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Let $X$ be a scheme over a field $K$ and $f\in\mathscr O_X(U)$ for some (say, affine) open $U\subseteq X$. For a $K$-rational point $P$, I can denote by $f(P)$ the image of $f$ under the map $$\mathscr O_X(U) \to \mathscr O_{X,P} \twoheadrightarrow \mathscr O_{X,P}/\mathfrak m_P = K.$$ This yields a map $f:U(K)\to K$. Giving $U$ the induced subscheme structure, when does this uniquely define a morphism $f:U\to\mathbb A_K^1$ of schemes? It certainly works when $X$ is a variety (and $K$ algebraically closed), so there should be some "minimal" set of conditions for this interpretation to make sense. Thanks a lot in advance!

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up vote 2 down vote accepted

The scheme $\mathrm{Spec}(k[X])=\mathbf{A}_k^1$ is the universal locally ringed space with a morphism to $\mathrm{Spec}(k)$ and a global section (namely $X$). What I mean by this is that for any locally ringed space $X$ with a morphism to $\mathrm{Spec}(k)$ (equivalently $\mathscr{O}_X(X)$) is a $k$-algebra) and any global section $s\in\mathscr{O}_X(X)$, there is a unique morphism of locally ringed $k$-spaces $X\rightarrow\mathbf{A}_k^1$ such that $X\mapsto s$ under the pullback map $f^*:k[X]=\mathscr{O}_{\mathbf{A}_k^1}(\mathbf{A}_k^1)\rightarrow\mathscr{O}_X(X)$. This is a special case of my answer here: on the adjointness of the global section functor and the Spec functor

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@everyone who answered; All of these were very helpful, thanks a lot. Had to accept just one of them. – user38451 Apr 23 '13 at 7:16

As $U$ is also a $K$-scheme, we can forget about $X$ and think only about $U = \mathrm{Spec} A$. $U$ being a $K$-scheme means that there's a structure morphism $U \to \mathrm{Spec} K$, which gives us a structure map of rings $K \to A$.

Now, every $K$-scheme morphism $U = \mathrm{Spec} A \to \mathrm{Spec} K[x] = \mathbb{A}_K^1$ is determined by a ring homomorphism $\phi: K[x] \to A$, such that $\phi|_K : K \to A$ is a structure morphism of a $K$-scheme $U$. Universal property of polynomial rings says that giving such homomorphism is equivalent to picking any element in $A$ and mapping $x$ to it. So, for any $f \in \mathcal{O}_U(U) = A$, we have a scheme morphism $U \to \mathbb{A}_K^1$, and all such morphisms arise in this way.

Note that you don't even need to assume that $U$ is affine: we use only the fact that scheme morphisms $U \to \mathrm{Spec} K[x]$ are determined by ring homomorphisms $K[x] \to \mathcal{O}_U(U)$.

Could you elaborate more on what is $U(K)$, and how does your composition define a morphism $U(K) \to K$?

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Another way to see it — $U$ doesn't even have to be affine, just a scheme really, over a ring $R$ — Then, $\mathscr O_U$ is an invertible sheaf on $U$ which is certainly generated by the global sections $\{\,1,f\;\}$. This defines a morphism $\phi: U\to\mathbb P^1_R$ which factors as a map $U\to\mathbb A^1_R$.

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