Multiple root

**xinglongdada** · December 11th 2012, 03:10 AM

Let $P$ be an arbitrary field, while $C$ is the complex field. $f(x)$ is a polynomial in $P$ with degree $>0$ . Suppose $f$ has no multiple factor in $P$ , prove then $f$ has no multiple root in $C.$

Here, we mean by $g(x)$ is a multiple factor of $f(x)$ , if $g(x)$ is irreducible, and for some $k>1$ , $g^k(x)\mid f(x).$

**Drexel28** · December 11th 2012, 01:04 PM

This doesn't make any sense as written? I'm going to assume that $P$ is a subfield of $\mathbb{C}$ . Recall then that $f$ having no multiple roots is equivalent to $(f,f')=1$ , which is trivially independent of what field you are discussing. Once again, this question dreally doesn't make sense.

Thread: Multiple root

LinkBack

Thread Tools

Display

Multiple root

Re: Multiple root

Similar Math Help Forum Discussions

n --> infinity, limit root(n-1) - root(n) = zero ?

square root + cube root of a number

Least common multiple of multiple numbers

Apply Newtown's method & shifting nth root algorithm to compute nth root

[SOLVED] Show that any root of the equation 5 + x - \sqrt{3+4x} = 0 is also a root...

Search Tags