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Show that the Discriminant of the product of two polynomials in $\mathbb{C} [x]$ is the product of the discriminants multiplied by the square of their resultant, Use the Usual definitions from Discriminant,

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I think it has little to do with Number Theory (except insofar as it may be applied there) and nothing to do with the Chinese Remainder Theorem. It just comes from rolling out the formulas for discriminant and resultant. Discriminant of $f$ involves a product of differences of roots of $f$, resultant of $f$ and $g$ involves a product of differences of roots of $f$ and roots of $g$. –  Gerry Myerson Oct 15 '12 at 22:46
    
@GerryMyerson Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Jun 12 '13 at 5:54
    
@Julian, done --- but I don't do chat. –  Gerry Myerson Jun 12 '13 at 9:19
    
@GerryMyerson No problem. –  Julian Kuelshammer Jun 12 '13 at 10:16

1 Answer 1

[comment promoted to answer, at suggestion of @Julian Kuelshammer]

This comes from rolling out the formulas for discriminant and resultant. Discriminant of $f$ involves a product of differences of roots of $f$, resultant of $f$ and $g$ involves a product of differences of roots of $f$ and roots of $g$, and the roots of a product are the union, with multiplicities, of the roots of the individual polynomials.

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