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For quadratic polynomials, say $x^2+bx+c$, the discriminant is given by $b^2-4c$. If this equals $1$ (or even just a square), then the polynomial has rational roots and is not irreducible. Does this trend continue to irreducible monics of greater degree?

More generally, is a lower bound known for the absolute value of the discriminant of irreducible, monic polynomials of a given degree? I know that the discriminant in this case must be an integer, and we have a trivial lower bound of $1$, by irreducibility. What more can be said?

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The discriminant can never be $\pm 1$. This follows from ramfication theory. If you let $\alpha$ be a root of your polynomial, then some prime will always ramify in the extension $\mathbb{Q}(\alpha)$ and such a prime must divide the discriminant. This is called Minkowski's theorem. Look it up in any book on algebraic number theory.

The same theory tells us that $3$ is the smallest absolute value of the discriminant of any number field besides ${\mathbb{Q}}$. This absolute value happens for ${\mathbb{Q}}(\sqrt{-3})$. Given any monic irreducible polynomial in $\mathbb{Z}[X]$ and a root $\alpha$ of the polynomial, the discriminant of the number field $\mathbb{Q}(\alpha)$ will always be less (in absolute value) than the discriminant of the polynomial (it's the discriminant of a polynomial divided by some of its square factors). Minkowki's theorem give us the following bound for the discriminant of a number field $K/\mathbb{Q}$ of degree $n$:

$\sqrt{|d_K|}\geq \frac{n^n}{n!}\left(\frac{\pi}{4}\right)^{n/2}$

The lower bound is strictly increasing for each $n$, so you can easily see that small discriminants are only possible for irreducible polynomials of small degree.

Since the discriminant of any number field beside ${\mathbb Q}$ has absolute value at least $3$ and there's an irreducible polynomial $X^2+X+1\in\mathbb{Z}[X]$ with discriminant $-3$, we see that $3$ is also the lower bound for the absolute value of the discriminant of any irreducible polynomial of degree greater than 1.

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    KCd, you are correct. @pki, please note that the answers (and comments) are not only for the OP (asker?), but also it is a collection of related discussion etc, for future use. Let me try to explain: say another user asked about the case $Q[t]$. Then we can just refer to this problem. So, it is useful.2011-11-08