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Let $V$ be a (classical) algebraic variety over an algebraically closed field $k$. The set $X$, which consists of all the irreducible closed sets of $V$, is a scheme on $k$. Hartshorne's proof doesn't seem to work unless $k$ is an algebraically closed field. Is $X$ a scheme when $k$ is a field that is not algebraically closed? If not, is there a way to recognize $V$ as a scheme?

In Hartshorne's book, the projective space $ \mathbb {P} ^ n_A $ on the ring $ A $ is defined using the fiber product. I don't know how this complex definition works.

Let $ k $ be a non-algebraically closed field. $ \mathbb {P} ^ n_k $ uses the same symbols as the classic projective space definition, but can they be recognized as the same?


addition. I came across the following description on wikipedia. In scheme theory, the $n$-dimensional projective space over $k$ is $$\mathbb{P}^n_k = \text{Proj} k [x_0, ..., x_n]$$ It is defined using a polynomial ring like this. The entire $k$-value point of $\mathbb{P}^n_k$ matches the classical projective space.

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  • $\begingroup$ What source do you have for classical varieties over non-algebraically closed fields? AFAIK, Hartshorne Chapter I assumes $k$ to be algebraically closed. $\endgroup$ Commented Oct 21, 2021 at 7:18

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I don't think that is easily done if $k$ is not algebraically closed. Consider for example $$V = \mathbb R$$ as a variety over $\mathbb R$. Then the associated scheme ought to be $X = \mathbb A^1_{\mathbb R} = \operatorname{Spec} \mathbb R[x]$, right? But the maximal ideal $$(x^2 + 1) \subset \mathbb R[x]$$ does not correspond to any irreducible closed subset of $V$.

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  • $\begingroup$ Thank you very much. I agree. Added related questions. $\endgroup$
    – Kazsugi
    Commented Oct 21, 2021 at 7:07

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