# Irreducible polynomial substitution

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 $F[a(x)][y]$.

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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

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.

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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 $F[a(x),y]$.

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