Factorization of schemes 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?
 A: 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.
