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

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

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 $\displaystyle a,b \in R$ with $\displaystyle b \ne 0$

... ... "

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 $\displaystyle a \in (b)$ if and only if $\displaystyle (a) \subseteq (b)$."

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

Can someone please confirm or clarify this for me?

Peter

2. ## Re: Principal Ideals - Need for a ring with identity or a unity

Originally Posted by Bernhard
My problem is this - I think D&F should have defined R as a commutative ring with identity since proving that $\displaystyle (a) \subseteq (b) \longrightarrow a \in (b)$ requires the ring to have an (multiplicative) identity or unity.

Peter
I think you're thinking to hard about this: $\displaystyle a\in \langle a \rangle \subseteq \langle b\rangle \implies a\in \langle b \rangle$.

3. ## Re: Principal Ideals - Need for a ring with identity or a unity

Originally Posted by Gusbob
I think you're thinking to hard about this: $\displaystyle a\in \langle a \rangle \subseteq \langle b\rangle \implies a\in \langle b \rangle$.
Hi Gusbob,

Yes you may be right .... but relevant to my thoughts is the following:

Previously I was working on a problem from Hungerford that I believe indicates strongly that $\displaystyle a \notin (a)$ where $\displaystyle (a) = \{ ra \ | \ r \in R \}$ and R is a commutative ring without multiplicative identity

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

4. ## Re: Principal Ideals - Need for a ring with identity or a unity

I see where your confusion came from.

Firstly, I'd like to point out that there are some conflicting notions of rings between authors. Some assert that a ring has to contain a multiplicative identity, others don't. The characterisation $\displaystyle \langle a \rangle=\{ra|r\in R\}$ comes from the former case: that is, all rings are unital. It is not true otherwise. In fact that is the reason Hangerford added the '$\displaystyle +na$' part of the definition of a principal ideal, as you have observed yourself. This should resolve the apparent contradiction with the exercise in Hungerford.

Secondly, if you look at the first definition of 'properties of ideals' (p251 ed3) presented in Dummit and Foote, it says that:

Let $\displaystyle \langle A \rangle$ denote the smallest ideal of $\displaystyle R$ containing $\displaystyle A$, called the ideal generated by $\displaystyle A$.

An ideal generated by a single element is called a principal ideal
Taking $\displaystyle A=\{a\}$, this says that $\displaystyle a$ always has to be an element of $\displaystyle \langle a \rangle$.

5. ## Re: Principal Ideals - Need for a ring with identity or a unity

Yes, thanks

I believe you have summarized the situation very accurately ... Most helpful ...

Peter

6. ## Re: Principal Ideals - Need for a ring with identity or a unity

Just a final note: for the definitions and discussion on page 251 D&F Section 7.4 we have (at the top of the section:

"Throughout this section R is a ring with identity 1 (not equql to 0)"

So I take that material to refer to unital rings ... And Hungerford's formulation referred to in the post above to refer to rings without a unity .. And so, if you understand things this way then a always belongs to the principal idea (a) ...

However, I am possibly being too pedantic

Peter

7. ## Re: Principal Ideals - Need for a ring with identity or a unity

Ah yes I see, you're completely right. It's good to get all your definitions sorted out when working through multiple books. From Hungerford's point of view (non-unital commutative rings), you define the ideal generated by a subset $\displaystyle A$ to be the intersection of all (two sided) ideals of $\displaystyle R$ containing $\displaystyle A$. The point being, $\displaystyle a\in \langle A \rangle$ if $\displaystyle a \in A$ is an inherent fact in either formulation of ideals, and not just for principal ideals.