# degree of a projective embedding

assume we have an irreducible (even toric in my case) variety $X$ with an ample line bundle $L$. Let d an integer such that $L^d$ is very ample and consider the projective embedding associated to $L^d$ in a $\mathbb{P}^N$. Assume I know how can compute $\chi(L^d)$ and $\chi(\mathcal{O}_X)$. Define $\deg L^d=\chi(L^d)-\chi(\mathcal{O}_X)$. What is the difference between $\deg L^d$ and the degree of $X$ inside $\mathbb{P}^N$? are they the same?

Thnks

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No, they are not the same, they are equal only in dimension one. There are lots of counterexamples. To name just one, if $X$ is an Abelian variety of dimension $n$ and $d=1$, then degree of $X$ inside $\mathbb{P}^N$ equals $(\chi(L)-\chi(\mathcal{O}_X))\cdot n!$.
There exists a formula relating $(\chi(L^d)-\chi(\mathcal{O}_X))$ to degree of the embedding, but it cannot be reduced to the equals sign as in the case of curves:) Cf. Riemann-Roch-Hirzebruch Theorem.