I was talking with a friend about if the set of two sided ideals of a ring have any interesting algebraic structure. And we without too much challenge got to a semi-ring structure. However, unsatisfied, we wanted to turn it into a ring by taking the "free ring of the semi-ring".
We however didn't find any such construction. Hence my question, does the inclusion functor from the category of Rings into the category of semi-rings have an adjoint?
Thanks in advance
If you understand a semiring as an additive commutative monoids and multiplicative monoid, where the operations are distributive, the free semiring is obtained by taking the Grothendieck group of underlying additive monoid. Distributivity of operations provides then a unique extension of multiplicative structure. Denoting the additive monoid as $R_+$ and its Grothendieck group as $GR_+$, suppose $a \in R_+$ and $[a]\in GR_+$ is its image under canonical inclusion. Then we define $[a][b] := [ab]$.
To show that it is independent the choice of representants.
$$ [ab] = [a'b'] \ \ \text{iff}\ \ \exists_{c \in R_+} ab + c = a'b' + c $$ Suppose we have two representant $[a'] = [a]$. Then $$ a' + c = a + c \\ b(a' + c) = b(a + c) \\ ba' + bc = ba + bc \\ $$
So we get $[ba'] = [ba]$. Similar calculation for the left multiplication then shows that it is indeed well define, as we have $[ab] = [a'b']$ whenever $[a] = [a'], [b] = [b']$