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Principal ideal in a ring without identity

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
(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
Mar 5, 2012
8,868
(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?
 
Last edited:

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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
Mar 5, 2012
8,868
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
Jun 22, 2012
2,918
Thanks again for your help

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
Mar 5, 2012
8,868
Thanks again for your help

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.

We start with a.
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. :eek:
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Thanks ... That post made things very clear!!!

Most helpful indeed

Peter