Embedding Ring into Ring Structure Induced by Ring Operations

Theorem
Let $\left({R, +, \circ}\right)$ be a ring.

Let $S$ be a non-empty set.

Let $\struct {R^S, +', \circ'}$ be the ring, where $+'$ and $\circ'$ are the operations induced on $R^S$ by $+$ and $\circ$.

For each $r \in R$, let $f_r: S \to R$ be the mapping defined by:
 * $\forall s \in S, f_r \paren {s} = r$

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:
 * $\quad \quad \quad \forall r \in R: \phi \paren {r} = f_r$

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:
 * $\quad \phi \paren {r+r' } = \phi \paren {r } +' \phi \paren {r' }$
 * $\quad \phi \paren {r \circ r' } = \phi \paren {r } \circ' \phi \paren {r' }$
 * $\quad r \neq r' \implies \phi \paren {r } \neq \phi \paren {r' }$

That is, for all $r, r' \in R$ it needs to be shown that:
 * $(1):\quad f_{r+r' } = f_r +' f_{r'}$
 * $(2):\quad f_{r \circ r' } = f_{r } \circ' f_{r' }$
 * $(3):\quad r \neq r' \implies f_{r } \neq f_{r' }$

$(1): f_{r+r' } = f_r +' f_{r'}$
For all $s \in S$ then:

The result follows.

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

The result follows.

$(3): r \neq r' \implies f_{r } \neq f_{r' }$
Let $s \in S$, then:
 * $f_r \paren {s} = r \neq r' = f_{r'} \paren {s}$

So $f_r \neq f_{r'}$