Induced Ring

Theorem

Let $\left({T, \oplus, \otimes}\right)$ be a ring.

Let $S$ be a set.

Let $\left({T^S, \oplus}\right)$ be the structure on $T^S$ induced by $\oplus$.

Let $\left({T^S, \otimes}\right)$ be the structure on $T^S$ induced by $\otimes$.

Then $\left({T^S, \oplus, \otimes}\right)$ is a ring.

If $\left({T, \oplus, \otimes}\right)$ has a multiplicative identity $1_R$ then $\left({T^S, \oplus, \otimes}\right)$ has multiplicative identity

$\forall x\in S,\ f_{1_T}(x) = 1_T$

If $\left({T, \oplus, \otimes}\right)$ is commutative, then so is $\left({T^S, \oplus, \otimes}\right)$.

Proof

• From Induced Group, $\left({T^S, \oplus}\right)$ is an abelian group.

• $\left({T^S, \otimes}\right)$ is patently closed, because from the definition of the composition of mappings:

$\forall f, g \in T^S: f \otimes g \in T^S$

• From Induced Structure Associative and Induced Structure Distributive, $\left({T^S, \oplus, \otimes}\right)$ is a ring.

• From Induced Structure Identity, if $\left({T, \oplus, \otimes}\right)$ has multiplicative identity, then the map $f_{1_T}$ above is a multiplicative identity for $\left({T^S, \oplus, \otimes}\right)$.

• From Induced Structure Commutative, if $\left({T, \oplus, \otimes}\right)$ is commutative, then so is $\left({T^S, \oplus, \otimes}\right)$.

This proves the result.

$\blacksquare$