Let $Ideal$ be the category whose objects are pairs $(R,I)$ consisting of a commutative ring $R$ and an ideal $I$ of $R$, and whose morphisms $(R,I) \to (S,J)$ are ring homomorphisms $f:R \to S$ for which $f(I) \subseteq J$.
Also, let $Module$ be the category whose objects are pairs $(R,M)$ consisting of a commutative ring $R$ and an $R$-module $M$, and whose morphisms $(R,M) \to (S,N)$ are pairs $(f_{ring}, f_{mod})$ consisting of a ring homomorphism $f_{ring}:R \to S$ and a homomorphism of additive groups $f_{mod}:M \to N$ for which $f_{mod}(rm)=f_{ring}(r)f_{mod}(m)$ for all $r \in R$ and $m \in M$.
Then, one can define an obvious functor $Ideal \to Module$ by observing that any ideal of a commutative ring $R$ is also an $R$-module.
My question then is: Does this functor $Ideal \to Module$ have a left adjoint, and if so, how is the left adjoint constructed?.
Given a commutative ring $R$ and an $R$-module $M$, one can consider the symmetric algebra $Sym(M)$, which is the free unital commutative associative $R$-algebra on $M$.
There is then an obvious unital $R$-algebra homomorphism $f:Sym(M) \to R$ that sends every element of $M$ to zero.
So, I think that there is a functor $Module \to Ideal$ that sends $(R,M)$ to $(Sym(M), Ker(f))$. Note that $Ker(f)$ is just the free non-unital commutative associative $R$-algebra on $M$.
But is this functor $Module \to Ideal$ really the left adjoint?