# 'Trivial' embeddings have the same degree?

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?

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.

• 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? – Tim May 20 '16 at 15:51
• @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. – hunter May 20 '16 at 17:47