Does an ideal necessarily obtain the distributive property of its ring?
Forgive me if the answer is obvious, I am new to ring theory. Also, any recommendation of a ring theory text would be appreciated.
Edit: I am wondering if an ideal is in some cases a ring, analogous to a normal subgroup being a group. Also, it appears to me that an ideal could be likened to a vector space, with the closed under scalar multiplication property being similar to an Ideal being closed under super-multiplication.
While both ideals and subrings are additive normal subgroups, it is Ideals that behave well in Rings similar to normal subgroups in Groups: every ideal $I$ is the kernel of the canonical ring homomorphism $R \rightarrow R/I$, something that subrings fails to do.
For a ring $R$ with a unit, ideals are not subrings except in the trivial case. For a subring must include the identity, and with the absoption property of ideals, $r.1 = r \in I$ for all $r \in R$, i.e. $I = R$.
This motivates the notion of R-modules where both rings and ideals belong to the same category. The interaction between an ideal and a ring can be seen through an action analogous to/generalizes scalar multiplication of vector spaces.
With this interpretation, the set of group endomorphisms of a normal subgroup $I$ becomes a ring, the ring $R$ acts on it with a ring homomorphism $\rho: R \rightarrow \text{End}_\text{AB}(I)$ giving $I$ its absorption property $\rho_r(a)= r a \in I$ for all $r \in R, a \in I$.
Thus, $I$ attains the distributive property in two ways: From the ring of endomorphisms of $I$, each element $\rho_r$ being an additive group homomorphism, that is $\rho_r(m + n) = \rho_r(m) + \rho_r(n)$, i.e. $r (m+n) = rm + rn$ for $r \in R$, $m, n \in I$; and from the action being a homomorphism so that $\rho_{r+s}(m) = \rho_r(m) + \rho_s(m) = rm + sm$.