Prove that: $B/A \triangleleft R/A$ if $A \subseteq B ;\ \ A, B \triangleleft R $(ring)

Question

Prove that: $B/A \triangleleft R/A$ if $A \subseteq B ;\ \ A, B \triangleleft R $(ring)

: $ \triangleleft $ means ideal. I need this proof to continue on, I'm told it's not that hard, but I just can;t seem to do this.

Answer 1

To show that $B/A$ is an ideal in $R/A$, or $B/A \triangleleft R/A$ as our OP Bozidar Vulicevic writes it, is really not at all difficult; we merely need to follow through the details inherent in the definitions. A modular approach is indeed possible, as our colleague Bernard has shown, but here I take a somewhat more first-principles, straight-from-the-definitions line.

That $B/A \triangleleft R/A$ is perhaps easiest to see if we recall that for any ring $\mathcal R$ and any subset $\mathcal J \subset \mathcal R$, we have $\mathcal J \triangleleft \mathcal R$ (that is, $\mathcal J$ is a two-sided ideal in $\mathcal R$) if and only if the following two assertions bind:

I1.) $\mathcal j_1 - \mathcal j_2 \in \mathcal J$ for any $\mathcal j_1, \mathcal j_2 \in \mathcal J$;

I2.) $\mathcal r \mathcal j, \mathcal j \mathcal r \in \mathcal J$ for any $\mathcal j \in \mathcal J$ and any $\mathcal r \in \mathcal R$.

(I1) and (I2) are in point of fact simply a standard formulation of the definition of an ideal; (I1) is indeed merely a formulation of the well-known criterion that a subset $H$ of a group $G$ is a subgroup provided that $h, k \in H$ implies $hk^{-1} \in H$ for all $h, k \in H$; in (I1) we see this principle cast in the customary notation for the abelian additive subgroup $\mathcal J$ of $\mathcal R$. (I2) is of course an essential aspect of any definition of ideal.

We apply (I1), (I2) to the present case. For cosets $b_1 + A, b_2 + A \in R/A$, where $b_1, b_2 \in B$, we evidently have, in $R/A$,

$(b_1 + A) - (b_2 + A) = (b_1 - b_2) + A \in B/A, \tag{1}$

since $b_1, b_2 \in B$. And for cosets $b + A, r + A \in R/A$ with $b \in B$ (i.e. $b + A \in B/A$), we have

$(r + A)(b + A) = rb + A \in B/A, \tag{2}$

since $rb \in B$; $(b + A)(r + A) \in B/A$ is similarly shown. (1) and (2) show that $B/A$ satisfies (I1) and (I2) in $R/A$, hence it is an ideal in same, or, in OP Bozidar's preferred notation,

$B/A \triangleleft R/A. \tag{3}$

QED.

Answer 2

This supposes $A$ is a two-sided ideal in $R$.

$B/A$ and $R/A$ are $R$-modules, $B/A$ is an $R$-submodule of $R/A$, and the inclusion of $B/A$ in $R/A$ is an $R$-homomorphism, since by definition, for all $x\in R$, $\,x(b+A)=xb+A$.

Now, $B/A$ is actually an $R/A$-module since, for any $b\in B$, $x,x'\in R$ such that $x+A=x'+A$, we have $\,x(b+A)=x'(b+A)$, as $\,xb-x'b=(x-x')b\in A$. It is clear that the inclusion morphism is actually $R/A$-linear.