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

Pinter's Abstract Algebra book on page 187-188:

Let $\displaystyle A$ be a ring, and $\displaystyle J$ an ideal of $\displaystyle A$. For any element $\displaystyle a \in A$, the symbol $\displaystyle J + a$ denotes the set of all sums $\displaystyle j + a$, as $\displaystyle a$ remains fixed and $\displaystyle j$ ranges over $\displaystyle J$. That is,

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

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

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

$\displaystyle (J + a) + (J + b) = J + (a + b)$

And

$\displaystyle (J + a)(J + b) = J + ab$

**My question:**

How do I prove that:

$\displaystyle (J + a) + (J + b) = J + (a + b)$

And

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

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?

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

Let $\displaystyle j_1, j_2 \in J$.

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

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

$\displaystyle j_1 + j_2 \in J$

And so:

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

And for the second proof:

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

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

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

And:

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

Q.E.D