For a fixed scheme $S$, a scheme over $S$ is defined to be a scheme $X$ together with a morphism $X \to S.$ I am following Hartshorne, and I cannot find a definition of scheme over a field $k$. Is it a scheme $X$ together with a morphism $X \to \operatorname{Spec}(k)$?
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Sorry, but what you wrote in answer could very well have been a comment, in which case my deletion would perhaps have been valid. – Aug 10 '16 at 20:20
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Yes, this is it. If your scheme $X$ is affine, this is the same thing as an algebra over the field $k$, hence the terminology.
Najib Idrissi
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I have a question to your answer. As I understand it, if $(X,\mathcal{O}X)$ is a $k$-scheme, then $(X,\mathcal{O}_X)$ is a scheme with a morphism of schemes $(X,\mathcal{O}_X)\rightarrow (\operatorname{Spec}(k),\mathcal{O}{\operatorname{Spec}(k)})$. But in my opinion since $k$ is a field $\operatorname{Spec}(k)={(0)}$ right? – user123234 Jan 06 '23 at 10:17
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But I would then have a morphism $X\rightarrow {(0)}$ wouldn't there be only one such morphism? – user123234 Jan 06 '23 at 10:31
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@user123234 A morphism of schemes is not just a continuous map, it's a map of ringed spaces. You also need to give a morphism between the structure sheaves. There can be multiple different ones, even with affine schemes. For example, complex conjugation is a ring morphism which is not the identity, so there are at least two different morphisms $\operatorname{Spec}\mathbb{C} \to \operatorname{Spec}\mathbb{C}$. – Najib Idrissi Jan 06 '23 at 15:46
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But there is only one continuous map and then several morphism of structure sheaves? – user123234 Jan 06 '23 at 15:48
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