995: Frobenius Endomorphism for Commutative Ring with Prime Characteristic

<The previous article in this series | The table of contents of this series | The next article in this series>

definition of Frobenius endomorphism for commutative ring with prime characteristic

---

Topics

About: ring

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

---

Starting Context

• The reader knows a definition of characteristic of ring.
• The reader knows a definition of %structure kind name% endomorphism.

---

Target Context

• The reader will have a definition of Frobenius endomorphism for commutative ring with prime characteristic.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$R$: $\in \{\text{ the commutative rings }\}$ with characteristic, $p\in \{\text{ the prime numbers }\}$
$\ast F$: $:R\to R,r\mapsto r^p$, $\in \{\text{ the ring endomorphisms }\}$
//

Conditions:
//

---

2: Note

$R$ needs to be commutative and $p$ needs to be prime in order for $F$ to be a ring endomorphism, as will be seen below.

Let us see that $F$ is indeed a ring endomorphism.

Let $r_1,r_2\in R$ be any.

$F(0)=0^p=0$.

$F(r_1+r_2)=(r_1+r_2{)}^p=r_1^p+r_2^p=F(r_1)+F(r_2)$: the binomial theorem holds for any commutative ring, because it is only about distributability and commutativity; so, $(r_1+r_2{)}^p=\sum _{j\in \{0,...,p\}}{}_p{C}_j{r}_1^{p-j}{r}_2^j$, where ${}_p{C}_j=p!/(j!(p-j)!)$; but for each $j\in \{1,...,p-1\}$, ${}_p{C}_j$ is a multiple of $p$, because the denominator does not contain any $p$ as a factor, which is because $p$ is a prime number, which implies that ${}_p{C}_j{r}_1^{p-j}{r}_2^j=0$: for each $r\in R$, $p\cdot r=p\cdot (1r)=1r+...+1r=(1+...+1)r=(p\cdot 1)r=0r=0$ and $(lp)\cdot r=l\cdot (p\cdot r)=l\cdot 0=0$.

$F(1)=1^p=1$.

$F(r_1{r}_2)=(r_1{r}_2{)}^p=r_1^p{r}_2^p=F(r_1)F(r_2)$, because $R$ is commutative.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>