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Suppose that $f:X\longrightarrow\text{Spec} K$ is a variety over $K$, namely $X$ is an integral, separated $K$-scheme of finite type. Now if $L$ is a subfield of $K$, it is clear that there exists a morphism of affine schemes $g:\text{Spec K}\longrightarrow\text{Spec L}$.

At this point one can consider the morphism $g\circ f: X\longrightarrow\text{Spec } L$ to conlude that $X$ is also a variety over $L$. By this reasoning, it seems that any variety over $K$ can be viewed as a variety oover any subfield, but this sentence is clearly false. Where is the mistake?

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    $\begingroup$ At the very least, to say that $X$ is a variety over $L$, we want $X\to\operatorname{Spec} L$ to be of finite type. Just having a morphism to $\operatorname{Spec} L$ doesn't say very much. $\endgroup$ Commented Oct 11, 2014 at 8:50
  • $\begingroup$ But (using the notations of my question), if $g$ is the morphism corresponding to the inclusion $L\subset K$, then $g$ is of finite type. $\endgroup$ Commented Oct 11, 2014 at 8:54
  • $\begingroup$ @magnetissimo No... It is of finite type, by definition, only if $K$ is a finitely generated algebra over $L$. $\endgroup$ Commented Oct 11, 2014 at 8:57
  • $\begingroup$ Mhh yes, you are right! I'm sorry for the stupid comment $\endgroup$ Commented Oct 11, 2014 at 8:58
  • $\begingroup$ To say that $X$ is a variety over $L$ means more than that there is a morphism $X \rightarrow \operatorname{Spec} L$... $\endgroup$
    – user64687
    Commented Oct 12, 2014 at 19:36

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A variety is not just the scheme $X$ but also the structural morphism $f:X\rightarrow Spec(K)$. To say that this variety can be defined over a subfield $L$ of $K$, as I understand, it means that there exists a variety $Y$ defined over $L$ such that $X=Y\times_{Spec(L)} Spec(K)$ and $f$ is the projection over $Spec(K)$.

Now, note that if you are considering $X$ as a variety over $L$ via $Spec(K)\rightarrow Spec(L)$ then it is not true in general that $X$ (as variety over $K$) is the fibered product $X\times_{Spec(L)}Spec(K)$.

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