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I have to proof that $f=X^3+aX+b\in F$, where $f$ is a polynomial which is a product of linear terms in a field $F$, has pairwise different roots iff $d=-4a^3-27b^2\neq0$.

Now what I have done is $f=(x-\alpha)(x-\beta)(x-\gamma)$. So I get $a=-\alpha\beta-\alpha^2-\beta^2$ and $b=\alpha^2\beta + \alpha\beta^2$ after using $-(\alpha + \beta+\gamma)=0$. Now for one direction I can plug in $\gamma=\alpha\Rightarrow\beta=-2\alpha$ or $\alpha=\beta$, so indeed $d=0$ if they arent pairwise different. Now I cant seem to find a quick argument. I did some calculations and I got (by brute force, plugging in $x\alpha$ for $\beta$): iff $g:=4x^6+12x^5-3x^4-26x^3-3x^2+12x+4=0$ then for $\beta=\alpha x$, $d$ will be zero (as $F$ is a field I will get all solutions with this substitution). So after I know $x=-2,1,-1/2$ I get through polynom division $g=(x-1)^2(x+2)^2(2x+1)^2$. So its proven now! Now this all seems way too much calculation for this exercise! Please tell me a more easy way!

frac
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1 Answers1

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Note that a polynomial $f(X)$ has a multiple root if and only if it has a common root with its derivative $f'(X)$ (can be checked by writing $f$ as a product of linear factors).

In your case, $f(X)=X^3+aX+b$, so $f'(X)=3X^2+a$. If $a=0$, the polynomials have a common root if and only if $b=0$, so the result is true in this case and we can thus assume $a\not=0$.

If there is a root of $f(X)$ and $f'(X)$, it is a root of $f(X)-\frac{X}{3}f'(X)=\frac{2a}{3}X+b$, so is $X=-\frac{3b}{2a}$.

Computing $$f'(-\frac{3b}{2a})=3\frac{9b^2}{4a^2}+a=\frac{27b^2+4a^3}{4a^2},$$ $$f(-\frac{3b}{2a})=-\frac{27b^3}{8a^3}-\frac{3b}{2}+b=\frac{-27b^3-12a^3b+8a^3b}{8a^3}=\frac{b(-27b^2-4a^3)}{8a^3},$$ you directly find that $f(X)$ has a multiple root if and only if $27b^2+4a^3=0$.

Jérémy Blanc
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