# Thread: Are all ideals of a ring rings themselves? A question

1. ## Are all ideals of a ring rings themselves? A question

I've two proofs that I'm having difficulty in proving.

Problem:

A nonempty subset $B$ of a ring $A$ is called an ideal of $A$ if $B$ is closed with respect to addition and negatives and $B$ absorbs products in $A$.

Pinter's Abstract Algebra book on page 187-188:
Let $A$ be a ring, and $J$ an ideal of $A$. For any element $a \in A$, the symbol $J + a$ denotes the set of all sums $j + a$, as $a$ remains fixed and $j$ ranges over $J$. That is,

$J + a = \{j + a \colon j \in J \}$

$J + a$ is called a coset of $J$ in $A$.

There is a way of adding and multiplying cosets which works as follows:

$(J + a) + (J + b) = J + (a + b)$
And
$(J + a)(J + b) = J + ab$

My question:

How do I prove that:
$(J + a) + (J + b) = J + (a + b)$

And

$(J + a)(J + b) = J + ab$?

How did the author figure out that left side is equal to right side for both of these equations? Is it possible to prove both of these equations?

2. ## Re: Are all ideals of a ring rings themselves? A question

Let $j_1, j_2 \in J$.

\begin{align*}(J + a) + (J + b) =& (j_1 + a) + (j_2 + b) \\ =&(a + j_1) + (j_2 + b)\text{......[by additive commutativity of a ring]} \\ =& a + (j_1 + j_2) + b\text{.........[by associativity of addition of a ring]}\\ =& (j_1 + j_2) + a + b\text{.........[by commutativity of addition of a ring]}\end{align*}

Because J is an ideal and ideals are subring. So by definition of subring:

$j_1 + j_2 \in J$

And so:
\begin{align*}(J + a) + (J + b) =& (j_1 + j_2)+ a + b \\ =& J + a + b............\text{[by definition of coset of a+b]}\\ =& J + (a + b)\text{.........[By associativity of addition of a ring]}\end{align*}

And for the second proof:

\begin{align*}(J + a)(J + b) =& (j_1 + a)(j_2 + b) \\ =& (j_1 + a)j_2 + (j_1 + a)b \text{......[by distributive property of addition]} \\ =& j_1 j_2 + aj_2 + j_1b + ab \text{........[by distributive property of addition of a ring]}\end{align*}

$j_1 j_2 \in J\text{.............[because J is a subring and so multiplication is closed]}$

$aj_2, j_1b \in J\text{.........[because J is an ideal]}$
And:

$j_1 j_2 + aj_2 + j_1b \in J\text{.........[because } j_1j_2 + aj_2 + j_1b \in J \text{ and addition is closed in a ring]}$

So $(J + a)(J + b) = J + ab\text{........[by definition of coset of ab]}$

Q.E.D