Find the degree of the \(d\)-uple embedding of \(\PP^n\) in \(\PP^N\).
Find the degree of the Segre embedding of \(\PP^r \times \PP^s\) in \(\PP^N\).
The answers are \(d^n\) for the \(d\)-uple embedding and \(\binom{r+s}{s}\) for the Segre embedding.
Solution
Recall that if a projective variety \(X\) has dimension \(m\) and Hilbert polynomial \(P_X(t)\), then \[ P_X(t)=\frac{\deg X}{m!}t^m+\text{terms of lower degree}. \]
1. The homogeneous coordinate ring of the \(d\)-uple embedding of \(\PP^n\) is the Veronese ring\[ S^{(d)}=\bigoplus_{q\ge0}S_{dq}, \qquad S=k[x_0,\ldots,x_n]. \]
The \(d\)-uple embedding is defined by all monomials of degree \(d\). Its homogeneous coordinate ring is therefore the subring of \(S\) generated by \(S_d\), with grading in which \(S_d\) has degree one. Its degree-\(q\) piece is exactly \(S_{dq}\), giving the displayed Veronese ring.
2. The degree of the \(d\)-uple embedding is\[ \boxed{d^n}. \]
For every \(q\ge0\), \[ \dim_k S_{dq}=\binom{dq+n}{n}. \] Hence the Hilbert polynomial of the embedded copy of \(\PP^n\) is \[ P(q)=\binom{dq+n}{n} =\frac{d^n}{n!}q^n+\text{terms of lower degree}. \] The image has dimension \(n\). Comparing the leading coefficient with the degree formula gives \[ \deg(\nu_d(\PP^n))=n!\cdot\frac{d^n}{n!}=d^n. \] This proves part (a).
3. The homogeneous coordinate ring of the Segre embedding has degree-\(q\) piece\[ k[x_0,\ldots,x_r]_q\otimes_k k[y_0,\ldots,y_s]_q. \]
The Segre embedding is defined by the bihomogeneous products \(x_i y_j\). A homogeneous polynomial of degree \(q\) in the Segre coordinates pulls back to a bihomogeneous polynomial of bidegree \((q,q)\). Conversely, every monomial of bidegree \((q,q)\) is a product of \(q\) terms \(x_i y_j\), so every such bihomogeneous monomial lies in the image of the degree-\(q\) piece of the Segre coordinate ring. Thus that piece identifies with \[ k[x_0,\ldots,x_r]_q\otimes_k k[y_0,\ldots,y_s]_q. \]
4. The degree of the Segre embedding is\[ \boxed{\binom{r+s}{r}=\binom{r+s}{s}}. \]
By step 3, its Hilbert function is \[ H(q) =\binom{q+r}{r}\binom{q+s}{s}. \] This is already a polynomial in \(q\), with leading term \[ \frac{1}{r!s!}q^{r+s}. \] The Segre image has dimension \(r+s\). Therefore its degree is \[ (r+s)!\frac{1}{r!s!} =\binom{r+s}{r} =\binom{r+s}{s}. \] This proves part (b).
5. Q.E.D.
Step 2 gives the degree of the \(d\)-uple embedding and step 4 gives the degree of the Segre embedding.