Recognize classical algebraic varieties as schemes

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.

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

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 $$V$$.