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Consider a domain $A$ and a non-zero element $f\in A$. That element $f$ is prime if and only if the subscheme $V(f)\subset \operatorname{Spec}(A)$ is integral and this is a completely satisfactory geometric interpretation of primeness.
However I have realized to my annoyance that I can't interpret geometrically what it means that $f$ is irreducible (i.e. not a product of two non-units), a concept that a beginning undergraduate student certainly finds clear and easy! Here is why some naïve guesses turn out to be false :

$f$ irreducible $\nRightarrow V(f)$ irreducible
Take $f=x$ in the ring $\mathbb R[X,Y]/\langle X^2+Y^2-1\rangle =\mathbb R[x,y]$ $V(f)$ irreducible $\nRightarrow f$ irreducible
Can't be right because $V(f)=V(f^2)$ as topological subspaces of $\operatorname{Spec}(A)$
The ideal $(f)$ is irreducible $\nRightarrow f$ is irreducible
(An ideal is irreducible if it is not the intersection of two strictly larger ideals) In the polynomial ring $k[X]$ over the field $k$ , take $f=X^2 \in k[X])$
$f$ is irreducible $\nRightarrow $ the ideal $(f)$ is irreducible
In $\mathbb Z[\sqrt {-5}]$ notice that $3$ is irreducible but $(3)=(3,1+\sqrt {-5})\cap(3,1-\sqrt {-5})$

So let me spell out two ( closely related) questions:

What is the geometric meaning of $f$ being irreducible ?
How do you show in/with Algebraic Geometry that an element is irreducible ?
Here I mean, for example, some analogue of the tricks used in Algebraic Number Theory, like saying that in $\mathbb Z[\sqrt {-5}]$, the number $3$ is irreducible, although not prime, because no element of the ring has norm $3$. I find it frustrating that I can't find an elegant argument proving that $x$ is irreducible in the ring $\mathbb R[X,Y]/\langle X^2+Y^2-1\rangle =\mathbb R[x,y]$ mentioned above...

Edit In relation with the comments below, let me add that in the first non-implication above the fact that $x\in \mathbb R[x,y] =\mathbb R[X,Y]/\langle X^2+Y^2-1\rangle $ is irreducible depends crucially on the ground field being $\mathbb R$: the corresponding irreducibility statement is false over $\mathbb C$.
Indeed in $ \mathbb C[x,y] =\mathbb C[X,Y]/\langle X^2+Y^2-1\rangle$ we have $$ x=1/2(x-iy+i)(-ix+y+1)$$

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Dear Qiaochu: no, it is (alas) not obvious at all. Take my question as a mixture of optimism and bias toward geometry... –  Georges Elencwajg May 4 '11 at 11:07
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An element is irreducible iff the ideal it generates is maximal among the principal ideals. But from the point of view of schemes we are considering all ideals not just principal ones, so it isn't clear there should be a geometric characterization... –  François Brunault May 4 '11 at 11:55
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Dear François, you are too modest: this is an interesting remark, which shows that the hypersurface $V(f)$ does not scheme-theoretically strictly contain a principal hypersurface . If you feel so inclined, I'm sure you can develop this into an answer that I'd be happy to upvote, since it would definitely be a step in the geometric direction. Anyway, thanks for your comment. –  Georges Elencwajg May 4 '11 at 12:36
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Dear Karl, unless I am misunderstanding something, this doesn't seem to be correct. Consider the real circle $x^2+y^2=1$ evoked in my question. The element $x\in \mathbb R[x,y]$ is irreducible. However its divisor $div(x)$ is the sum of the two divisors corresponding to the two points with coordinates $(0,1)$ and $(0,-1)$ (since the circle is a regular variety, there is no distinction between Cartier and Weil divisors). –  Georges Elencwajg May 4 '11 at 16:19
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You are right, I was being dumb. Here's a correct statement (hopefully). For what it's worth, if R is normal and local then I think that $f$ is irreducible if and only if the associated divisor $\text{div}(f)$ cannot be written as a sum of effective non-zero Cartier divisors. This is basically the geometric phrasing of François's statement in the local case. –  Karl Schwede May 4 '11 at 17:09
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3 Answers

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).

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Dear Sándor, thank you for your interesting and insightful comments. Be assured that I would certainly never call non-singular a curve "all of whose singular points are not (individually) visible over $\mathbb R$" ( I suppose you mean something like like $y^2+(x^2+1)^2=0$ ?). It is unfortunately definitely possible that, as you say, there is no good geometric interpretation of irreducibility, but perhaps (my secret wish !) you, or someone else, will find one and pleasantly surprise me with it.. –  Georges Elencwajg May 6 '11 at 12:16
    
Dear Georges, when I wrote "you" I did not mean "you personally", I meant "one" indefinitely. Also, I don't think it is inherently wrong to use that terminology, but one has to keep in mind that over non-algebraically closed fields "non-singular" and "smooth" are not synonyms. –  Sándor Kovács May 6 '11 at 16:19
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In the dictionary in my head, I think of elements or $R$ as functions on $\mathop{\mathrm{Spec}} R$ -- i.e. as being analogous to scalar fields on manifolds. In my mind, this lessens the expectation that their properties should be directly interpretable in terms of subschemes of $\mathop{\mathrm{Spec}} R$.

That said, there is still correlation between the reducilibty of an element of $R$ and the reducibility of its zero set; the game just becomes cataloging the differences.

One of them is that we should be looking at the subscheme, not the subspace. For example, in the plane with coordinate ring $k[x,y]$, the origin $V(x,y)$ and the scheme $V(y, y-x^2)$ are different; the latter is the "double point" over $k$. The geometry remembers the difference between primary and prime ideals, even though the topology forgets.

That said, the real missing ingredient is the Picard group -- your reducibility criterion is not merely splitting $\mathop{\mathrm{Div}}(f)$ into a sum of nonzero effective divisors, but those divisors must also vanish into the Picard group.

Anyways, your analog of algebraic number theory for the circle is to consider the norm from $\mathbb{R}(x,y)$ down to $\mathbb{R}(x)$. If $(x,1-y)$ was principal with generator $f(x) + g(x) y$, then for some nonzero $a \in \mathbb{R}$: $$ a x = N(f(x) + g(x) y) = \frac{ (x^2 - 1) g(x)^2 + f(x)^2 }{g(x)^2} $$ From which you argue $g(x) = 1$, and $f(x)$ must be linear with leading coefficient $\mathbb{i}$, an impossibility.

Aside: I feel like there should be some sort of Galois descent argument to compare $\mathbb{C}[x,y]$ with $\mathbb{R}[x,y]$, but it's beyond my expertise to see how it should go.

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Thank you for your answer, Hurkyl. –  Georges Elencwajg May 6 '11 at 11:58
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Dear Georges,

I don't have an answer for this interesting question but here are some remarks (which raise more questions than answers...).

1) If $A$ is a UFD then $f$ is irreducible iff $(f)$ is a nonzero prime ideal. So in this case we have a geometric characterization. Unfortunately this equivalence already fails for Dedekind domains. One could first investigate the case $A=k[C]$ where $C$ is a nonsingular affine curve over an algebraically closed field $k$ : is there a characterization of the irreducibility of $f$ in terms of $\operatorname{div}(f)$ ?

2) As I mentioned in my comment, for any domain $A$, a nonzero element $f \in A$ is irreducible iff the ideal $(f)$ is maximal among the principal ideals of $A$. This is equivalent to saying that the hypersurface $V(f) \subset \operatorname{Spec}(A)$ is minimal among the hypersurfaces of $\operatorname{Spec}(A)$ (endowed with their scheme structure). However, I find this characterization a little bit artificial, and I don't really know what to do further with it.

3) As you mention in the question, the topological space $V(f) \subset \operatorname{Spec}(A)$ is not enough to tell whether $f$ is irreducible (because $V(f)=V(f^2)$). One can further see that the full scheme structure of $V(f)$ is not sufficient to decide the irreducibility of $f$. An example (already given) is $A=\mathbf{R}[X,Y]/(X^2+Y^2-1)$, $f=\overline{X}$ and $A'=\mathbf{R}[Y]$, $f'=Y^2-1$. We have $A/(f) \cong A'/(f')$ but $f$ is irreducible in $A$ while $f'$ is reducible in $A'$.

So it seems that one needs further information on how $V(f)$ is embedded in $\operatorname{Spec}(A)$ in order to decide the irreducibility of $f$. Again giving just the continuous map $V(f) \to \operatorname{Spec}(A)$ isn't sufficient (think of $V(f^2)$). Of course the embedding of schemes $V(f) \to \operatorname{Spec}(A)$ enables one to recover $(f)$ and thus the nature of $f$, but we would like to use less information. In this direction I think Sándor's idea of looking what happens locally is a very good idea.

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Cher François, thank you for fleshing out your comment and turning it into a thoughtful answer. In particular, I was first baffled by your point 3) but now that (I think) I understand it, I find it quite subtle and intriguing. –  Georges Elencwajg May 7 '11 at 13:09
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