TITLE: What does it mean geometrically that an element in a domain is irreducible?
QUESTION [37 upvotes]: 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
$$ x=1/2(x-iy+i)(-ix+y+1)$$

REPLY [7 votes]: 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.