definition of Frobenius endomorphism for commutative ring with prime characteristic
Topics
About: ring
The table of contents of this article
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, \(p \in \{\text{ the prime numbers }\}\)
\(*F\): \(: R \to R, r \mapsto r^p\), \(\in \{\text{ the ring endomorphisms }\}\)
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Conditions:
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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, \((r_1 + r_2)^p = \sum_{j \in \{0, ..., p\}} {}_pC_j r_1^{p - j} r_2^j\), where \({}_pC_j = p! / (j! (p - j)!)\); but for each \(j \in \{1, ..., p - 1\}\), \({}_pC_j\) 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 \({}_pC_j r_1^{p - j} r_2^j = 0\): for each \(r \in R\), \(p \cdot r = p \cdot (1 r) = 1 r + ... + 1 r = (1 + ... + 1) r = (p \cdot 1) r = 0 r = 0\) and \((l p) \cdot r = l \cdot (p \cdot r) = l \cdot 0 = 0\).
\(F (1) = 1^p = 1\).
\(F (r_1 r_2) = (r_1 r_2)^p = r_1^p r_2^p = F (r_1) F (r_2)\), because \(R\) is commutative.
