# Example where prime spectrum suits better than the maximal spectrum

in a lot of algebraic geometry books I've heard that working over $\mathrm{Spec}(A)$ is better than working over $\mathrm{Spmax}(A)$ in the case where you consider a variety over a non-agebraically closed field $k$. Of course, the prime spectrum suits better because of the functoriality, but in a lot of books dealing with arithmetic of curves (in affine charts, without sheaves) there are discussions and proofs of facts dealing with non-algebraically closed fields (even zeta functions) using only maximal ideals of reduced $k$-algebras (zero sets of polynomials).

Therefore I'm searching for a simple example where prime ideals are better than the maximal ideals. Furthermore I would like to know how should I think about non-maximal prime ideals (points that are not geometric). I think that it would be useful, too, if someone could explains what's the difference between $\mathrm{Spec}(k)$ and $\mathrm{Spec}(k')$.

• What is $k'$? Is it the algebraic closure of $k$? – Manos Jan 23 '14 at 21:54
• @Manos any extension – user40276 Jan 23 '14 at 22:00
• The difference between $Spec(k)$ and $Spec(k')$ is exactly the same as that between $k$ and $k'$ ! But this is a tautology and proves nothing for or against the use of nonmaximal prime ideals... – Georges Elencwajg Jan 23 '14 at 22:34
• @GeorgesElencwajg but this is just a point – user40276 Jan 23 '14 at 22:36
• Dear user: $spec(k)$ is not just a point but a point plus a sheaf of local rings on that point. In this trivial situation this sheaf is just the datum of the field $k$, but the beauty of scheme theory is that the scheme $Spec(k)$ remembers the field $k$ in its structure. – Georges Elencwajg Jan 23 '14 at 23:09

If $k$ is an algebraically closed field and $A=k[T_1,...,T_n]/\mathfrak p\,$ (with $\mathfrak p$ a prime ideal) is a finitely generated domain over $k$, you should think of the affine scheme $\mathrm{Spec}(A)$ as the variety $V=V(\mathfrak p)\subset \mathbb A^n(k)$ to which you adjoin formally a point $\eta_W$ for every irreducible subvariety $W\subset V$.
If $k$ is not algebraically closed the advantage of schemes over varieties is more convincing: the schemes $\mathrm{Spec}(\mathbb R[X,Y]/(X,Y))$ and $\mathrm{Spec}(\mathbb R[X,Y]/(X^2+Y^2))$ are completely different (they don't even have the same dimension) whereas the varieties $V(X,Y)\subset \mathbb R^2$ and $V(X^2+Y^2)\subset \mathbb R^2$ are both equal to a single point.
• I know this is an old question, but I'm not sure the middle paragraph answers the question here, since for nonalgebraically closed fields the variety as a subset of affine space is completely different than the max spec. Max spec of $\Bbb{R}[X,Y]/(X^2+Y^2)$ has a pile of complex points that distinguish it from the single point of the max spec of $\Bbb{R}\cong \Bbb{R}[X,Y]/(X,Y)$. – jgon Jan 8 '19 at 0:27