$x^3 - 3x - 1$ irreducible in $\mathbb Z[x]$ by Gauss Lemma

Question

In Dummit & Foote, they claim this can be shown to be irreducible by Gauss Lemma and applying it to show it has no rational root.

But this doesn't make sense to me since Gauss Lemma says:

Let $F$ be a field of fraction for UFD $R$, then if $p(x) \in F[x]$ is reducible, then $p(x) \in R[x]$ is reducible. Contrapositive say $p(x) \in R[x]$ irreducible, then $p(x) \in F[x]$ irreducible.

So they implies that $P \to Q$ is equivalent to $\sim P \to \sim Q$?

Answer 1

It's almost trivial that irreducible over $\Bbb Q$ implies irreducible over $\Bbb Z$ (as long as the polynomial is monic). That's not what Gauss' lemma says.

The way I read the D&F quote, you definitely don't need Gauss' lemma for this. You just need its converse, which is the above mentioned triviality, along with propositions 10 and 11 (which presumably is the rational root theorem).

Answer 2

It seems to me that you don’t need Gauss Lemma. Indeed you want to prove that $x^3-3x-1=0$ is irreducible in $\Bbb Z[x]$.

Here another approach: Consider the polynomial $$x^3-3x-1=0$$ and take the reduction modulo $2$. It becomes $$x^3+x+1=0.$$ Since it has not solution in $\Bbb Z_2$, no solution exists in $\Bbb Z$. Hence the original polynomial is irreducible in $\Bbb Z[x]$