# Elementary proof of "irreducible polynomials remain irreducible over transcendental extension"?

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 $K$.

• Would using the Nullstellensatz instead of quantifier elimination be more "elementary"? Apr 25, 2018 at 18:56
• Definitely already a step in the right direction. Apr 25, 2018 at 20:31

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 $K$.)

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.

• Interesting approach to use this apparently stronger statement on arbitrary field extensions to circumvent the use of quantifier elimination. Apr 26, 2018 at 8:24
• @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! Apr 26, 2018 at 9:09
• 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? Apr 26, 2018 at 19:15
• @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. Apr 27, 2018 at 3:55
• @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. Apr 27, 2018 at 4:45