Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F$ be a field. Suppose that the polynomial $p(x,y)$ is irreducible in $F[x,y]$. Let $a(x)\in F[x]$ be a polynomial of positive degree. Prove that $p(a(x),y)$ is irreducible in $F(a(x))[y]$.

I have tried using the Tower Theorem, but that doesn't seem to get me anywhere. Does anyone else have any ideas?

Edit: Jacob Schlather raise a good point. But the original question was posted as $F(a(x))[y]$. So let's assume it is $F[a(x)][y]$.

share|cite|improve this question
Should that be $F[a(x),y]$ otherwise if you have $p(x,y)=x$ and $a(x)=x$ then $p(a(x),y)=x$ is a unit in $F(a(x))[y]=F(x)[y]$? – JSchlather Oct 11 '12 at 2:29
Use the Prof Lawrence's notes! – user44327 Oct 11 '12 at 3:11
Units are irreducible; after all, they can't have any non-unit factors, let alone be factored into a product of two non-units. – Hurkyl Dec 24 '12 at 9:01
@Hurkyl: No, units are neither reducible nor irreducible, just like the number $1$ is neither composite nor prime. Simlarly the number/polynomial $0$ is excluded from both the reducible and irreducible classes. – Marc van Leeuwen Dec 24 '12 at 9:41
@Marc: Odd: I'm used to constant polynomials being considered irreducible in contrast with the general notion of an irreducible element of a domain, although wikipedia agrees with you. – Hurkyl Dec 24 '12 at 9:44
up vote 1 down vote accepted

I'll consider the original question as posted, with irreducibility of a polynomial in $y$ over the field $F(a(x))$ as goal. The statement is not true as given: one needs the additional hypothesis that $p(x,y)\notin F[x]$, since otherwise $p(a(x),y)\in F(a(x))$ is a unit in $F(a(x))[y]$, and therefore neither reducible nor irreducible.

The first step is to show that $p(x,y)$ is irreducible as a polynomial in $y$ over the field $F(x)$, in other words irreducible as element of $F(x)[y]$. With the assumption that $p(x,y)\notin F[x]$ this amounts to showing that $p(x,y)$ is not reducible in $F(x)[y]$. This is the hardest part, but it is a standard result: a lemma attributed to Gauss says that is a polynomial (here in $y$) with coefficients in a Unique Factorization Domain (here $F[x]$) is irreducible, then its is also irreducible over its field of fractions (here $F(x)$).

The second step is to show that from $p(x,y)$ irreducible in $F(x)[y]$ one may conclude that $p(a(x),y)$ is irreducible in $F(a(x))[y]$. But this is mostly a formality, as in the answer by Lior B-S, once one realises that $a(x)$, like any non-constant polynomial, is transcendental over $F$. This transcendence allows defining a ring morphism $F(x)\to F(x)$ that fixes $F$ and sends $x\mapsto a(x)$ (it more generally sends $\frac{P(x)}{Q(x)}\mapsto\frac{P(a(x))}{Q(a(x))}$) and whose image is by definition the subfield $F(a(x))\subseteq F(x)$; since $F(x)$ is a field, this morphism is necessarily injective, and therefore defines an isomorphism of fileds $F(x)\to F(a(x))$. Applying this isomorphism to the coefficients induces an isomorphsims of polynomial rings $F(x)[y]\to F(a(x))[y]$ that sends of $p(x,y)\mapsto p(a(x),y)$ and preserves irreducibility.

For completeness I'll fix the loose ends in the second step. Any non-constant polynomial $a(x)\in F[x]$ is transcendental over $F$ because for any nonnzero $P(x)\in F[x]$ one has $\deg P(a(x))=\deg P\times \deg a$ and in particular $P(a(x))\neq0$: the contribution of the leading term of $P$ (which has that degree) cannot be cancelled by the contribution of any other term of $P$. And this allows defining the indicated morphism $F(x)\to F(x)$, because it ensures that $Q(a(x))\neq0$ for the denominator $Q(x)$: the image of every rational function is well defined.

share|cite|improve this answer

Since $x,y$ are variable they are algebraically independent over $F$. Hence so are $a(x),y$. Therefore there exists an isomorphism of rings $F[x,y]\to F[a(x),y]$ defined by $x\mapsto a(x)$ and $y\mapsto y$.

Any isomorphism of rings preserves the irreducibility property (since one can move factorization from one side to the other). Hence since $p(x,y)$ is irreducible in $F[x,y]$ its image under the isomorphism, namely $p(a(x),y)$, is irreducible in $F[a(x),y]$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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