Power to Characteristic of Field is Monomorphism

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let the characteristic of $F$ be $p$ where $p \ne 0$.

Let $\phi: F \to F$ be the mapping on $F$ defined as:

$\forall x \in F: \map \phi x = x^p$


Then $\phi$ is a (field) monomorphism.


Proof

Let $a, b \in F$.

First note that:

\(\ds \forall k: 0 < k < p: \, \) \(\ds \binom p k\) \(\equiv\) \(\ds 0\) \(\ds \pmod p\) Binomial Coefficient of Prime
\(\ds \leadsto \ \ \) \(\ds \binom p k\) \(=\) \(\ds r p\) for some $r \in \Z$
\(\ds \leadsto \ \ \) \(\ds \binom p k a^k b^{p - k}\) \(=\) \(\ds r p \cdot a^k b^{p - k}\)
\(\ds \leadsto \ \ \) \(\ds \binom p k a^k b^{p - k}\) \(=\) \(\ds 0_F\) Characteristic of Field by Annihilator: Prime Characteristic
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 1}^{p - 1} \binom p k a^k b^{p - k}\) \(=\) \(\ds 0_F\)

So:

\(\ds \map \phi {a + b}\) \(=\) \(\ds \paren {a + b}^p\)
\(\ds \) \(=\) \(\ds a^p + \sum_{k \mathop = 1}^{p - 1} \binom p k a^k b^{p - k} + b^p\) Binomial Theorem
\(\ds \) \(=\) \(\ds a^p + 0_F + b^p\)
\(\ds \) \(=\) \(\ds a^p + b^p\)
\(\ds \) \(=\) \(\ds \map \phi a + \map \phi b\)


Multiplication is more straightforward:

\(\ds \map \phi {a b}\) \(=\) \(\ds \paren {a b}^p\)
\(\ds \) \(=\) \(\ds a^p b^p\) Power of Product of Commutative Elements in Monoid
\(\ds \) \(=\) \(\ds \map \phi a \map \phi b\)


Thus, $\phi$ is a (field) homomorphism.


It is clear that $\phi$ is not a zero homomorphism, since:

\(\ds \map \phi {1_F}\) \(=\) \(\ds 1_F^p\)
\(\ds \) \(=\) \(\ds 1_F\) Power of Identity is Identity
\(\ds \) \(\ne\) \(\ds 0_F\)

Hence, from Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, it follows that $\phi$ must be a monomorphism.

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Power_to_Characteristic_of_Field_is_Monomorphism&oldid=544239"