3

In the middle of a homework problem I realized I proved this lemma: an irreducible polynomial $q \in \mathbb{Z}[x]$ with $\deg q > 1$ has no rational roots. This seems... odd, since many well-known theorems on irreducibility and existence of roots for polynomials with integer coefficients (like the rational root theorem and Eisenstein's criterion) do not go this far. What went wrong in this proof? Or otherwise, am I missing a super easy corollary of the definition of an irreducible polynomial that makes the lemma trivial? Any hints are appreciated.

Here is the proof:

Assume $\beta$ is a root of $q$. Since $\mathbb{Z}[x]$ is a PID we have that $I = (q)$ is a maximal ideal of $\mathbb{Z}[x]$ and so $K = \mathbb{Z}[x]/I$ is a field. Define $f: K \to \mathbb{Q}$ by $f(r + I) = r(\beta)$; this map is well defined because if $r + I = s + I$ then $r - s \in I = (p)$, so $r(x) - s(x) = t(x) p(x)$ for some $t \in \mathbb{Z}[x]$ and thus $r(\beta) = s(\beta) + t(\beta) p(\beta) = s(\beta)$. This map is also a field homomorphism since $(r + s)(\beta) = r(\beta) + s(\beta)$ and $(rs)(\beta) = r(\beta) s(\beta)$. Being a homomorphism between fields, $f$ is injective. If $r(\beta) = 0$ then $r + I \in \ker f = \{I\}$ and so $r \in I = (q)$; this means, in particular, that $\deg r \geq \deg q > 1$ (I'm adopting the convention that $\deg 0 = \infty$), so any polynomial in $\mathbb{Z}[x]$ with $\beta$ as a root has degree greater than $1$. But if $\beta = a/b \in \mathbb{Q}$ then the polynomial $r(x) = bx - a \in \mathbb{Z}[x]$ has $r(\beta) = 0$ and $\deg r = 1$, a contradiction. Hence $\beta \notin \mathbb{Q}$ and so $q$ has no rational roots.

Lorenzo Riva
  • 389

1 Answers1

3

“Since $\mathbb{Z}[x]$ is a PID...”

Ehr... no, it isn’t. $(2,x)$ is not principal.

It’s also not true that $\mathbb{Z}[x]/(q)$ is a field when $q$ is irreducible: $\mathbb{Z}[x]/(x)\cong\mathbb{Z}$ is not a field (this because $\mathbb{Z}[x]$ is not a PID, so $(q)$ need not be maximal). I don’t think your proof can be salvaged at this point.

Note that maximal ideals of $\mathbb{Z}[x]$ have the form $(q(x),p)$, where $q$ is an irreducible polynomial of degree greater than or equal to $1$, and $p$ is a rational prime. The quotient will be a field of positive characteristic.

Your result is true: it follows from Gauss’s Lemma: Gauss’s Lemma says that if $p(x)\in\mathbb{Z}[x]$ is such that the gcd of its coefficients is $1$, and $p(x) = r(x)s(x)$ with $r(x),s(x)\in\mathbb{Q}[x]$, then there exist polynomials $R(x),S(x)\in\mathbb{Z}[x]$, $\deg(R)=\deg(r)$, $\deg(S)=\deg(s)$, such that $p(x) = R(x)S(x)$.

In particular, if a polynomial $p(x)$ is irreducible in $\mathbb{Z}[x]$, then it remains irreducible in $\mathbb{Q}[x]$. In particular, it cannot have rational roots if $\deg(p)\gt 1$.

Also, note that

  1. The Rational Root Theorem is not about irreducible polynomials; it has nothing to say about it, and it just gives you a finite list of possible roots. I don’t think it is related to the result.

  2. Eisenstein’s Criterion allows you to deduce irreducibility in $\mathbb{Q}[x]$: as such, it necessarily implies no roots when the polynomial has degree greater than $1$. So I do not understand why you say its conclusion “does not go as far” as the result at hand. Having no roots in $\mathbb{Q}[x]$ is weaker than being irreducible.

Arturo Magidin
  • 398,050