The affine scheme of Spec(R)

The construction of the affine scheme on Spec(R) hinges on the fact that you can you localize at the points of the spectrum: the irreducible radical ideals. This is possible because irreducible radical ideals are prime, which means that their complement is multiplicatively closed. This provides a link between the lattice of radical ideals and the lattice of multiplicative sets, which isn't entirely obvious. As a preliminary, I will therefore explain why irreducible radical ideals are prime ideals, and why you can indeed localize over them. This doesn't require much more then the proof that the nilradical is the intersection of all prime ideals and the lattice theorem.

Radical ideals are the complement of multiplicative-iteration closed sets and prime ideals are the complements of multiplicatively closed sets, so prime ideals are quite obviously radical. Prime ideals are also clearly irreducible (not the intersection of any other ideals) because if they are reducible then we can form elements from each set not in P whose product is in P. It remains only to show that irreducible radical ideals are prime. The proof is based upon Zorn's lemma, which is applicable because the lattice of ideals is a complete lattice.

Lemma. the nilradical $N$ is the intersection of all prime ideals

The nilradical is the smallest radical ideal, so it is contained in all prime ideals. We will show the converse. Let $x \not\in N$ be a non-nilpotent. Let $F$ be the family of all ideals not containing $x^n$ for $n \ge 0$. We can apply Zorn's lemma to get a maximal element $I_{max} \in F$. If $g,h \not\in I_max$ and $gh \in I_{max}$ we have $I_{max} + (g), I_{max} + (h) \not\in F$. Which implies that there exists $n,m \in \mathbb{N}$ such that $x^n \in I_{max} + (g)$ and $x^m \in I_{max} + (h)$ so $x^{n+m} \in I_{max}$ which contradicts the supposition that $I_{max}$ contains no powers of $x$. Non-nilpotents are all contained in prime ideals, so elements contained in all prime ideals are nilpotent. $\square$

The proof that the nilradical is the intersection of all prime ideals pays off, because everything else to do with the intersection between prime and radical ideals follows from it. The rest of the proofs are suprisingly easy.

Lemma. let $R$ be a commutative ring then if $(0)$ is an irreducible radical ideal it is prime.

Proof. If $(0)$ is a radical ideal it is the nilradical and the intersection of all prime ideals. By the fact that $(0)$ is irreducible it is contained in this set of all prime ideals which means it is a prime ideal. $\square$

Theorem. irreducible radical ideals are prime

Proof. let $I$ be an irreducible radical ideal of a commutative ring $R$. By the lattice theorem $(0)$ is an irreducible radical ideal in $\frac{R}{I}$ which means it is prime. The inverse image of prime ideal is prime, so this $(0)$ ideal in $\frac{R}{I}$ can be reflected back to get a prime ideal $I$. $\square$

With these preliminaries out of the way, we can now construct the affine scheme associated with a commutative ring $R$. In summary, the lattice of radical ideals is dual to a locale whose points are prime ideals. That the cotopology $Spec(R)$ is order-dual to the lattice of radical ideals is relevant, as for example coatomic ideals (which are called maximal ideals for some reason, although I have already gotten used to it) are translated into atomic sets in the cotopology which are of course singletons $\{x\}$.

The stalk of the affine scheme is defined by the localisation, which is a morphism from the lattice of multiplicative sets $Sub(R,*,1)$ to the objects of the category of rings $Ob(Rings)$. In this case of integral domains, the larger the zero free multiplicative set the larger the resulting localisation, as the localisation approaches the field of fractions, which means in certain cases localisation is a functor on a thin category. \[ \ell : Sub(R,*,1) \to Ob(Rings) \] By their very definition, the complement of a prime ideal is a multiplicative. This is formalized by the map $f : PrimeIdeals(R) \to Sub(R,*,1)$. The input action of this map on the localisation map $\ell$ produces a new morphism in the category of sets: \[ \ell : PrimeIdeals(R) \to Ob(Rings) \] We can now apply the image functor, to turn this into a map from sets of prime ideals (including open sets of our topology $Spec(R)$) to sets of rings. \[ \ell : \wp(PrimeIdeals(R)) \to \wp(Ob(Rings)) \] We now have a set of rings associated to any open set in the spectrum $Spec(R)$, but we want to get a single ring. There is a very obvious way to get a single object from a set of objects in a category: the product. This is precisely what we are going to use to construct the presheaf of rings. \[ \times : \wp(Ob(Rings)) \to Ob(Rings) \] We now have an adjusted localisation map $\ell_*$ which assigns a ring to any set of prime ideals by the product of localisations. \[ \ell_* : \wp(PrimeIdeals(R)) \to Ob(Rings) \] This is functorial on open sets because given any product of a set of objects we can form restriction morphisms to any product of a subset of objects, so the restriction maps merely restrict products of localisations to their subsets, which forms a ring homomorphism as it would in any category. So we have a presheaf of rings $\ell_*$ which we constructed from the localisation. This isn't necessarily a sheaf, so all that remains is to use sheafification to get a sheaf $\ell_*^{\#}$.

The sheafification, which is adjoint to the inclusion functor from the category of presheaves to the category of sheaves, is very convenient because it is much easier to construct presheafs then it is sheafs. Presheafs are abound in mathematics and in category theory, and all we need to do turn them into sheafs is to apply a single functor. This sheaf is a scheme because it is defined on the spectrum $Spec(R)$ of a topological space.

It is amazing that this scheme construction is even possible, which is why I spent so much time at the beginning describing why it is. There are obviously limitations of the traditional set-theoretic approach to algebraic geometry, such as that it doesn't take care of multiplicities which is responsible for the utility of this scheme construction. I will examine some of the details of that later.

Source:
Algebraic geometry by Robin Hatshorne