Composition of Ring Automorphisms is Ring Automorphism
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Theorem
Let $R$ be a set.
Let:
- $\left({R, +_1, \circ_1}\right)$
- $\left({R, +_2, \circ_2}\right)$
- $\left({R, +_3, \circ_3}\right)$
be rings.
Let:
- $\phi: \left({R, +_1, \circ_1}\right) \to \left({R, +_2, \circ_2}\right)$
- $\psi: \left({R, +_2, \circ_2}\right) \to \left({R, +_3, \circ_3}\right)$
Then the composite of $\phi$ and $\psi$ is also a (ring) automorphism.
Proof
A ring automorphism is a ring isomorphism $f$ from a set to itself.
That is:
- $\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\phi}\right)$
- $\operatorname{Dom} \left({\psi}\right) = \operatorname{Cdm} \left({\psi}\right)$
From Composition of Ring Isomorphisms is Ring Isomorphism, $\psi \circ \phi$ is a ring isomorphism.
By definition of composition of mappings:
- $\operatorname{Cdm} \left({\phi}\right) = \operatorname{Dom} \left({\psi}\right)$
Thus:
- $\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\psi}\right) = R$
and so $\psi \circ \phi$ is a ring automorphism.
$\blacksquare$
Sources
- 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): $\S 2.2$: Homomorphisms