Given a primary ideal I in a ring A, we can consider the subscheme V(I) of Spec(A).
It is a nilpotentification (?) of the integral subscheme V(rad(I)) given by the radical rad(I) of I.
My question is what kind of nilpotentifications you get that way.
It is a vague question, but I'm asking because i am completely unable to understand the algebraic meaning of "primary ideal" (altho I learned it by heart)
[Math] What does primary mean geometrically
ac.commutative-algebra
Related Solutions
Theorem 1: Let $R$ be an integral domain with field of fractions $K$, and $R \to A$ a homomorphism. Then $Spec(A) \to Spec(R)$ is an open immersion if and only if $A=0$ or $R \to K$ factors through $R \to A$ (i.e. $A$ is birational over $R$) and $A$ is flat and of finite type over $R$.
Proof: Assume $Spec(A) \to Spec(R)$ is an open immersion and $A \neq 0$. It is known that open immersions are flat and of finite type. Thus the same is true vor $R \to A$. Now $R \to K$ is injective, thus also $A \to A \otimes_R K$. In particular, $A \otimes_R K \neq 0$. Open immersions are stable under base change, so that $Spec(A \otimes_R K) \to Spec(K)$ is an open immersion. But since $Spec(K)$ has only one element and $Spec(A \otimes_R K)$ is non-empty, it has to be an isomorphism, i.e. $K \to A \otimes_R K$ is an isomorphism. Now $R \to A \to A \otimes_R K \cong K$ is the desired factorization.
Of course, the converse is not as trivial. It is proven in the paper
Susumu Oda, On finitely generated birational flat extensions of integral domains Annales mathématiques Blaise Pascal, 11 no. 1 (2004), p. 35-40
It is available online. In the section "Added in Proof." you can find some theorems concerning the general case without integral domains. In particular, it is remarked that in E.G.A. it is shown that
Theorem 2: $Spec(A) \to Spec(R)$ is an open immersion if and only if $R \to A$ is flat, of finite presentation and an epimorphism in the category of rings.
More generally, in EGA IV, 17.9.1 it is proven that a morphism of schemes is an open immersion if and only if it is flat, a (categorical) monomorphism and locally of finite presentation.
There are several descriptions of epimorphisms of rings (they don't have to be surjective), see this MO-question.
Dear Georges,
this is a very interesting question. I can't quite answer it, but I have some ideas that may be worth sharing.
1
I think that if you want a geometric meaning it would be sensible to restrict the question to finitely generated algebras over an algebraically closed field. After all the usual meaning of geometric is that it holds over the algebraic closure. So, perhaps one could say that if $A$ is a $k$-algebra, then $f\in A$ is geometrically irreducible if $f$ stays irreducible in $\overline A=A\otimes_k \overline k$.
I feel that your line intersecting the circle example is a bit misleading. In that example $x$ is only irreducible because we can't see its decomposition over $\mathbb R$. If you consider a singular curve defined over $\mathbb R$, all of whose singular points are not (individually) visible over $\mathbb R$, one may call that curve non-singular (over $\mathbb R$) but it is definitely not smooth.
Nevertheless, even though at first I thought this would help make some headway into the question, I have not been able to make much of this stronger condition either.
2
It would also be reasonable to assume that $A$ is integrally closed, just so speaking about divisors is safe.
3
Based on Francois's and Karl's comments one could try to look for locally irreducible elements, that is, $f\in A$ such that $f$ remains irreducible after localization at any maximal ideal. Their condition does give something geometric, but interestingly, local irreducibility seems to be a pretty strong condition. For obvious irreducible but not prime elements such as $x$ in an algebra where $xy=zt$ but nothing else non-obvious holds, localizing at a prime that contains $x,z,t$ but not $y$ exhibits $x$ as a product, so it is not locally irreducible.
4
Another idea is to look for Cartier divisors contained in the divisor defined by $f$. You give an example where this global assumption fails for an irreducible element. In case somebody (like me at first) thinks that this description might work for geometrically irreducible elements, here is an example for that:
Let $A$ be a regular ring (i.e., all of whose localizations are regular local rings, e.g., the coordinate ring of a smooth affine variety) that is not a UFD.
(Here is one way to construct such a ring: Take an arbitrary smooth projective variety $X$ and let $Z\subset X$ be a hyperplane section. Assume that $Z$ does not generate the Picard group of $X$. This can be achieved if for example $\mathrm{Pic}X$ has rank $\geq 2$ or taking a power of a generator. Then $X\setminus Z$ will be a smooth affine variety with a non-trivial Picard group, so its coordinate ring cannot be a UFD.)
Now let $\mathfrak p\subset A$ be a height $1$ prime ideal that is not principal. This exists by the choice of $A$. Finally let $f\in \mathfrak p$ be a non-zero irreducible element. Then $V(f)\supseteq V(\mathfrak p)$ where the latter is a Cartier divisor. By choice $f$ is not a prime element, so there is a product, say $gh$ that it divides but it does not divide either $g$ or $h$. Now if $V(g)$ and $V(h)$ do not share an irreducible component, then it follows that $V(f)$ is not irreducible (as in the case of $f=x$ when $xy=zt$) which implies that $V(\mathfrak p)\subsetneq V(f)$, so $V(f)$ strictly contains a Cartier divisor even though it is irreducible. (I realize that I have not given a complete example for this behaviour, but I think it is plausible that this can happen).
5
Conclusion (?)
Well, as I said at the beginning I can't really answer your question, but perhaps the answer is that there is no good geometric interpretation. There are other reasons that one may use to argue that geometry corresponds more to ideals than to elements.
One could also take the locally irreducible elements as the irreducible elements corresponding to a geometric meaning. I kind of like that solution as many things in geometry are local. If one wanted to talk about a larger class of elements, one could say that for an $f\in A$, the locus of irreducibility is the subset of $\mathrm{Spec A}$ such that for any point in this locus $f$ is irreducible in the local ring at that point. This should be an open set and then at any point the local irreducibility could be checked by the Francois-Karl criterion.
Remark: The example in the previous point would likely fail to be irreducible somewhere along $V(f)\setminus V(\mathfrak p)$ (not necessarily at the entire set as $f$ would be contained in other height $1$ primes and it seems possible that $V(f)$ is actually the union of Cartier divisors).
Best Answer
As Harry suggests in his answer, it is probably more intuitive to work with associated primes, rather than the slightly older language of primary decompositions.
If $I$ is an ideal in $A$, an associated prime of $A/I$ is a prime ideal of $A$ which is the full annihilator in $A$ of some element of $A/I$. A key fact is that for any element $x$ of $A/I$, the annihilator of $x$ in $A$ is contained in an associated prime.
The associated primes are precisely the primes that contribute to the primary decomposition of $I$. Geometrically, $\wp$ is an associated prime of $A/I$ if there is a section of the structure sheaf of Spec $A/I$ that is supported on the irreducible closed set $V(\wp)$. E.g. in the example given in Cam's answer, the function $x^2 - x$ is not identically zero on $X:=$ Spec ${\mathbb C}[x,y]/(x y, x^3-x^2, x^2 y - xy),$ but it is annihilated by $(x,y)$, and so is supported at the origin (if we restrict it to the complement of $(0,0)$ in $X$ then it becomes zero).
The non-minimal primes of $I$ that play a role in the primary decomposition of $I$ (i.e. appear as associated primes of $A/I$) are the generic points of the so-called embedded components of Spec $A/I$: they are irreducible closed subset of Spec $A/I$ that are not irreducible components, but which are the support of certain sections of the structure sheaf.
An important point is that if $I$ is radical, so that $A/I$ is reduced, then there are no embedded components: the only associated primes are the minimal primes (for the primary decomposition of $I$ is then very simple, as noted in the question: $I$ is just the intersection of its minimal primes).
There is a nice criterion for a Noetherian ring to be reduced: Noetherian $A$ is reduced if and only if $A$ satisfies $R_0$ and $S_1$, i.e. is generically reduced, and has no non-minimal associated primes. Geometrically, and applied to $A/I$ rather than $A$, this says that if $A/I$ is generically reduced, then the embedded components are precisely the irreducible closed subsets of Spec $A/I$ over which the nilpotent sections of the structure sheaf are supported. This may help with your ``nilpotentification'' mental image.