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Let $S$ be a irreducible scheme over a field $k$ (for example a smooth projective curve over algebraically closed field). Denote by $k(S)$ its field of fractions. Let $K$ be a(n algebraically closed) field. Given an embedding $k(S) \hookrightarrow K$ we can construct a $K$-point of $S\otimes K$ in the following way: choose an affine open set in $S$ with the coordinate ring $R$. Then we can restrict the embedding to the embedding $R \hookrightarrow K$. After tensor product with $K$ it becomes a $K$-point in $S\otimes K$.

My question is: which points arise as the image of this map?

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    $\begingroup$ The generic point of $S$ arises. $\endgroup$ – Snakeboi Mar 14 '18 at 3:20
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So I am at the airport with some time to kill, so I will answer this question when $S$ is the spectrum of $k[t]$, i.e., is the affine line. The base change is the spectrum of $K[t]$. A $K$-point of this is given by an element of $K$, namely the image of $t$ under the corresponding $K$-algebra map. Now I leave it up to you to decide when this comes from an embedding of $k(t)$ into $K$.

There's a trivial generalization to higher dimensional affine spaces and then to arbitrary varieties.

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Let $\eta \to S$ be the generic point of $S$. The function field of $S$ is exactly the residue field of $\eta$. Let $k(S) \subset K$ be a field extension (no need to assume that $K$ is algebraically closed).

The map $\mathrm{Spec}\, K \to S$ factors as $\mathrm{Spec}\, K \to\mathrm{Spec}\, k(S) \to S$. The image of $\mathrm{Spec}\, K \to S$ is thus just $\eta$.

But now you apply a base-change to $K$ and $S \otimes K$ need no longer be irreducible. The image of $\mathrm{Spec}\, K \to S \otimes K$ is a point of $S \otimes K$ which lies above $\eta$; such a point is exactly a generic point of an irreducible component of $S \otimes K$. Different choices of embeddings $k(S) \subset K$ will give rise to different irreducible components, and all the irreducible components will arise this way.

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