Embedding Ring into Ring Structure Induced by Ring Operations

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 $s \in S$:

The result follows.

$(2): f_{r \circ r'} = f_r \circ' f_{r'}$
For all $s \in S$:

The result follows.

$(3): r \ne r' \implies f_r \ne f_{r'}$
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.