# Factoring polynomials with prime discriminant

I was busy doing a homework exercise in which I had to compute the discriminant $\Delta(f)$ of the polynomial

$$f(X) = X^4+X^2+X+1$$ which turned out to be the prime $257$. Subsequently, I was asked to show that $f$ is irreducible over $\Bbb Q[X]$ (straightforward to do directly since only linear and quadratic factors need to be considered).

However, we also have the general identity: $g \mid f \implies \Delta(g) \mid \Delta(f)$ so that any proper factorisation of $f$ needs a factor with discriminant $1$. I was trying out a few values and came to the conjecture that the only irreducible monic polynomials in $\Bbb Z[X]$ having discriminant $1$ are the linear monic polynomials (i.e. the trivial examples).

I was however unable to prove or disprove this conjecture, or find any information about it. I'd like to know whether it holds; if possible with proof/counterexample and/or references. Thanks in advance.

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However, it is true that, if $f(x)$ is an irreducible monic polynomial with discriminant $\pm 1$, then $f(x) = x + c$ for some constant $c$. This argument uses notions from algebraic number theory, and particularly the relationship between ramification and discriminants; see this blurb by our own KCd. If you are not ready to read this yet, consider it an incentive to learn algebraic number theory!
Proof: If $\Delta(f) = \pm 1$, then $\mathbb{Z}[x]/f(x)$ is the ring of integers in the number field $\mathbb{Q}[x]/f(x)$, and this number field is unramified over $\mathbb{Q}$. The only unramified extension of $\mathbb{Q}$ is $\mathbb{Q}$ itself, so $f$ has degree $1$. $\square$
We are using that $f$ is irreducible to know that $\mathbb{Q}(x)/f(x)$ is a field.
The ANT lecture today was about the geometric methods to obtain finiteness of $\operatorname{Pic}(R)$. A corollary was the statement that any nontrivial number field ramifies. Thanks again. – Lord_Farin Nov 6 '12 at 12:59