Posted by Jason Polak on 16. August 2012 · 4 comments · Categories: commutative-algebra · Tags: , ,

Let us ponder a familiar property of any integral domain $ R$. Suppose that $ a,b\in R$ are such that $ (a) = (b)$, where $ (a)$ denotes the ideal generated by $ a$. Then $ a = rb$ and $ b = sa$ for some elements $ r,s\in R$. Combining these two facts together give $ a = rsa$, or $ (1 – rs)a = 0$. If $ a\not = 0$, this means that $ 1 – rs = 0$ or that $ r$ and $ s$ are units. Thus $ a$ is a unit multiple of $ b$. Of course if $ a = 0$ then $ b= 0$ and $ a$ is still a unit multiple of $ b$.

Now, the question is, what happens if we drop the assumption that $ R$ is an integral domain? If $ (a) = (b)$, can we still prove that $ a$ is a unit multiple of $ b$? Let us still assume that $ R$ is commutative. Of course, the above proof for integral domains can’t work because we used the defining property of integral domains to conclude that $ 1 – rs = 0$. In this post we shall see a counterexample to show that two generators of the same principal ideal do not necessarily have to be unit multiples of each other.

I first heard of this question in Hershy Kisilevsky‘s algebraic number theory class. I forgot about it for some time until I came across Question 101989 on MathOverflow that gave a reference for a paper that was purported to have a counterexample showing that one cannot prove this statement for arbitrary commutative rings.

The paper was D.D. Anderson’s and Silvia Valdes-Leon’s “Factorization in Commutative Rings with Zero Divisors“. The title of the paper acted as a magnet since I am rather found of curious algebraic constructions that push the boundaries of sanity. So I shall present their wonderful example and urge any readers to take a glance at the paper, being available at Project Euclid under open-access. The paper of course gives much more than this example, and is definitely a good place to start for anyone interested in factorization in commutative rings with zero divisors.

In fact, the authors present various definitions of “associate”, which an an integral domain are all the same. In many commutative rings that are not integral domains, these different definitions of associate in fact are different, and lead to different theories of factorization. The paper is written in enough detail so that it can be understood by anyone with a knowledge of elementary algebra, such as a typical first course in abstract algebra.

Show Me The Example!

We start with the commutative ring $ R = k[x,y,z]/(x-xyz)$ where $ k$ is a field (or even just an integral domain) and use $ [.]$ notation to denote elements of the quotient.

Observe that $ ([x]) = ([xyz])$. Now the string of inclusions $ ([xy])\subseteq ([x])\subseteq ([xyz])\subseteq ([xy])$ show that $ ([x]) = ([xy])$. Thus we have an equality of two nonzero ideals. Suppose now that $ [f][x] = [xy]$ for some $ f\in k[x,y,z]$. We wish to show that $ [f]$ is not a unit in $ R$.

To this end, assume for the sake of contradiction that $[f]$ is a unit. Then $ fx – xy \in (x-xyz)$ in $k[x,y,z]$. Thus, there is an $ h\in k[x,y,z]$ such that $ (f-y)x – h(1-yz)x = 0$. Now $ k[x,y,z]$ is an integral domain (!), so $ f-y = h(1-yz)$ or that $ f = y + h(1-yz)$.

Now, $[f]$ being a unit means that the ideal $(f,x-xyz)$ is the entire ring $ k[x,y,z]$. And $(f,x-xyz) = k[x,y,z]$ implies $(f,x) = k[x,y,z]$ so that if we can show that $ (f,x) = (y+h(1-yz),x) \not= k[x,y,z]$, then we will have found our contradiction.

So, consider the surjective ring homomorphism that is the $ k$-algebra homomorphism $ k[x,y,z]\to k[t]$ given by $ x\mapsto 0,y\mapsto t,z\mapsto t$. The surjective image of an ideal is an ideal and the image of an ideal is generated by the image of its generators, so to say that $ (y+h(1-yz),x)$ is $ k[x,y,z]$ would imply that $ (t + h(1-t^2)) = k[t]$, which is impossible, since no matter what $ h$ is, the polynomial $ t + h(1-t^2)$ is not constant.

Conclusion

This very nice example of a commutative ring in two generators of a principal ideal don’t necessarily have to be unit-associate shows that zero divisors are tricky, and that quotients and other constructions from polynomial rings are a handy source of counterexamples for that innocuous seemingly plausible statement. Even in commutative ring theory, trust nothing!

4 Comments

  1. Hi Jason… what about in a Boolean ring, i.e., where x^2 = x for every element x? These can have zero divisors. It’s easy to prove Boolean rings are principal ideal rings and that they only have one unit… is it easy to prove the generator of each ideal is unique? (Or isn’t it true?) Put another way, if a = sb and b = ta, does a = b?

    • Hello! One could use a version of the Stone representation theorem (which in general depends upon Zorn’s lemma): every Boolean ring is actually isomorphic to a subring of P(X) for some set X, where P(X) is the power set. It has the operations of addition being $latex A + B = (A\cup B) – (A\cap B)$ and multiplication being $latex AB = A\cap B$.

      If one uses this embedding, we see that the generator of a principal ideal must be unique, for if $latex A$ and $latex B$ are elements in $latex R$ generating the same principal ideal in a subring of P(X), then for some subset $latex C$, we have $latex C\cap A = B$ so $latex B\subseteq A$. Similarly, $latex A\subseteq B$ so actually $latex A = B$.

  2. I think in your third-to-last paragraph (second-to-last of the example) you want “Now, if $(f,x?xyz)=k[x,y,z]$ then $(f,x)=k[x,y,z]$, so that if we can show that …”.

Leave a Reply to Jason Polak Cancel reply

Your email address will not be published. Required fields are marked *