I presume this is a GAGAstyle result, but I cannot find a reference.
2 Answers
Also like affine varieties, we have:
Theorem. A complex manifold is Stein if and only if it embeds into some $\mathbb{C}^N$ as a closed complex submanifold.
For the "only if" direction, see Hörmander, An Introduction to Complex Analysis in Several variables, Theorem 5.3.9. For the converse, an argument is contained on pp 109110 of Hörmander, immediately after the definition of Stein manifold.

3$\begingroup$ In particular, the analytification of a complex algebraic variety is a Stein space. $\endgroup$– EarthliŋMay 23, 2019 at 10:31

1$\begingroup$ @Earthliŋ: You probably mean affine algebraic variety. $\endgroup$ May 25, 2019 at 20:45

$\begingroup$ @JérômePoineau Yes, of course. Somehow "affine" got lost when I added "complex"... $\endgroup$– EarthliŋMay 25, 2019 at 21:50
I believe the reason for this is Cartan's 'Theorem B': for a Stein manifold $\mathrm{X}$ sheaf cohomology vanishes, $\mathrm{H}^n(\mathrm{X},)=0$ for $n\geqslant 1$, and this property characterises Stein manifolds among complex manifolds. In the same way affine schemes are characterised among (nice) schemes by cohomology vanishing (this is a theorem of Serre). This comes up in the proof of the cohomological comparison result which is part of GAGA, see for example SGA 1, Exposé XII.

5$\begingroup$ There is no way to be completely tactful here; the vanishing condition is for coherent analytic coefficients only. It is certainly not true for all coefficients, e.g. $X= \mathbb{C}^*$ and cohomology with coefficients in $\mathbb{Z}$ is not zero, but $X$ is Stein. $\endgroup$ May 25, 2019 at 13:29

2$\begingroup$ @DonuArapura this was understood... on the scheme side you are also not going to consider sheaves which are not even $\mathcal{O}_X$modules or which have no finiteness properties. $\endgroup$– ssxMay 25, 2019 at 13:38

6$\begingroup$ OK, but I'll leave the comment, since not everyone would understand the hidden assumptions. $\endgroup$ May 25, 2019 at 13:45