8
$\begingroup$

Consider the following statement

Let $K$ be a field, $L/K$ a purely transcendental extensions of fields (i.e. $K$ is relatively algebraically closed in $L$). Let $F \in K[X_1, \ldots, X_n]$ be an irreducible polynomial over $K$. Then $F$ remains irreducible over $L$.

Is there an elementary way to show this statement, accesible to undergraduate students? I will give a proof of which I consider at least part 1 to be non-elementary. (source: Irreducibility of Polynomials over Global Fields is Diophantine, Philip Dittmann, arxiv)

  1. Special case $K$ and $L$ algebraically closed. In this special case, the statement follows by quantifier elimination in algebraically closed fields. Indeed, irreducibility of a polynomial $F$ can be written as a firs-order formula with the coefficients of $F$ as parameters. By quantifier elimination this formula is equivalent to a formula without quantifiers, which then holds in $L$ as soon as it holds in $K$.

  2. General case. Let $\overline{L}$ be an algebraic closure of $L$ and $\overline{K}$ the algebraic closure of $K$ in $\overline{L}$. After a change of coordinates we may assume the constant coefficient of $F$ is $1$. Suppose $F$ factors as a product of irreducible polynomials $F_1, \ldots, F_n$ over $\overline{K}$ with constant coefficient $1$. By the special case each of these factors (in $\overline{K}[X_1, \ldots, X_n]$) remains irreducible over $\overline{L}$. Suppose for the sake of a contradiction that $F$ were reducible over $L$, then we would have $F = G \cdot H$ for certain $G, H \in L[X_1, \ldots, X_n]$ with constant coefficient $1$. By unique factorisation in $\overline{L}[X_1, \ldots, X_n]$, both $G$ and $H$ would have to be products of certain $F_i$'s, whereby their coefficients would lie in $\overline{K} \cap L = K$, contradicting the assumption that $F$ is irreducible over $K$.

$\endgroup$
2
  • $\begingroup$ Would using the Nullstellensatz instead of quantifier elimination be more "elementary"? $\endgroup$ Apr 25, 2018 at 18:56
  • $\begingroup$ Definitely already a step in the right direction. $\endgroup$
    – Bib-lost
    Apr 25, 2018 at 20:31

2 Answers 2

2
$\begingroup$

This may not be as elementary as you would like, but since you commented that it would be a step in the right direction, here is how you can replace the use of quantifier elimination with the Nullstellensatz.

Suppose that $K$ is algebraically closed, and that $F$ has a nontrivial factorization $F=GH$ over $L$. Let $A\subseteq L$ be the $K$-subalgebra generated by the coefficients of $G$ and $H$. Then $A$ is a finitely generated reduced commutative $K$-algebra. By the Nullstellensatz, for any nonzero $a\in A$, there exists a $K$-algebra homomorphism $\varphi:A\to K$ such that $\varphi(a)\neq 0$. In particular, pick nonzero coefficients of $G$ and $H$ other than the constant coefficients and let $a$ be their product, and let $\varphi:A\to K$ be such that $\varphi(a)\neq 0$.

Now, let $G'$ and $H'$ be the polynomials obtained by applying $\varphi$ to the coefficients of $G$ and $H$. Since $\varphi$ is a $K$-algebra homomorphism and $F$ has coefficients in $K$, we have $G'H'=F$. Moreover, since $\varphi(a)\neq 0$, $G'$ and $H'$ are both nonconstant. This contradicts the irreducibility of $F$.

(You can rephrase this to use more concrete version of the Nullstellensatz explicitly in terms of solving polynomial equations. Consider the equation $F=GH$ as a system of polynomial equations over $K$, with the variables being the coefficients of $G$ and $H$ and having one equation for each coefficient of $F$. By the Rabinowicz trick, you can add some more equations and variables that imply that $G$ and $H$ are nonconstant (pick a nonzero coefficient of each that is not the constant coefficient, and add a new variable which is its inverse). Since this system of equations has a solution in $L$, these polynomial equations cannot generate the unit ideal. By the weak Nullstellensatz, that implies they have a solution in $K$, which gives a factorization of $F$ over $K$.)

$\endgroup$
2
$\begingroup$

One way to see it is as follows: if $F$ is a field, $f_i = 0$ a system of polynomial equations with coefficients in $F$ that has finitely many solutions in any extension of $F$. Then any solution in an extension of $F$ has all components algebraic over $F$. Indeed, consider one solution in an extension $K$ of $F$. Let $F'$ the field generated by the components of that solution. It's enough to show that $F'$ is algebraic ( and so finite) over $F$. Indeed, if it were not, we would have infinitely many $F$ morphisms of $F'$ into some conveniently large extension $L$ of $F$. Each such morphism would produce a different solution of the system over $L$, contradiction.

Note that "finitely many solutions in any extension of $F$" is apriori stronger than "finitely many solutions in (some extension of )$\bar F$ ", although elimination of quantifiers clarifies they are equivalent statements. But we only use the weaker implication, that only uses basic field theory.

Now consider a factorization
$$P\cdot Q = R$$ where $R\in F[x_1, \ldots, x_n]$ and $P$, $Q \in K[x_1, \ldots, x_n]$. Let $I$, $J$ the largest monomials in the supports of $P$, $Q$ for some monomial order. Then for the coefficients $a_I$, $b_J$, $c_{I+J}$ of $P$, $Q$, $R$ we have $$a_I \cdot b_J = c_{I+J}$$ So let's choose a rescaling of the coefficients of $P$, $Q$ so that $$a_I = 1\\ b_J = c_{I+J}$$

Note that the system of equations for the coefficients of $P$, $Q$ ( polynomial, coefficients in $F$) that translates the equalities $$P\cdot Q = R\\ a_I= 1\\ b_J= c_{I+J}$$ has finitely many solutions in any extension $L$ of $F$ ( it follows from the fact that $L[x_1, \ldots,x_n]$ is UFD). We now conclude that the coefficients of $P$, $Q$ are algebraic over $F$.

The rescaling seems a bit unnatural. In fact one can show the following:

We have integral dependencies of any $a_I\cdot b_J$ over the ring $\mathbb{Z}[c_K]$. One can show this by induction on $n$, using Ex 8, 9 chap v in Atiyah-Macdonald Commutative Algebra. One can produce explicitely such dependencies for a given type of factorizations $P\cdot Q =R$, using Groebner bases for elimination.

$\endgroup$
5
  • $\begingroup$ Interesting approach to use this apparently stronger statement on arbitrary field extensions to circumvent the use of quantifier elimination. $\endgroup$
    – Bib-lost
    Apr 26, 2018 at 8:24
  • $\begingroup$ @Bib-lost: Think of it: if a system of polynomials with rational coefficients has finitely many complex solutions then those solutions are from $\bar{ \mathbb{Q}}$. Same trick works because transcendence degree of $\mathbb{C}$ over $\mathbb{Q}$ is infinite, so we have enough space inside, no quantifiers needed in this case.More interesting: if it has finitely many real solutions, those solutions are from $\mathbb{Q}$. Of course we can real quantifier elimination, but one can also do "by hand". The problem is extending morphisms, which means solving "real equations". Hope it helps. Cheers! $\endgroup$
    – orangeskid
    Apr 26, 2018 at 9:09
  • $\begingroup$ I am still a bit confused by the part "Indeed, if it were not [algebraic], we would have infinitely many $F$ morphisms of $F′$ into some conveniently large extension $L$ of $F$." Is this easily justified? $\endgroup$
    – Bib-lost
    Apr 26, 2018 at 19:15
  • $\begingroup$ @Bib-lost: The extension is finitely generated. So it is a finitely generated transcendental,and on top of it a transcendental one. It is the transcendental part that provides infinitely many starts: for instance, $(t_1, \ldots, t_m) \mapsto (t_1 + t_2^N, t_2,\ldots,t_m)$. Then we need to extend to the algebraic extension on top of it. So the large field we take is the algebraic closure of the field generated by $(t_1, \ldots, t_m)$. That is the only subtlety, the existence of an algebraitc closure of any field. We can certainly appreciate it. $\endgroup$
    – orangeskid
    Apr 27, 2018 at 3:55
  • $\begingroup$ @Bib-lost: We may not need the algebraic closure, we only need to enlarge enough the extension on top of the transcendental one so that a bunch of equations have solutions. It will be a finite extension in fact, it we take into account Hilbert basis theorem. $\endgroup$
    – orangeskid
    Apr 27, 2018 at 4:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .