When do polynomial rings admit extensions over which every polynomial factorizes?

Question

Given a polynomial ring $F[x_1...x_n]$ over a field $F$, I wanted to ask about when we can factor some $f \in F[x_1...x_n].$ I was curious to see whether a fundamental theorem of algebra-esque notion exists for every polynomial ring (i.e., we can always devise a field extension wherein all polynomials of our given field can be factored).

My thought process is inspired by the case of $\mathbb{Q}= F$ - here, we cannot factor polynomials such as $f(x) = x^2 - 2$, so we devise the algebraic extension $\mathbb{Q}(x)/x^2-2$, quotienting by the ideal generated by $f$.

Similarly, it seems that for an arbitrary case, we can simply consider $\mathbb{F}[x_1...x_n]/f$, which is a field whenever $f$ is irreducible. Therefore, perhaps we get the restriction that we can always construct field extensions wherein all irreducible polynomials are factorable. This is Idea 1.

However, Idea 1 is the same as Idea 2, wherein we take some wonderful object $a$ such that it satisfies the relation $f(a) = a^2 - 2=0$, adjoin it to $\mathbb{Q}$, and yield $\mathbb{Q}(a)$ - a field wherein $f$ as above now factors. We know that Idea 1 is isomorphic to Idea 2, but somehow in Idea 2, we bring out a strange element and (assuming $f$ is irreducible), derive a field by adjoining it to $\mathbb{Q}$.

Now, I imagine that when we have $f$ reducible, then we cannot do this if the polynomial ring is not a UFD - but in that case, what stops us from taking our field $F$, and simply adjoining it to it objects $a_i$ defined by the relation that $f(a_i) = 0$? I guess it would not be a field any longer, but what exactly breaks down? Moreover, are there examples of fields which don't admit any extension over which every polynomial splits? What do such examples tell us about our ability to devise some object, call it a root, and attach it to our field?

Answer 1

You may wish to consider the notion of absolute irreducibility. Wikipedia (https://en.wikipedia.org/wiki/Absolute_irreducibility) says,

In mathematics, a multivariate polynomial defined over the rational numbers is absolutely irreducible if it is irreducible over the complex field. For example, $x^2+y^2-1$ is absolutely irreducible.

More generally, a polynomial defined over a field $K$ is absolutely irreducible if it is irreducible over every algebraic extension of $K$.

Some references are given at that Wikipedia page.

See also $x^2+y^n-1$ is absolutely irreducible

Answer 2

I want to add the following to Gerry's nice answer.

Several things break down when you have at least two variables. The questions becoming geometrical in nature, and hopefully a tag expert on algebraic-geometry can find a suitable older thread.

1. A problem in looking at the quotient ring $F[x_1,x_2,\ldots,x_n]/\langle f\rangle$ is that this is not field. Recall the relevant results from a first course on abstract algebra: A) if $R$ is a commutative ring and $M$ its ideal, then the quotient ring $R/M$ is a field if and only if $M$ is a maximal ideal. B) If $F$ is a field, and $f(x)\in F[x]$ is a non-constant polynomial, then the principal ideal $\langle f\rangle$ is maximal if and only if $f(x)$ is irreducible. We need both of these to get extension fields using polynomial rings. But part B only works when we have a single variable. You may recall that the root cause is the long division algorithm of univariate polynomials. But this is available only when we have a single variable.
2. I said that geometrical ideas enter the scene, so I add the following remark to drive the point home. You may be wishing that there might exist an extension field $L$ such that a given polynomial, say $f(x,y)\in F[x,y]$, factors into a product of linear factors over $L$. Like $f(x,y)=g_1(x,y)g_2(x,y)\cdots g_k(x,y)$ with all the factors linear, so $$g_i(x,y)=a_ix+b_iy+c_i$$ with $a_i,b_i,c_i\in L.$ Let's think about the zero loci of $f(x,y)$. We clearly have $f(u,v)=0$ if and only if $g_i(u,v)=0$ for some $i$. Here $u,v$ may come from any extension field of $L$. But, the zero locus of each and every factor $g_i(x,y)$ is a line in the plane $L^2$. So it would follow that the zeros of $f(x,y)$ should fall on a finite union of lines. This is obviously not the case for example with $f(x,y)=y^2-x^3+x\in\Bbb{R}[x]$. More generally, a factorization of a bivariate polynomial $f(x,y)$ means that its zero locus (usually a curve of some kind) is a union of two simpler curves. This usually does not happen.
3. But some things familiar from extensions of $\Bbb{Q}$ do survive in the bivariate world. If $F$ is any field, the ring $F[x]$ is a Euclidean domain. It follows that the analogues of Eisenstein's irreducibility criterion go through. Recall how Gauss's lemma implied the connection between irreducibility of polynomials over $\Bbb{Z}$ and over $\Bbb{Q}$, the field of fractions. Here the field of fractions of $F[x]$ is the field $F(x)$ of rational functions. So, for example, as $x$ is an irreducible polynomial, it follows that for all natural numbers $n$ $$ g(y)=y^n-x $$ is irreducible as a polynomial in $F(x)[y]$. It is hopefully believable (I skip the argument) that this implies the irreducibility of $G(x,y)=y^n-x$ over any extension field of $F$. This is a useful method for studying the (absolute) irreducibility of bivariate polynomials in particular.
4. The usual process of adjoining zeros does work, if we isolate one of the variables. Then we need to think of the other variables as "constants" that happen to be transcendental over $F$. For example, when we think of the absolutely irreducible bivariate polynomial $f(x,y)=x^2+y^n-1$ as being a univariate polynomial in $x$ and with coefficients in the field $K=F(y)$ of rational functions in the indeterminate $y$, the usual process of adjoining zeros works, and leads to the factorization $$f(x,y)=(x-\sqrt{1-y^n})(x+\sqrt{1-y^n})$$ over a splitting field $L=K(\sqrt{1-y^n})$. On the other hand, if we think of $f(x,y)$ as a polynomial in $y$ with coefficients in the field $E=F(x)$, then the splitting field is $M=E(\zeta,\root n\of{1-x^2})$, where $\zeta$ is an $n$th root of unity (which may or may not reside in $F$). Actually there are issues with separability here should it happen that the characteristic of $F$ is a factor of $n$, but I ignore those for now.

Answer 3

You can ask this: why is it that every monic polynomial $f\in \mathbb{C}[x]$ of degree $m+n$ is a product of a monic polynomial $g\in \mathbb{C}[x]$ of degree $m$ and a monic polynomial $h\in \mathbb{C}[x]$ of degree $n$

$$f= g\cdot h$$

We know that already. But let's look at the dimensions:

$$\mathbb{C}^m \times \mathbb{C}^n \ni (g,h) \mapsto g\cdot h =\colon f \in \mathbb{C}^{m+n}$$ surjective. Sure, why not.

But if you consider polynomials in two variables the dimension on LHS is smaller. Conclusion: most of the polynomials in two or more variables are irreducible over $\mathbb{C}$. OK.

Now, let $F$ be a field, $K$ a field extension of $F$. Assume that some polynomial in $f\in F[x_1, \ldots, x_n]$ has a non-trivial factorization $$f= g\cdot h$$ with $g$, $h \in K[x_1, \ldots, x_n]$. Non-trivial means the polynomials $g$, $h$ are not constants. But then we may assume that at least one of them has a coefficients (not free term) that is $1$. OK

Now, the above equality $f=g\cdot h$ is equivalent to a system of polynomial equations ( of degree $\le 2$ but that is not relevant) with coefficients in $F$ that has a solution in some $K^N$. But then ( Hilbert Nullstellensatz) the system will have a solution in $\bar F^N$ ( if it didn't, then some combination gets us $0=1$ in $K$, not possible). Note also that the decomposition obtained over $\bar F$ is not trivial, since some coefficient will be $1$.

So now we see why people define "absolutely irreducible" to mean irreducible over $\bar F$.

Now, another question: maybe we could extend $K$ to some ring $R$ ( not a domain) such that some polynomials become "reducible". The problem now is that we are working in a ring $R[x_1, \ldots, x_n]$ that is not a domain, so reducibility may not behave so well. For instance, if $\delta^2 = 0$ then $$1 = (1+\delta x)(1-\delta x)$$ Are we interested in such decompositions ( surely both factors are invertible, )? Probably the statements in this case are something that we have seen before ( when is a polynomial a unit). I'll leave it at that.

$\bf{Added:}$ The dimension argument for polynomials over $\mathbb{C}$. Let's calculate the dimension over $k$ of the space of polynomials in $k[x_1, \ldots, x_N]$ of degree $\le d$. For each integer $k$, $0\le k\le d$, we have exactly $\binom{k+N-1}{N-1}$ monomials of degree $k$ in $x_1$, $\ldots$, $x_n$. So in total there are

$$\sum_{k=0}^d \binom{k+N-1}{N-1} = \binom{d+N}{N}$$

( you can also see the above identity as : every polynomial in $x_1$, $x_N$, of degree $\le d$ "is" a homogeneous polynomial in $x_1$, $\ldots$, $x_{N+1}$ of degree $d$)

In other works,

$$\dim \mathcal{P}_{d,N } =\binom{d+N}{N}$$

However, if $k+l = d$, $k$, $l$ integers $>0$, and $N > 1$ then we have

$$\dim \mathcal{P}_{k,N }+ \dim \mathcal{P}_{l,N }< \dim \mathcal{P}_{d,N }$$

This means: the (some) measure of the set of polynomials in $\mathcal{P}_{d,N }$ that decompose into a product of pols of deg $k$, $l$ is $0$.