Embedding Ring into Ring Structure Induced by Ring Operations
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Theorem
Let $\struct {R, +, \circ}$ be a ring.
Let $S$ be a non-empty set.
Let $\struct {R^S, +', \circ'}$ be the ring of mappings, where $+'$ and $\circ'$ are the pointwise operations induced on $R^S$ by $+$ and $\circ$.
For each $r \in R$, let $f_r: S \to R$ be the mapping defined by:
- $\forall s \in S, \map {f_r} s = r$
That is, $f_r$ is the constant mapping from $S$ to $r$.
Let $\phi: R \to R^S$ be the mapping from the ring $R$ to the ring $R^S$ defined by:
- $\forall r \in R: \map \phi r = f_r$
Then:
- $\phi$ is a ring monomorphism.
Proof
By the definition of a ring monomorphism it is sufficient to prove for all $r, r' \in R$ that:
- $\quad \map \phi {r + r'} = \map \phi r +' \map \phi {r'}$
- $\quad \map \phi {r \circ r'} = \map \phi r \circ' \map \phi {r'}$
- $\quad r \ne r' \implies \map \phi r \ne \map \phi {r'}$
That is, for all $r, r' \in R$ it needs to be shown that:
- $(1): \quad f_{r + r' } = f_r +' f_{r'}$
- $(2): \quad f_{r \circ r' } = f_r \circ' f_{r'}$
- $(3): \quad r \ne r' \implies f_r \ne f_{r'}$
$(1): f_{r + r'} = f_r +' f_{r'}$
For all $s \in S$:
\(\ds \map {f_{r + r'} } s\) | \(=\) | \(\ds r + r'\) | Definition of $f_{r + r'}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {f_r} s + \map {f_{r'} } s\) | Definition of $f_r$ and $f_{r'}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {f_r +' f_{r'} } } s\) | Definition of Pointwise Operation $+'$ |
The result follows.
$\Box$
$(2): f_{r \circ r'} = f_r \circ' f_{r'}$
For all $s \in S$:
\(\ds \map {f_{r \circ r'} } s\) | \(=\) | \(\ds r \circ r'\) | Definition of $f_{r \circ r'}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {f_r} s \circ \map {f_{r'} } s\) | Definition of $f_r$ and $f_{r'}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {f_r \circ' f_{r' } } } s\) | Definition of Pointwise Operation $\circ'$ |
The result follows.
$\Box$
$(3): r \ne r' \implies f_r \ne f_{r'}$
Let $s \in S$, then:
- $\map {f_r} s = r \ne r' = \map {f_{r'} } s$
So:
- $f_r \ne f_{r'}$
and the proof is complete.
$\blacksquare$