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Let K be an algebraically closed field. Does a non-constant polynomial with coefficients in K necessarily define a surjective function from K to itself? Does a polynomial with coefficients in K defining an injective function from K to itself necessarily have degree 1? GeoffreyT2000 ( talk, contribs) 22:58, 24 September 2016 (UTC)
- Yes to both questions: (1) if f(x) is a polynomial and a is in K, then f(x)-a splits. (2) A degree n>1 polynomial that is one-to-one would have the property that f(x)-a is a perfect nth power for all a in K. In particular, f(x)=x^n, and x^n-1=(x-1)^n. This implies that n is a power of the characteristic of K, so f (x) is either the identity or is the Frobenius automorphism of a finite field. No finite field is algebraically closed, a contradiction.
Sławomir Biały (
talk)
23:44, 24 September 2016 (UTC)
- f(x) can also be the Frobenius automorphism of an infinite field of positive characteristic. And you can multiply it by a nonzero constant, and add a constant to it -- 77.125.79.22 ( talk) 21:22, 30 September 2016 (UTC)
| Mathematics desk | ||
|---|---|---|
| < September 23 | << Aug | September | Oct >> | Current desk > |
| Welcome to the Wikipedia Mathematics Reference Desk Archives |
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| The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
Let K be an algebraically closed field. Does a non-constant polynomial with coefficients in K necessarily define a surjective function from K to itself? Does a polynomial with coefficients in K defining an injective function from K to itself necessarily have degree 1? GeoffreyT2000 ( talk, contribs) 22:58, 24 September 2016 (UTC)
- Yes to both questions: (1) if f(x) is a polynomial and a is in K, then f(x)-a splits. (2) A degree n>1 polynomial that is one-to-one would have the property that f(x)-a is a perfect nth power for all a in K. In particular, f(x)=x^n, and x^n-1=(x-1)^n. This implies that n is a power of the characteristic of K, so f (x) is either the identity or is the Frobenius automorphism of a finite field. No finite field is algebraically closed, a contradiction.
Sławomir Biały (
talk)
23:44, 24 September 2016 (UTC)
- f(x) can also be the Frobenius automorphism of an infinite field of positive characteristic. And you can multiply it by a nonzero constant, and add a constant to it -- 77.125.79.22 ( talk) 21:22, 30 September 2016 (UTC)