What is the degree of a variety after the Segre embedding? Construction: Let $k$ be a field and let $X:Proj(S) \subseteq \mathbb{P}^N_k$ be a projective variety. The degree $deg_S(X)$ is defined in terms of the graded ring $S$ and the Hilbert polynomial $P_S(t)$ of $S$, hence $deg_S(X)$ depends on the embedding $X\subseteq \mathbb{P}^N_k$. The Segre embedding
$$\psi_{n,m}:Y:=\mathbb{P}^n_k \times_k \mathbb{P}^m_k \rightarrow \mathbb{P}^{nm+n+m}_k$$
is a closed embedding and given a product $X:=X_1\times_k X_2 \subseteq Y$ we may ask the following question: What is $deg(\psi_{n,m}(X))$?
Partial answer: I just finished exercise about calculating the degree of Segre variety $\Sigma_{m,n}$ which is $(m + n) \choose n$. I was thinking about generalization of this exercise. Suppose we start with a variety $X_1 \subset \mathbb{P}^n$ and $X_2 \subset \mathbb{P}^m$. We have the Segre embedding $\Sigma : \mathbb{P}^n \times \mathbb{P}^m \rightarrow \mathbb{P}^{(n + 1)(m + 1) - 1}$. Suppose that $deg(X_1) = x_1$ in $\mathbb{P}^n$ and $deg(X_2) = x_2$ in $\mathbb{P}^m$. What is the degree of $X_1 \times X_2$ in $\mathbb{P}^{(n + 1)(m + 1) - 1}$.
 A: Quewstion: "What is the degree of $X_1 \times X_2$ in $\mathbb{P}^{(n + 1)(m + 1) - 1}$."
Answer/reduction: There is a formula (Hartshorne, Ex.IV.3.3.2) saying that if $\phi:X \rightarrow \mathbb{P}^n_k$ is a closed immersion corresponding to a linebundle $L(D)$ and if $m=1$ you get $deg(\phi(X))=deg(L(D))$.
More generally if $dim(X)=m$ you get the formula
$$deg(X)=deg([X])=\int_{\mathbb{P}^n_k} c_1(\mathcal{O}(1))^m \cap [X].$$
The product $\cap$ is defined in the Chow group $CH^*(\mathbb{P}^n_k)$. The first Chern class $c_1(\mathcal{O}(1))$ and its "powers" acts on the Chow group.
The Segre embedding is the embedding
$$\phi: \mathbb{P}^n_k \times_k \mathbb{P}^m_k \rightarrow \mathbb{P}^N_k$$
corresponding to $\mathcal{O}(1,1):=p^*\mathcal{O}(1)\otimes q^*\mathcal{O}(2)$ and if $X:=X_1\times_k X_2$ and if $L(D):=\mathcal{O}(1,1)_X$ you get
$$deg(\phi(X))=D^m.$$
This reduces the problem to calculating this degree/intersection. Are you able to do this?
Note: If $k$ is algebraically closed and if $D:=\sum_i n_i [Y_i]$ with $Y_i \subseteq X$ an integral subscheme, we define $deg(D):=\sum_i n_i deg(Y_i)$ (HH.Ex.II.6.2), hence you must calculate the divisor $D$ corresponding to $L$.
Example: If $C:=\mathbb{P}^1_k$ and $L:=\mathcal{O}(d)$ it follows the corresponding divisor $D$ is $D=\sum_{i=1}^r n_i[P_i]$ with $P_i \in C(k)$ a closed point and $\sum_i n_i=d$. Hence $deg(L):= deg(D):= \sum_i n_i=d$. The divisor $D$ corresponds to the $d$-uple embedding
$$v_d: C \rightarrow \mathbb{P}^d_k$$
and hence $deg(v_d(C))=d$. If $d=2$ it follows $v_d(C)=Z(F)$ with $F:=x_0x_2-x_1^2$, and hence $deg(Z(F))=deg(F)=2$. For $d \geq 3$ it follows $v_d(C)$ is no longer a hypersurface hence the calculation does not follow from HH.CH.I.7.
https://mathoverflow.net/search?q=degree+of+a+divisor
Note: A more elementary approach using the "projective bundle formula" is outlined as an exercise in Harris "Algebraic geometry - a first course" (Ex. 19.2, pg 240).
You write the fundamental class
$$[X]= \sum_i d_i[\mathbb{P}^i_k\times_k \mathbb{P}^{l-i}_k]$$
and the formula says
$$(*) deg(\phi(X))= \sum_i \binom{m}{i}d_i.$$
Note: If $k$ is the complex number field there is the Chern character isomorphism
$$Ch: K_0(\mathbb{P}(E^*))_{\mathbb{Q}} \cong H_*(\mathbb{P}(E^*),\mathbb{Q})$$
and there is the projective bundle formula
$$K_0(\mathbb{P}(E^*)) \cong K_0(X)[t]/(t^{e+1})$$
where $rk(E)=e+1$. Hence if $E:=\mathcal{O}_{\mathbb{P}^n_k}^{m+1}$ it follows
$$K_0(\mathbb{P}^n_k \times_k \mathbb{P}^m_k)_{\mathbb{Q}} \cong 
 \mathbb{Q}[t_1,t_2]/(t_1^{n+1}, t_2^{m+1}).$$
There is an equality of classes $t_1^{i}t_2^{j}=[\mathbb{P}^{n-i}_k \times_k \mathbb{P}^{m-j}_k]$
and you must express the fundamental class $[X_1\times_k X_2]$ in this basis. In the Grothendieck group you write
$$[X_1\times_k X_2]:= \sum_{i,j} d_{i,j}t_1^it_2^j$$
and use $(*)$. The numbers $d_{i,j}$ should be integer functions of the numbers $x_1,x_2$.
Remark: I have not done this exercise but it is a good place to start.
