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Let $f(x) = x^3 +ax+b$ contained in $Q[x]$ prove that $f(x)$ has a multiple root in $Q$ if and only if $disc(f) =0$

This is what I've so far since $f(x) = (x-A_1)(x-A_2)(x-A_3), A_1,A_2, A_3$ are roots of $f(x)$ and $discriminant(f) = (A_1-A_2)(A_1-A_3)(A_2-A_3)$. We prove that $f(x)$ has multiple root if and only if $disc(f(x))=0$.

$(=>)$;Assume $f(x)=0$ has a multiple root, so at least 2 roots are same WLOG, $A_1 =A_2$ then this gives $disc(f)=0$

$(<=)$;Assume $disc(f(x))=0$, so $(A_1-A_2)^2(A_1-A_3)^2(A_2-A_3)^2 =0$, $R =0$-divisior-free this gives one of the factor is zero. Then f(x) has a multiple roots.

Am I doing this proof right? Any help will be appreciated!

etet112
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1 Answers1

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The discriminant of a polynomial f(x) is (up to a scaling factor depending on the leading coefficient) defined to be $$ \prod_{i<j}(r_i-r_j)^2 $$ where the $r_i$ are the roots of $f(x)$ in some splitting field. It's clear from this definition that the discriminant vanishes if and only if $f$ has a multiple root.

carmichael561
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