# Why does the discriminant of an integral polynomial have integral coefficients?

Ok, Ok, I know that in fact the discriminant is defined (up to sign) as a product of differences of the roots of the polynomial.

But why does it then have integral coefficients, if the polynomial you started with had integer coefficients?

• The title of this question should probably be modified, since the body asks a different question. Commented Jul 27, 2010 at 23:07
• @Qiaochu: True enough. Modified.
– user218
Commented Jul 28, 2010 at 13:26

An example will probably make this clearer. Suppose I have a quadratic polynomial $x^2 + bx + c$ with two roots $r_1, r_2$. Then $x^2 + bx + c = (x - r_1)(x - r_2)$, so $b = - r_1 - r_2$ and $c = r_1 r_2$. These are the elementary symmetric functions in two variables, and the theorem above implies that every polynomial function of $r_1$ and $r_2$ which is invariant under switching the two is actually a polynomial in $b$ and $c$. For example, the discriminant is
$$(r_1 - r_2)^2 = r_1^2 - 2r_1 r_2 + r_2^2 = (r_1 + r_2)^2 - 4r_1 r_2 = b^2 - 4c.$$