# Principal Ideals - Need for a ring with identity or a unity

#### Peter

##### Well-known member
MHB Site Helper
On page 273 of Dummit and Foote the last sentence reads: (see attachment - page 273)

"The notion of the greatest common divisor of two elements (if it exists) can be made precise in general rings." (my emphasis)

Then, the first sentence on page 274 reads as follows: (see attachment - page 274)

"Definition. Let R be a commutative ring and let [TEX] a,b \in R [/TEX] with [TEX] b \ne 0 [/TEX]

... ... "

In this definition D&F go on to define multiple, divisor and greatest common divisor in a commutative ring.

D&F then write the following:

"Note that b|a in a ring if and only if [TEX] a \in (b) [/TEX] if and only if [TEX] (a) \subseteq (b) [/TEX]."

My problem is this - I think D&F should have defined R as a commutative ring with identity since proving that [TEX] (a) \subseteq (b) \longrightarrow a \in (b) [/TEX] requires the ring to have an (multiplicative) identity or unity.

Can someone please confirm or clarify this for me?

Peter

#### Siron

##### Active member
Can you give the proof?

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Can you give the proof?
Hello Siron,
Of what do you demand a proof?

#### Peter

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MHB Site Helper
I imagined the proof went as follows:

$$\displaystyle (a) \subseteq (b) \longrightarrow ra \in (b)$$ for all $$\displaystyle r \in R$$

$$\displaystyle \longrightarrow$$ in particular that choosing r = 1 that $$\displaystyle 1.a \in (b)$$ i.e. $$\displaystyle a \in (b)$$

I did look for another proof that did not involve assuming $$\displaystyle 1 \in R$$ but could not find one.

Peter

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I imagined the proof went as follows:

$$\displaystyle (a) \subseteq (b) \longrightarrow ra \in (b)$$ for all $$\displaystyle r \in R$$

$$\displaystyle \longrightarrow$$ in particular that choosing r = 1 that $$\displaystyle 1.a \in (b)$$ i.e. $$\displaystyle a \in (b)$$

I did look for another proof that did not involve assuming $$\displaystyle 1 \in R$$ but could not find one.

Peter
Interesting! It depends on how an ideal generated by a set is defined exactly.
Does $(a)$ contain $a$ or not?

From wiki:
Let R be a (possibly not unital) ring. Any intersection of any nonempty family of left ideals of R is again a left ideal of R. If X is any subset of R, then the intersection of all left ideals of R containing X is a left ideal I of R containing X, and is clearly the smallest left ideal to do so. This ideal I is said to be the left ideal generated by X. Similar definitions can be created by using right ideals or two-sided ideals in place of left ideals.​

I draw the implication that the generated ideal (a) does contain a, even if there is no element r in the ring such that ra=a.

Since $a \in (a)$ and $(a) \subseteq (b)$, it follows from the definition of a subset that $a \in (b)$.

#### Peter

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MHB Site Helper
Seems that the question still remains - does $$\displaystyle a \in (a)$$?

Are there any other clarifying thoughts?

I must say I am inclined to the idea that on page 274 D&F should have specified the ring R as a commutative ring with identity $$\displaystyle 1 \ne 0$$

Peter

#### caffeinemachine

##### Well-known member
MHB Math Scholar
I draw the implication that the generated ideal (a) does contain a, even if there is no element r in the ring such that ra=a.
Hello ILikeSerena,

Let $R$ be a commutative ring and $a\in R$.
Then $(a)=\{ra:r\in R\}$.

Take $R=2\mathbb Z$. Then the ideal $(2)$ of $R$ doesn't have $2$ in it.
So I don't think that in general $a\in (a)$.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Seems that the question still remains - does $$\displaystyle a \in (a)$$?

Are there any other clarifying thoughts?

I must say I am inclined to the idea that on page 274 D&F should have specified the ring R as a commutative ring with identity $$\displaystyle 1 \ne 0$$

Peter
The wiki section seems to state clearly that an ideal generated from a set X, is the smallest ideal that contains X.

Let's consider the commutative ring $2\mathbb Z = \{0, \pm 2, \pm 4, ...\}$.
When we look at the generated ideal (2), do we mean the multiples of 2, or the multiples of 4?

Googling a bit I found here:
Definition
Let (R,+,∘) be a ring.
Let S⊆R.
The ideal generated by S is the smallest ideal of R containing S.​

Here:
Defi nition: The ideal generated by a, denoted (a), is defi ned to be the smallest ideal containing
the element a.​

I do not have Dummit and Foote.
Can you quote how they define an ideal generated by a set please?

- - - Updated - - -

Let $R$ be a commutative ring and $a\in R$.
Then $(a)=\{ra:r\in R\}$.
Currently I believe this is only true for commutative rings with unity and not in general for rings.
In particular what you mention is for a left ideal, while I believe $(a)$ is intended for 2-sided ideals.

#### Peter

##### Well-known member
MHB Site Helper
Just tried to upload 2 pages from D&F on ideal generated by a set but had a problem - as follows:

I have two uploads that I have used previously but I cannot delete them. When I try to make a new upload I cannot because the previous old and used files nearly make up my limit - but I cannot delete them - this is a real problem. How can I delete them.

(Thus I have a real problem with the way the MHB deals with uploads. )

Will try to find a way around this problem

Peter

#### caffeinemachine

##### Well-known member
MHB Math Scholar
- - - Updated - - -

Currently I believe this is only true for commutative rings with unity and not in general for rings.
In particular what you mention is for a left ideal, while I believe $(a)$ is intended for 2-sided ideals.
I used a commutative ring. For a commutative ring I think the left ideals, right ideals and two-sided ideals are the same things. Isn't it?
Keeping this in mind don't we have good reason to believe that $a\in (a)$ doesn't hold in general, where $(a)$ is the 2-sided ideal generated by $a$.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I used a commutative ring. For a commutative ring I think the left ideals, right ideals and two-sided ideals are the same things. Isn't it?
Yes.

Keeping this in mind don't we have good reason to believe that $a\in (a)$ doesn't hold in general, where $(a)$ is the 2-sided ideal generated by $a$.
Not as I see it.
As far as I know, ideals have been invented to generalize properties of numbers.
So we can generalize saying 2 divides 4, by saying (2) divides (4), which actually means that (4) is a subset of (2).
This only makes sense if (2) is really the set of multiples of 2 in $2\mathbb Z$ and (4) the multiples of 4.

More to the point, in the wiki section your definition is given as well, but it is preceded by "If R has unity, then ...".

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#### Peter

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MHB Site Helper

Previously I was working on a problem from Hungerford that I believe indicates strongly that $$\displaystyle a \notin (a)$$.

The problem is from Hungerford - Abstract Algebra: An Introduction - Section 6.1 Ideals and Congruences, Exercise 31 on page 143:

"Let R be a commutative ring without identity and let $$\displaystyle a \in R$$.

Show that $$\displaystyle A = \{ ra + na \ | \ r \in R, n \in \mathbb{Z} \}$$ is an ideal containing a and that every ideal containing a also contains A. A is called the principal ideal generated by a."

I know this does not go as far as to assert that $$\displaystyle I = \{ ra \ | \ r \in R \}$$ does not contain a, but given the special construction involved in A above, it seems this is likely to be the case. The statement of the exercise seems to indicate that when R does not have an identity we have to add in the elements a, a + a, a + a + a, etc to ensure that A is an ideal containing a. (would it even be an ideal without this? Maybe it would?)

Mind you I would really like to see a proof that $$\displaystyle a \notin I$$ when R is a commutative ring without identity.

Peter

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#### Klaas van Aarsen

##### MHB Seeker
Staff member

Previously I was working on a problem from Hungerford that I believe indicates strongly that $$\displaystyle a \notin (a)$$.

The problem is from Hungerford - Abstract Algebra: An Introduction - Section 6.1 Ideals and Congruences, Exercise 31 on page 143:

"Let R be a commutative ring without identity and let $$\displaystyle a \in R$$.

Show that $$\displaystyle A = \{ ra + na \ | \ r \in R, n \in \mathbb{Z} \}$$ is an ideal containing a and that every ideal containing a also contains A. A is called the principal ideal generated by a."

I know this does not go as far as to assert that $$\displaystyle I = \{ ra \ | \ r \in R \}$$ does not contain a, but given the special construction involved in A above, it seems this is likely to be the case.

Mind you I would really like to see a proof that $$\displaystyle a \notin I$$ when R is a commutaitve ring without identity.

Peter
It's the other way around.
We have $a \in A$ and furthermore "A is called the principal ideal generated by a", so the principal ideal does contain a.
This confirms that $a \in (a)$.

Also note that A is defined slightly different from $\{ ra \ | \ r \in R \}$ to allow for $a$ being in there, even though R has no unity.

#### Peter

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MHB Site Helper
Hi ILikeSerena,

$$\displaystyle a \in A$$, certainly, but then A is especially constructed to contain a.

I think that the issue is that $$\displaystyle a \notin I$$ (I is just the standard definition of a left ideal - which is same as right and two sided ideals since R is commutative.

What do you think?

Peter

#### Peter

##### Well-known member
MHB Site Helper
Hi ILikeSerena,

It is very likely we are both actaully saying the same thing - that is when R has no identity the definition of a principal ideal has to be adjusted in the way Hungerford indicates.

Thanks so much for your help in this matter!

Peter