Prove that $A \subseteq B$ and $B$ is an ideal of $R$.

Let $R$ be a commutative ring with unity, let $A$ be an ideal of $R$ and let $a \in R$. Define $B = \{b \in R ~|~ ab \in A \}$.
Prove that $A \subseteq B$ (where $A$ does not have to be the proper subset of $B$) and $B$ is an ideal of $R$.

$\textbf{My Attempt:}$
Prove of $A \subseteq B$:
Since, $A$ is an ideal of $R$.
Then, by definition of ideal, $A$ is a subring $R$ and $A \subseteq R$.
Also, by definition of ideal, $A$ is an ideal of $R$, if for all $r \in R$ and for all $x \in A$ both $rx$ and $xr$ are in $A$.
Since, $R$ is a commutative ring with unity. Then, $A$ is a commutative ring with unity as well.
So, $rx = xr$ is in $A$.
Since, $A \subseteq R$ then by def of subset, for all $x \in A$ $\implies$ $x \in R$.
So, for all $r \in R$ and for all $x \in R$ such that $rx = xr$ is in $A$.
Since, we defined $B = \{ b \in R ~|~ ab \in A \}$. So, $rx = xr$ is also in $B$.
So, this proves that elements that in $A$ is also in $B$. Which means that $A \subseteq B$.

Prove of $B$ is an ideal of $R$: Let $y \in B$ and since we let $a \in R$
Since, we define $B = \{ b \in R ~|~ ab \in A \}$. So, $y \in R$ and $ay \in A$.
Since, $A \subseteq R$. So, $ay \in R$.
So, we have that $B \subseteq R$.
Then, we need to apply the ideal test.
ideal test case (1):
Let $u,v \in B$. WTP: $u - v \in B$.
Since, $u, v \in B$. Then, $u, v \in R$ and $au \in A$ and $av \in A$.
Since, $R$ is a commutative ring with unity.
Then, by def of commutative ring, $u - v \in R$ as well.
Since, $A$ is an ideal of $R$. Then, by definition of ideal, $A$ is a subring $R$.
Which means $A$ is a commutative ring with unity as well.
Since, $au \in A$ and $av \in A$.
So, $au - av = a(u - v)$ which $a(u - v) \in A$.
Since, we prove that if $u, v \in B$ both $u - v \in R$ and $a(u - v) \in A$.
So, $u - v \in B$ which means ideal test case (1) passed.
ideal test case (2):
Let $\beta \in B$ and $\gamma \in R$. WTP: both $\gamma\beta$ and $\beta\gamma$ are in $B$.
Since, $\beta \in B$. Then, $\beta \in R$ and $a\beta \in A$.
Since, both $R$ and $A$ are commutative ring with unity. So, $\gamma \beta = \beta \gamma \in R$.
Since, $A$ is an ideal of $R$ and $\gamma \in R$ and $a\beta \in A$.
So, $a\beta\gamma = a\gamma\beta \in A$.
So, we proved that both $\gamma\beta$ and $\beta\gamma$ are in $B$. Which means ideal test case (2) passed.
Therefore, we proved that $B$ is an ideal of $R$.

$\textbf{Questions:}$
I don't feel my prove of both $A \subseteq B$ (where $A$ does not have to be the proper subset of $B$) and $B$ is an ideal of $R$ are quite right. Are there any other ways to prove better than mine ? Or are there anyting I can fixes for my prove to make it better ?

2 Answers

It’s correct, but it’s much wordier than necessary. For instance, the proof that $A\subseteq B$ could be reduced to this:

Let $x\in A$. $A$ is an ideal, so $ax\in A$, and that by definition means that $x\in B$. Thus, $A\subseteq B$.

In fact it can even be reduced to this:

$aA\subseteq RA=A$, so $A\subseteq B$.

Here $RA=\{rx:r\in R\text{ and }x\in A\}$. $RA\subseteq A$ because $A$ is an ideal, and $RA=A$ because $R$ is a ring with unity.

The proof that $B$ is an ideal can also be drastically shortened.

Let $b_0,b_1\in B$. Then $a(b_0-b_1)=ab_0-ab_1$, where $ab_0,ab_1\in A$. $A$ is a subring of $R$, so $ab_0-ab_1\in A$, and therefore $b_0-b_1\in B$, i.e., $B$ is closed under subtraction.

Now let $b\in B$ and $r\in R$. Then $a(br)=(ab)r$, where $ab\in A$, and $A$ is an ideal of $R$, so $(ab)r\in A$ as well. Thus, $br\in B$, and since $R$ is commutative, $rb$ is also in $B$, and we’ve shown that $B$ is an ideal of $R$.

Indeed, if I were writing it up for myself, I’d shorten it even further:

If $b_0,b_1\in B$, then $a(b_0-b_1)=ab_0-ab_1\in A-A=A$, so $b_0-b_1\in B$. If $b\in B$ and $r\in R$, then $a(br)=(ab)r\in Ar\subseteq A$, so $rb=br\in B$. Thus, $B$ is an ideal.

When you’re just learning, it’s generally better to err on the side of giving more detail than necessary, but it’s also a good idea to work on cutting out parts that really can safely be left to the reader: including too much detail makes it harder to read. With practice one learns to strike an appropriate balance for a given audience.

Your proves look quite complicated. I would do something like that.

As $R$ is supposed to be commutative, it is enough for proving that $I \subseteq R$ is an ideal to prove that $I$ is a left ideal.

Take $x \in A$. As $A$ is a (left) ideal, we have $ax \in A$. Which implies $x \in B$ by definition of $B$. This proves that $A \subseteq B$.

Let's now prove that $B$ is a left ideal of $R$.

$b,b^\prime \in B$ mean that $ab, ab^\prime \in A$ and therefore $$ab + ab^\prime = a(b + b^\prime) \in A$$ as $A$ is an ideal. And bydefinition of $B$, $b+b^\prime$ belongs to $B$.

Also if $b \in B$ and $x \in R$, we have $ab \in A$ (by definition of $B$) and $rb \in B$ as

$$a(rb)= r(ab) \in A$$ because $R$ is supposed to be commutative and $A$ to be an ideal.

This concludes the proves of the requested questions.