(f)-\sum_{j=1}^{m}h(f_j)\right| \le d\ln 2 \]

where \(d\) is the sum of the partial degrees of \(f\).

Segre Embedding And Heights