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Anyone know how to show the following interesting statement (maybe it is not true):

Problem: Let $K$ be an algebraically closed field with char $K=0$, and $f(X),g(X)\in K[X]$ be such that $f(X)$ and $g(X)$ are relatively prime. For any $a\in K$, we define $h_a(X)=f(X)-ag(X)\in K[X]$. Then the discriminant $\Delta(a)$ of $h_a(X)$ is a nonzero polynomial of $a$.

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Is it the discriminant you mean by any chance? – Andreas Caranti Feb 25 '13 at 15:38
Consider the case $f(X)=X^3+bX^2+cX+d$ with $d\ne 0$ and $g(X)=X^3$. Then $$\Delta(a)=b^2c^2-4(1-a)c^3-4b^3d+18(1-a)bcd-27(1-a)^2d^2\quad\text{if }a\ne 1.$$ But $\Delta(1)=c^2-4bd$, which differs from the above by a factor of $b^2$. Maybe you want to add the condition that $\deg g<\deg f$? – Hagen von Eitzen Feb 25 '13 at 16:57
If the two polynomials are relatively prime over an algebraically closed field they have no roots in common. Do you see how this helps? – user38268 Feb 25 '13 at 22:35
Yes, Caranti is right. It should be discriminant. I solved the problem. Just look at $H(X)=\frac{f(X)}{g(X)}$, so the zeros of $h_a(X)$ is the same as the zeros of $H(X)$. Notice that any multiplicity(means at least two) roots of $f(X)-ag(X)$ must be zeros of $f'(X)g(X)-f(X)g'(X)$. – Mingfeng Zhao Feb 25 '13 at 22:35
To BenjaLim, for example, let $f(X)=X^2$, $g(X)=X^2$, for any $a\in K$, we have $h_a(X)=(1-a)X^2$, then $\Delta(a)=0$ for all $a\in K$. – Mingfeng Zhao Feb 25 '13 at 22:39

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