I'm planning lectures for my intro algebraic geometry course, and I noted something awkward that is coming up. We're starting projective varieties soon. Of course, we'll prove that projective maps are closed.

Then, I want to talk about Grassmannians. I want to show that the Grassmannian is closed in the Plucker embedding. In other words, I want to know that the set of rank $1$ tensors is closed in $\mathbb{P} \left( \bigwedge\nolimits^k \mathbb{C}^n \right)$. This sounds like it should be a great application of the theorem that projective maps are closed, right? Send $\mathbb{P}^{n-1} \times \mathbb{P}^{n-1} \times \cdots \times \mathbb{P}^{n-1}$ to $\mathbb{P} \left( \bigwedge\nolimits^k \mathbb{C}^n \right)$ by $(v_1, v_2,\ldots, v_k) \mapsto v_1 \wedge v_2 \wedge \cdots \wedge v_k$? Except that this is only a rationally defined map -- it isn't defined when the $v_i$ are linearly dependent.

Of course, I can prove that the rank $1$ tensors are closed by brute force.

I also know some conceptual explanations that are not appropriate as the primary proof for this class (though I may well comment on some of them):

Working topologically over $\mathbb{C}$, every $k$-plane has an orthonormal basis, so $U(n)$ acts transtively on $G(k,n)$ and $G(k,n)$ is compact.

In terms of the valuative criterion for closedness, it is enough to take a $k \times n$ matrix with entries valued in a dvr and rank $k$ at the generic point and write down its limit in $G(k,n)$. We can do this explicitly in terms of Smith normal form.

When a reductive group $G$ acts on an affine variety $X$, the map from the semistable points to Proj of the invariant ring is surjective. Apply this with $G = GL_k$ and $X$ the $k \times n$ matrices. I suspect I am also implicitly using that the ring of invariants is generated by the $k \times k$ minors, which is already nontrivial.

Is there some clever algebraic proof I'm missing, ideally one which uses that projective morphisms are closed?

For context, this is a mixed grad-udergrad course, taught out of Shavarevich volume 1. Everything is over an algebraically closed field and done in a fairly concrete way.

AlgebraIII, §11, no. 12. $\endgroup$13more comments