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freshman's dream
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(Theorem)
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When $k$ is finite then it is indeed an automorphism. A field $k$ is called a perfect field when the map is surjective.
The theorem is so named because it is a common mistake for freshman math students to make over the real numbers. However, as the characteristic of the real numbers is 0, this does not apply in any interesting way to that setting.
It should also be noted that the result applies only to powers of the characteristic, and not all exponents.
Proof. The proof is an application of the binomial theorem. We prove it for $p$ first. $$ (x+y)^{p}=\sum_{i=0}^p \binom{p}{i} x^{i} y^{p-i} $$ Now observe $$ \binom{p}{i}=\frac{p!}{(p-i)!i!}=p\!\cdot\!\frac{(p-1)!}{(p-i)! i!} $$ As $p$ is prime and $1\leq i\leq p-1$ it follows $i!$ and $(p-i)!$ do not divide $p$ . As the field $k$ has characteristic $p$ , $\frac{(p-1)!}{(p-i)!i!}$ is an integer $m$ where $$ \binom{p}{i}=pm\equiv 0 $$ Thus $(x+y)^p=x^p+y^p$ .
Now for $p^i$ simply use induction: $$ (x+y)^{p^i}=((x+y)^p)^{p^{i-1}}=(x^p+y^p)^{p^{i-1}} =x^{p^i}+y^{p^i} $$ 
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"freshman's dream" is owned by . [ full author list (3) ]
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See Also: formal congruence
| Other names: |
Frobenius Automorphism |
| Keywords: |
Field automorphism, finite characteristic, positive characteristic |
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Cross-references: induction, integer, divide, binomial theorem, application, proof, exponents, characteristic, real numbers, theorem, surjective, map, perfect field, automorphism, finite, monomorphism, field monomorphism, prime, field
There are 3 references to this entry.
This is version 15 of freshman's dream, born on 2006-04-17, modified 2007-05-18.
Object id is 7839, canonical name is FreshmansDream.
Accessed 4713 times total.
Classification:
| AMS MSC: | 11T30 (Number theory :: Finite fields and commutative rings :: Structure theory) | | | 11T23 (Number theory :: Finite fields and commutative rings :: Exponential sums) |
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Pending Errata and Addenda
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