Let $k$ be a ring (perhaps a field). Let $M$ be the "set" of isomorphism classes of $k$-algebras and regard it as a commutative monoid with multiplication $\otimes_k $ and unit element $k$. There is an obvious notion of "irreducible" and "factorization into irreducibles" in any commutative monoid. Is $k[x]$ irreducible in $M$? That is, is there any nontrivial factorization $k[x] \cong A \otimes B$? What about more variables, is $k[x_1,...,x_n] \cong k[x_1] \otimes ... \otimes k[x_n]$ essentially the unique factorization of the polynomial ring into irreducibles?

More generally, we may consider the "set" of isomorphism classes of $k$-schemes, consider it as a monoid with respect to $\times_k$ and may ask if $\mathbb{A}^1$ is irreducible in this monoid (this is a more general question since the factors don't have to be affine, a priori, EDIT: however, under suitable finiteness conditions it follows).

Are questions of this nature studied in the literature?

  • 1
    $\begingroup$ If $k$ is a field, then I believe that $k[x]$ is indeed irreducible. If $k[x]\cong A\times_kB$, then it is fairly easy to see that both $A,B$ must be finitley generated over $k$ and then we have, $1=\dim k[x]=\dim A+\dim B$, making one of them to be zero. Rest is easy. $\endgroup$ – Mohan May 19 '11 at 14:27
  • $\begingroup$ I guess this is vaguely related to Zariski's cancellation problem: If $X\times \mathbb{A}^1 \simeq Y \times \mathbb{A}^1$, does it follow that $X \simeq Y$? I think there are counterexamples to this problem for $X,Y$ affine surfaces. In any case, it seems that problems of this type tend to be very difficult. $\endgroup$ – J.C. Ottem May 19 '11 at 14:33
  • $\begingroup$ @J.C. Ottem: Why do you think that factorization is related to cancellation? @Mohan: Thanks, this works! $\endgroup$ – Martin Brandenburg May 19 '11 at 15:00
  • 2
    $\begingroup$ It is related to cancellation. For example, if $X\times \mathbb{A}^1\cong \mathbb{A}^{n+1}$, one can ask whether $X\cong \mathbb{A}^n$. If not, it gives you `interesting' factorizations of polynomial rings in your sense. For $n=1$, the answer is in the affirmative and easy. For $n=2$, the answer is much deeper and shown by Miyanishi and Sugie. Many of these results can be found (if not the proofs) in Freudenberg's book. $\endgroup$ – Mohan May 19 '11 at 15:21

The question of $\mathbb A^1$ being irreducible in the larger category can still be decided using dimension. If $\mathbb A^1=X\times_kY$ then $\dim_kX+\dim_kY=1$ so one of them has to be zero dimensional. Since $\mathbb A^1$ is connected, it has to be a point, etc.

In a way you are asking a question of classification theory, but I think that at the end you will find that this may not be the best notion of irreducibility.

Nevertheless, there are a few things one can say. For instance, if you restrict to smooth projective varieties then on any product there are semi-ample line bundles whose Kodaira dimension is positive but not maximal. In other words, if for example there is at most one non-trivial semi-ample line bundle which is not ample, then the variety is irreducible in this sense.

This happens for example if the Picard number is $1$. In particular, $\mathbb P^n$ is irreducible in this sense. For an example with Picard number larger than $1$ consider a ruled surface. Its Picard number is $2$ and hence there are two potential semi-ample divisors, the fiber of the ruling morphism and the negative section. The latter is clearly not semi-ample, so the only non-trivial semi-ample line bundle corresponds to the ruling and hence a non-trivial ruled surface is irreducible in this sense.

A perhaps more interesting question is whether this can be done birationally, that is, if something is birational to a product. For instance, $\mathbb P^n$ is birational to $\mathbb P^k\times \mathbb P^{n-k}$ for any $0<k<n$ and any ruled surface over $C$ is birational to $C\times \mathbb P^1$.

So, how do you find a (birational) decomposition of a variety?

Pick a line bundle $L$. Then there exists a birational model and fibration $\pi:X\to B$ such that $\dim B$ is the Kodaira dimension of $L$. So, if if you have a line bundle with positive Kodaira dimension that is less then $\dim X$, then there exists, at least birationally, a non-trivial fibration.

And if there is a non-trivial fibration, then there exists a line bundle with positive Kodaira dimension that is less then $\dim X$. So, this is pretty much settled.

In general, fibrations, even smooth families are far from being products, because first of all the fibers do not have to be isomorphic and even if they are, the fibration is not necessarily trivial, just think of $\mathbb P^1$-bundles (see ruled surfaces above).

So, you can study these fibration and try to find criteria for when the fibers are isomorphic. If that happens and say the fibers have finite automorphism groups, then a finite étale cover is a product. Not exactly birational, but pretty close.

One could write a lot more on this issue. In fact, a lot of people have already written a lot more.

  • $\begingroup$ I know the equation $\dim(X \times_k Y) = \dim(X) + \dim(Y)$ only if $X,Y$ are of finite type over $k$. $\endgroup$ – Martin Brandenburg May 20 '11 at 8:34
  • $\begingroup$ Anyway, thank you for this answer and interesting variants! $\endgroup$ – Martin Brandenburg May 20 '11 at 8:36
  • $\begingroup$ Actually one does not need the dimension formula in the case of one variable: let $C$ be an integral $k$-algebra such that the fraction field of $C$ has transcendence degree $1$ over $k$, and $k$ is algebraically closed in $C$. Then $C$ is irreducible: it is clear that in a factorization $C\cong A\otimes B$ one of the factors, $A$ say, must contain an element $x$ transcendental over $k$. It remains transcendental over $B$. Hence $B$ must be algebraic over $k$. The natural map $B\rightarrow A\otimes B$ is injective, thus $B=k$ by assumption. $\endgroup$ – Hagen May 20 '11 at 9:46
  • $\begingroup$ @Martin: if $A\simeq X\times Y$ and $A$ is of finite type over $k$, then so are $X$ and $Y$: Let $U\subseteq X$ and $V\subseteq Y$ be non-empty open affine subschemes and let $W=U\times V\subseteq A$. Then $W$ is an open affine subscheme of $A$, so it is of finite type. Let $p\in U$ be a closed point. Then $V\simeq \{p\}\times V\subseteq W$ is a closed subscheme, so $V$ and similarly $U$ are of finite type. This proves that $X$ and $Y$ are locally of finite type. Now take all of these affines, take their product, so you get an open covering of $A$. (cont'd) $\endgroup$ – Sándor Kovács May 20 '11 at 17:34
  • $\begingroup$ (cont'd) As $A$ is noetherian, finitely many of these suffice to cover it and the corresponding open sets on $X$ and $Y$ show that they both are of finite type. $\endgroup$ – Sándor Kovács May 20 '11 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.