# Principal ideal in a ring without identity

#### Peter

##### Well-known member
MHB Site Helper
(Hungerford exercise 31, page 143)

Let R be a commutative ring without identity and let [TEX] a \in R [/TEX]

Show that [TEX] A = \{ ra + na \ | \ r \in R, n \in \mathbb{Z} \} [/TEX] is an ideal containing a and that every ideal containing a also contains A. (A is called the prinicipal ideal generated by a)

#### Klaas van Aarsen

##### MHB Seeker
Staff member
(Hungerford exercise 31, page 143)

Let R be a commutative ring without identity and let [TEX] a \in R [/TEX]

Show that [TEX] A = \{ ra + na \ | \ r \in R, n \in \mathbb{Z} \} [/TEX] is an ideal containing a and that every ideal containing a also contains A. (A is called the prinicipal ideal generated by a)
Hi Peter!

It would be helpful if you showed something about how far you got and/or what is puzzling you.

Anyway, you need to prove:
1. $(A,+)$ is a subgroup of $(R,+)$
2. $A$ absorbs multiplication from both sides
3. $a \in A$
4. If $I$ is an ideal with $a \in I$, then $A \subseteq I$

To prove a subgroup, it suffices that A is non-empty and that for any 2 elements x and y in A holds: $x-y \in A$.
A is non-empty, since $(0a+0a) \in A$.
When we pick arbitrary $x=r_1 a + n_1 a$ and $y=r_2 a + n_2 a$, it follows that $x-y=(r_1 - r_2)a + (n_1-n_2)a \in A$.
Therefore $(A,+)$ is a subgroup of $(R,+)$.

How far can you get with 2?

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#### Peter

##### Well-known member
MHB Site Helper
Thanks for the help so far - I just lacked the confidence to start this problem - my issue was the nature of na where $$\displaystyle n \in \mathbb{Z}$$ and not in R

But thinking about it na probably equals a + a + a .... (n times in all) which is in R since R includes an additive group in its structure.

But perhaps you can clarify your use of $$\displaystyle 0_R a + 0_{ \mathbb{Z} } a$$

I am uneasy about this ... is $$\displaystyle 0_R$$ the same as $$\displaystyle 0_{ \mathbb{Z} }$$ and does it matter? But how do we proceed with a sum with different elements in it - perhaps it does not matter since the sum is clearly in a set that contains ra + na and we do not have to add them???

Can you clarify?

Peter

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Thanks for the help so far - I just lacked the confidence to start this problem - my issue was the nature of na where $$\displaystyle n \in \mathbb{Z}$$ and not in R

But thinking about it na probably equals a + a + a .... (n times in all) which is in R since R includes an additive group in its structure.

But perhaps you can clarify your use of $$\displaystyle 0_R a + 0_{ \mathbb{Z} } a$$

I am uneasy about this ... is $$\displaystyle 0_R$$ the same as $$\displaystyle 0_{ \mathbb{Z} }$$ and does it matter? But how do we proceed with a sum with different elements in it - perhaps it does not matter since the sum is clearly in a set that contains ra + na and we do not have to add them???

Can you clarify?

Peter
Good!
You have it totally right.

The symbol $0_{ \mathbb{Z} }$ and any number in $\mathbb{Z}$ has to be carefully distinguished from elements in R, since it is not given that they are elements of R.
The expression $na$ is merely a shorthand for repeatedly adding $a$ as you already surmised.
Keeping that in mind, and expanding it when in doubt, you can address the problem.

So when I wrote $0 a + 0a$, I meant $0_R \cdot a$ with $a$ added zero times.
The result is $0_R$ (did you know that $0_R \cdot a = 0_R$?), which is therefore an element of $A$.

#### Peter

##### Well-known member
MHB Site Helper

Yes, I was aware that $$\displaystyle 0_R \cdot a = 0_R$$ but was worried about $$\displaystyle 0_{ \mathbb{Z} } \cdot a$$ which I took to result in $$\displaystyle 0_{ \mathbb{Z} }$$ leaving us the sum $$\displaystyle 0_R +0_{ \mathbb{Z} }$$ and I did not really know how one should interpret this.

It seems that you have (qualitatively) interpeted $$\displaystyle 0_{ \mathbb{Z} } \cdot a$$ as "a added zero times" and hence dropped the term. I lacked your confidence to make what I take to be a qualitative interpretation.

What do you think about my concerns?

Peter

#### Klaas van Aarsen

##### MHB Seeker
Staff member

Yes, I was aware that $$\displaystyle 0_R \cdot a = 0_R$$ but was worried about $$\displaystyle 0_{ \mathbb{Z} } \cdot a$$ which I took to result in $$\displaystyle 0_{ \mathbb{Z} }$$ leaving us the sum $$\displaystyle 0_R +0_{ \mathbb{Z} }$$ and I did not really know how one should interpret this.

It seems that you have (qualitatively) interpeted $$\displaystyle 0_{ \mathbb{Z} } \cdot a$$ as "a added zero times" and hence dropped the term. I lacked your confidence to make what I take to be a qualitative interpretation.

What do you think about my concerns?

Peter
Consider how A is constructed.

The ring contains both multiplication and addition as operations.
So to construct an ideal, we need to multiply each element (starting with only a) by any number r, generating a set of new elements for A.
Then we add any element that is already in A (starting with a) to the known elements, which are $r\cdot a$ by now.
This generates $r\cdot a +a$ as the first new element, but then we can add $a$ again, generating $r \cdot a + a + a$.
Now we need to start to think how to write this down in a comprehensible way.
So let's introduce $r\cdot a + na$ to denote this.
And then....

Oh wait! Now I've almost answered the 4th part, proving that any ideal that contains a must also contain A.

#### Peter

##### Well-known member
MHB Site Helper
Thanks ... That post made things very clear!!!