The Department of Mathematics at IIT Bombay is running a Popular Lecture series this semester, and the first talk in the series was deliverd by Prof. Sudarshan Gurjar on 6th August. His talk was titled ‘Applications of Analysis and Topology to Commutative Algebra’. The talk was an hour long and also fairly elementary. He picked up two attractive examples of such applications, one each from Analysis and Topology.
Topology in Algebra
Let be a commutative ring with identity and an ideal of . Then, we have the natural projection map
Similarly, one can talk about matrix groups over these rings, for example, the special linear groups and . Then, the projection map from onto extends naturally to a map from into .
Prof. Gurjar asked the question, “Does every element in lift to some element in ?” One does not expect this to be true in general. After all, if a matrix is in , then it is only the image of some matrix with entries in whose determinant is where . There is no reason to expect that we can always find such a matrix with .
And indeed, the question has a negative answer, and a proof for the case uses a very neat application of topology.
Before proceeding to the proof, it is worth mentioning that there are important cases where it is true that one can always find such a lift. The example I know comes up in the study of modular forms, or more precisely, when working with discrete subgroups of : the natural map from into the congruence subgroup is a surjective map. The interested reader can refer to Chapter 2 of Murty, Dewar & Graves’ “Problems in the Theory of Modular Forms”.
Returning to our claim, let and . We claim that the element
has no lift to .
The proof is by contradiction. We will show that if there is a lift, then there is a retract of onto (the circle group), which is known to not exist.
So, suppose that there is a lift to some element in . Then, that element would look like
where the polynomials must satisfy as well as the equations
for some polynomials . Now, we have a map given by
We also have a map , the space of orthogonal matrices, given by Gram-Schmidt orthogonalisation. But now observe that Gram-Schmidt orthogonalisation is a continuous map that does not change the determinant. Since is connected, this is actually a map into , the space of orthogonal matrices with determinant one. This space has a natural identification with , using the map
Hence, we have a the following sequence of maps:
where is the canonical inclusion map. Observe that because of the equations that satisfy. So, is actually what one calls a retract of onto . However, it is a well-known result in topology that there is no retract of onto for any , and so we have arrived at a contradiction.
Analysis in Algebra
Let . Consider the maximal ideal of . (Exercise: show that maximal!) It is an interesting question to ask whether is a principal ideal. We shall see that the answer is that it is not principal, and the proof neatly applies ideas from analysis.
It makes sense to view as the space of polynomials on modulo the equivalence relation
An important idea that goes into the proof is localisation. Let and be the localisations of and , respectively, at the element . Then, is also an ideal of . But, is a unit in and it is also one of the generators of . So, there is a unit in , which implies that .
Similarly, consider the localisations of and at the element . Since , we see that is the principal ideal in .
Now, let us prove that is not maximal by contradiction. Suppose . Then, and are also generated by the same element as ideals of and , respectively. By the above examination of these localisations, we can say that there exist elements and such that
This implies that
for some .
Now, the analysis kicks in. Observe that is a continuous function from into . But is nonzero everywhere on this domain, and so it is actually a map into . So, has the same sign everywhere on this domain. Similarly, has the same sign everywhere on the domain . Hence, the product has the same sign everywhere on . But this product equals whose signs differ on the upper and lower arcs, which is a contradiction.