I realize there have been plently of discussions about this, but most of them are over my head and I never understand the geometric intuition behind them.

I'm trying to make a big list of "justifications" for using the prime spectrum of a ring. So far, I have two fairly pleasing formal ones:

  • Preimages of prime ideals are prime ideal, so we have functoriality.
  • The prime spectrum is the "free local ring" on a ring in the following sense: in passing from $\mathsf{Set}$ to the sheaf topos, we find that in terms of the internal language of sheaves over the prime spectrum, all rings are local.

These are formally satisfying but I don't have any feel for them. For instance, it makes sense that maximal ideals don't pull back to maximal ideals because points needn't pull back to points, however why should I expect irreducible varieties to pull back to irreducible varieties? I don't know topos theory so I don't know how or why $\mathrm{Spec}(R)$ falls out of the sky.

Now I'm looking for geometric justifications to look at the space whose points are prime ideals. The people I asked have all mentioned generic points; something along the lines of: "Instead of looking at particular points of a variety, we should really be studying a 'generic point' of each irreducible component, which only has the properties shared by all of its points."

I don't understand how prime ideals magically produce this vague notion... The ideal-variety correspondence simply says they corrspond to irreducible subvarieties, nothing more.

So, in painstaking detail, what are generic points and why is the prime spectrum the space we should be looking at?

  • $\begingroup$ It's true that geometrically the fiber over a point doesn't have to be a point, but in the classical setting of finitely generated $k$-algebras it is true that the preimage of a maximal ideal is maximal, so I'm a little worried that we have things backwards here. $\endgroup$
    – Hoot
    Oct 26, 2015 at 22:27
  • $\begingroup$ @Hoot you are probably right, which just goes to show I don't have a clue about the geometric picture. $\endgroup$
    – Arrow
    Oct 26, 2015 at 22:32
  • $\begingroup$ Will the downvoter explain the downvote? $\endgroup$
    – Arrow
    Nov 1, 2015 at 20:14

1 Answer 1


A point of a geometric object $X$ should be, up to equivalence, a morphism $P\to X$, where $P$ is a one-point space. So we can approach this question by asking what kinds of one-point spaces we are likely to encounter in an algebraic setting.

Euclid defined a point as "that which has no part". We can interpret this as follows: $P$ should be a locally ringed space such that any sheaf of ideals $\mathcal{I}\subsetneq \mathcal{O}_P$ that is locally generated by sections, is identically zero. Otherwise, the zero set of $\mathcal{I}$ would cut out a nontrivial "part" of $P$.

With some thought, this implies that $P$ has only one point as a topological space, and its associated ring of functions is a field. In other words, $P=\operatorname{Spec} K$, with $K$ a field.

To be clear, there is nothing very fancy going on here: I am arguing from abstract principles that the underlying topological space of $P$ should have only one point, which is intuitive.

Then the ring $K$ associated to $P$ should also have no nontrivial ideals. This is less intuitive, but follows from the idea that the vanishing of a non-invertible element of $K$ should cut out a non-empty subobject of $P$.

It remains to classify morphisms $\operatorname{Spec} K\to \operatorname{Spec} R$. Since the whole point of the spectrum is to make the category of affine schemes dual to the category of commutative rings, it is equivalent to classify morphisms $R\to K$.

But, with the right notion of equivalence, a morphism from a commutative ring $R$ into a field is just a prime ideal of $R$.

  • $\begingroup$ "some thought" is a wall of technical results that does not even mention the word 'spectrum'. I also don't understand why "that which has no part" should be interpreted as a locally ringed space satisfying your condition. Finally, maximal and prime ideals coincide for fields, and they don't have nontrivial ideals, so what's the point of $\mathrm{Spec}(K)$? $\endgroup$
    – Arrow
    Oct 27, 2015 at 8:07
  • $\begingroup$ @Arrow Very little of the wall is really important. If you're prepared to accept that $P$ should really be a one-point topological space, then the only content of my claim is that $P$ should be reduced. $\endgroup$ Oct 27, 2015 at 8:15
  • $\begingroup$ @Arrow The point is that ideals (of rings of functions) cut out closed sub-objects, and so things without nontrivial ideals—fields—are exactly our models for points. As for why we're working with locally ringed spaces in the first place, we start with topological spaces, and add a sheaf of rings (the regular functions) so there's some connection to algebra. We should work with locally ringed spaces because if we use general ringed spaces, then $\operatorname{Spec} R$ ends up being a single point with $R$ attached to it, which is quite boring. $\endgroup$ Oct 27, 2015 at 8:18
  • $\begingroup$ @Arrow By the way, I thought of a more categorical and intuitive definition of a point: it should be the case that any morphism $P\to X$, where $X$ is not the empty space, is a monomorphism. Reversing the arrows, this once again gives us morphisms to fields. This amounts to the same thing, but is probably easier to digest. $\endgroup$ Nov 9, 2015 at 4:37
  • $\begingroup$ @Slade Reversing the arrows, we would search for a commutative ring $B$ such as any ring morphism $A\to B$ is epic (because in the geometric dual category you started with mono). But not all morphisms to fields are epic, for instance the inclusion $\mathbb{R}\to\mathbb{C}$ is not epic. $\endgroup$
    – V. Semeria
    May 27, 2022 at 13:47

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