Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I found this link. It says that we can define a Noetherian ring $R$ that need not have unity element.

Note that, in my book, a Noetherian ring is a commutative ring $R$ with the property that every ideal in $R$ is finitely generated. How could we define the concept "Ideal is generated by a finite set" on a ring which don't have unity? (The definition of "Ideal is generated by a set" in Roman's Advanced Linear Algebra require that $R$ is a ring with unity element.)

My question is how can we define a Noetherian ring $R$ that need not have unity element?

Thanks a lot.

share|improve this question
add comment

2 Answers 2

A Noetherian ring, as you probably know, is a ring which satisfies the ascending chain condition. That is, every chain of ideals with $I_n \subset I_{n+1}$ has some $m$ above which all $I_n = I_{n+1}$ for $n > m$

This has nothing to do with the existence of a neutral element, and can be satisfied by rings without unity.

share|improve this answer
    
But I read Rotman's Advanced Modern algebra, the definition of Noetherian Ring: A commutative ring R is called noetherian if every ideal in R is finitely generated. Tou can check it here: books.google.co.id/… –  user59705 Jan 26 '13 at 4:37
    
@user59705 those two facts are equivalent for rings with identity. Here is the wikipedia page for Noetherian rings. And I'd imagine this alternate characterization is covered in Rotman's book. –  Sam DeHority Jan 26 '13 at 4:45
    
Looking at proposition 5.33 in your book, you see that the two characterizations are equivalent. –  Sam DeHority Jan 26 '13 at 4:55
    
But the definition of ring in Rotman's also require unity element –  user59705 Jan 26 '13 at 4:59
    
The same definitions hold without unity. You can have an ideal being generated by a finite set without there being a unity element. As a matter of fact, unless that ideal is R, it won't even contain the unit element. While the definition in your book uses unity, the properties we are discussing don't require it. –  Sam DeHority Jan 26 '13 at 5:02
add comment

When $R$ is a rng, then the ideal $\langle X \rangle$ generated by a subset $X \subseteq R$ is defined as the intersection of all ideals of $R$ containing $X$. In other words, it is the smallest ideal of $R$ (with respect to inclusion) that contains $X$. Explicitly, it is the $\mathbb{Z}$-span of $X \cup R \cdot X \cup X \cdot R \cup R \cdot X \cdot R$. Thus, elements are finite sums of elements of the form $x$, $r \cdot x$, $x \cdot r$, and $r \cdot x \cdot s$, where $r,s \in R$ and $x \in X$. Thus, the element structure gets a little bit more complicated as for rings, where we only need the $\mathbb{Z}$-span of $R \cdot X \cdot R$ (contrary to what has been claimed in the comments so far), only the abstract definition stays the same.

A rng is called noetherian when every ascending chain of ideals is stationary. By the usual proof, this is equivalent to the condition that every ideal is finitely generated. You can use the abstract definition which I have given above, you won't need elements.

The left ideal generated by $X$ is the smallest left ideal containing $X$, and it is the $\mathbb{Z}$-span of $X \cup R \cdot X$. Again a rng is called left noetherian when the ascending chain condition holds for left ideals, and this is satisfied if and only if every left ideal is finitely generated. By duality, the same statements hold with "left" replaced by "right".

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.