1
$\begingroup$

We can define the degree of a projective variety $X\subseteq\mathbb{P}^n$ in terms of the maximal number of intersections with projectivisations $L=\mathbb{P}(\hat{L})$ of linear varieties $\hat{L}\in\mathrm{Gr}(n+1-\dim X,n+1)$, or in terms of the leading coefficient $(\deg X/(\dim X )!)\lambda^{\dim X}$ of the Hilbert polynomial $p_X(\lambda)$.

Either way, the result depends on the embedding of $X\subseteq\mathbb{P}^n$. My question is, if we take some $X$ that comes with an embedding into $\mathbb{P}^n$ and then embed it into $\mathbb{P}^{n+1}$ in a 'trivial' way, does this leave the degree the same? For example, consider $$v_2(\mathbb{P}^1)=\big\{[\lambda^2:(\lambda\mu)^2:\mu^2]\mid\lambda,\mu\neq0\big\}\cup\big\{[1:0:0],[0:0:1]\big\}\subset\mathbb{P}^2.$$ If we embed this into $\mathbb{P}^3$ as $$v_2(\mathbb{P}^1)\cong\big\{[\lambda^2:(\lambda\mu)^2:\mu^2:\mu^2]\mid\lambda,\mu\neq0\big\}\cup\big\{[1:0:0:0],[0:0:1:1]\big\}\subset\mathbb{P}^3$$ i.e. by repeating a coordinate, is the degree unchanged?

$\endgroup$
1
$\begingroup$

If the homogeneous coordinate ring of your projectively embedded $X$ was $k[x_0, \ldots, x_n]/(f_1, \ldots, f_m)$, then your "trivially embedded $X$" has homogeneous coordinate ring $k[x_0, \ldots, x_n, x_{n+1}]/(f_1, \ldots, f_m, x_{n+1})$, which is obviously isomorphic to the ring you started with. This will be true whenever you take a linear embedding of $\mathbb{P}^n$ into a higher $\mathbb{P}^N$ (you may need a change of variables, depending on your embedding).

So the Hilbert polynomial doesn't change (since the $d$th graded piece has the same rank for all $d$) and hence the degree doesn't change.

$\endgroup$
  • $\begingroup$ Ah ok, when you say it like that it seems almost obvious! Although the degree depends on the embedding, it is the same on varieties with isomorphic (in a graded sense) homogeneous coordinate rings – yes? $\endgroup$ – Tim May 20 '16 at 15:51
  • 1
    $\begingroup$ @Tim right... the homogeneous coordinate ring itself depends on the embedding, but in the explicit case of a P^n sitting linearlly in a higher P^N the isomorphism class doesn't change. $\endgroup$ – hunter May 20 '16 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.