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Consider a field extension $L\subseteq K$ and two $L$ schemes $X_L$ and $Y_L$ with an embedding $j:X_L\longrightarrow Y_L$. Now take the base changes $$X:=X_L\times_{\text{Spec L}}\text{Spec} K$$ $$Y:=Y_L\times_{\text{Spec L}}\text{Spec} K$$

We have the following commutative diagram:

enter image description here

If we know by hypotesis that $j\times\text{id}:X\longrightarrow Y$ is a closed embedding, then can we conclude that also $j$ was a closed embedding?

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Yes.

Being a closed embedding is a local question so basically you're asking if $A \to B$ is an $L$-algebra map and $A\otimes K \to B \otimes K$ is surjective then is $A \to B$ surjective? The answer to that is yes, because field extensions $L \to K$ are faithfully flat.

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