Let f(x) be an irreducible polynomial over Q of degree n>1. Prove f(x) has no multiple roots.
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Let f(x) be an irreducible polynomial over Q of degree n>1. Prove f(x) has no multiple roots.
f(x) has no multiple roots in the algebraic closure of Q?Quote:
Originally Posted by apple2009
Then, use the fact an irreducible polynomial over the field of characteristic 0 is separable.
I think OP wants to prove that any extension of Q is separable?
If a is a repeated root, then you can show that a is a root of both f(x) and f'(x), where f' is the derivative of a polynomial, defined not by limit but by (x^n)'=nx^(n-1) directly. Suppose g(x) is the minimal polynomial of a over Q, then g|f and g|f'. If n>1, then f' is not zero polynomial, so deg(f')=deg(f)-1, so g will give a proper factor of f, contradicting that f is irreducible
add: this proof works for any field of characteristic 0. The reason that it can't apply to fields with characteristic p is that f' can be zero polynomial even if f is not a constant, e.g. (x^p)'=px^(p-1)=0 if char(k)=p