Embedding Ring into Ring Structure Induced by Ring Operations

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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:

\(\displaystyle f_{r+r' } \paren {s }\) \(=\) \(\displaystyle r+r'\) Definition of $f_{r+r' }$
\(\displaystyle \) \(=\) \(\displaystyle f_r \paren {s } + f_{r' } \paren {s }\) Definition of $f_{r }$ and $f_{r' }$
\(\displaystyle \) \(=\) \(\displaystyle \paren {f_r +' f_{r' } } \paren {s }\) Definition of induced operation $+'$

The result follows.

$\Box$


$(2): f_{r \circ r' } = f_{r } \circ' f_{r' }$

For all $s \in S$ then:

\(\displaystyle f_{r \circ r' } \paren {s }\) \(=\) \(\displaystyle r \circ r'\) Definition of $f_{r \circ r' }$
\(\displaystyle \) \(=\) \(\displaystyle f_r \paren {s } \circ f_{r' } \paren {s }\) Definition of $f_{r }$ and $f_{r' }$
\(\displaystyle \) \(=\) \(\displaystyle \paren {f_r \circ' f_{r' } } \paren {s }\) Definition of induced operation $\circ'$

The result follows.

$\Box$


$(3): r \neq r' \implies f_{r } \neq f_{r' }$

Let $s \in S$, then:

$f_r \paren {s} = r \neq r' = f_{r'} \paren {s}$

So $f_r \neq f_{r'}$

$\blacksquare$


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