Background: Let $k$ be a commutative ring. In general for a $k$-functor $X : \textbf{Alg}_k \to \textbf{Set}$ we define the ring of functions on $X$ to be $\textbf{Nat}(X , \mathbb{A}^1)$ with the ring operations given term-wise (where $\mathbb{A}^1$ denotes the forgetful functor).

For any $n \in \mathbb{N}$ the Grassmannian $\text{Gr}_n$ is defined to be the $k$-functor $\textbf{Alg}_k \to \textbf{Set}$ which assigns to every $k$-algebra $R$ the set of module direct summands of $R^n$, and to every $k$-algebra homomorphism $\varphi: R \to S$, the function which sends a direct summand $N \subseteq R^n$ to the image of $N \otimes_R S$ in $S^n$ under the obvious inclusion map (or equivalently the $S$-submodule of $S^n$ generated by $\varphi^{\oplus n}(N)$). It is well known that the Grassmannian is represented by a scheme.

Question: Is there a nice description of the ring of functions on the Grassmannian? I'm sure this is standard, but I couldn't find a reference. Thanks!!

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    $\begingroup$ The Grassmannian is a projective variety, so its ring of functions is just the constant functions $k$. $\endgroup$ – Qiaochu Yuan Sep 21 at 18:57
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    $\begingroup$ Dear Nate: do you assume $k$ is algebraically closed? Because if so, then the response of @QiaochuYuan is complete and correct (if not, you may need to an $\epsilon$ more work to prove it, depending on your definition of "variety" - the statement "all integral subschemes of $\Bbb P^n_k$ have global functions $k$" is equivalent to $k$ algebraically closed, but the Plucker relations which generate the ideal of the grassmanian are particularly nice and thus the global functions are always $k$ no matter whether $k$ is algebraically closed or not.) $\endgroup$ – KReiser Sep 21 at 19:33
  • $\begingroup$ Thank you both for the great comments! I was actually thinking of the case when $k$ is just a commutative ring, rather than a field. Does this change the results you stated (or the proofs)? $\endgroup$ – Nate Gallup Sep 21 at 20:01
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    $\begingroup$ Let me just comment that using $k$ for a commutative ring that's not assumed to be a field is sort of unconventional. Even though you explicitly stated "[l]et $k$ be a commutative ring", you managed to get two of the five gold tag-badge holders on this one! I might suggest using $R$ for a ring and $A$ for an algebra in the future. $\endgroup$ – KReiser Sep 21 at 20:52
  • $\begingroup$ @KReiser Lol! Good point! I've been reading Jantzen's "Representations of Algebraic Groups" and Milne's "Affine Group Schemes" and they both use $k$ for a general commutative ring, which I actually did think was strange at first. It's good to know that this isn't standard! $\endgroup$ – Nate Gallup Sep 22 at 2:23

Claim: For any flat and proper morphism of schemes $X\to S$ with $S$ connected and having geometrically connected, geometrically reduced fibers, we have that $\Gamma(X,\mathcal{O}_X)\cong\Gamma(S,\mathcal{O}_S)$.

Proof: By properness and flatness, we have that $X\to S$ is both closed and open, so because $S$ is connected $X\to S$ is surjective. By Stein factorization, we can write $X\to S$ as $X\to S'\to S$, where $S'\to S$ is finite and has geometrically connected and reduced fibers. A finite surjective morphism with geometrically connected and geometrically reduced fibers is an isomorphism, so $S'\cong S$ and since the structure sheaf of $S'$ is the pushforward of the structure sheaf on $X$, we have that the global sections are equal. $\blacksquare$

To apply this in our scenario, we note that the Grassmanian over a ring $R$ is just the base change of the Grassmanian over $\Bbb Z$ along the spectrum of the canonical map $\Bbb Z\to R$. Since the Grassmanian over the integers is smooth and proper and the conditions about the fibers are preserved by base change, we have that the Grassmanian over any base satisfies our conditions, so the lemma applies and we have that the regular functions are just $R$.

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  • $\begingroup$ Excellent, thanks so much for this clear and detailed answer! $\endgroup$ – Nate Gallup Sep 22 at 2:25

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