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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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up vote 8 down vote accepted

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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The question is about discriminants of polynomials, not discriminants of number fields, so you should remark on the relation between the two. If you allow rational coefficients you can get discriminant 1: $x^3 - x + 1/3$ and $x^3 - x - 1/3$ both have discriminant 1. Up to a change of variables $x' = x + c$ these are the only monic cubics with Q-coeff. and discrim. 1 (Kronecker). –  KCd Nov 7 '11 at 5:19
The theorem of Kronecker I cited at the end of my previous comment uses Fermat's Last Theorem for exponent 3. See pp. 411--412 in Uspensky and Heaslet's Elementary Number Theory (McGraw-Hill, 1939). –  KCd Nov 7 '11 at 5:21
The subject was about polynomials in $\mathbb{Z}[t]$, for rational coefficients it makes no sense. –  pki Nov 7 '11 at 5:24
pki: The point of the later part of my comment was to illustrate that if you omit the condition of the coefficients being integers then you can get cubics with discriminant 1 which are irreducible (a contrast to the quadratic case) and moreover the complete description of when this happens is closely related to FLT for exponent 3. I think that is interesting! Even if a question is asked just for ${\mathbf Z}[t]$, it can be worthwhile to see what happens if you look at it in ${\mathbf Q}[t]$ too. –  KCd Nov 8 '11 at 0:01
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. –  Tapu Nov 8 '11 at 0:53
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