TITLE: What do you do if you believe a problem is undecidable?
QUESTION [22 upvotes]: While the title of this question is subjective, I hope to make what I'm looking for quite concrete.  The first, and main question is this: If you believe that a problem you are working on is formally undecidable, but you are not a logician, what are some ways to begin learning the tools necessary for you to try to prove undecidability?  Where should one begin?
Now, for an explicit example.  In ring theory one of the big open problems is called Koethe's conjecture.  There are many equivalent ways to state the conjecture such as "The sum of two nil left ideals is still nil."  Let me give another equivalent description.
Let $R=\mathbb{Z}\langle a,b\rangle$ be the free ring on two non-commuting generators.  Suppose that $I$ is a left ideal of $R$, containing some power of $b$, and containing some power of every element of $Ra$.  Then the conjecture asserts $I$ contains some power of $a+b$.  Note that $R$ is countable, and in fact we can easily enumerate the elements of $Ra$ as $f_1,f_2,\ldots$.
Thus, given any sequence of positive integers $\vec{n}=\{n_0,n_1,n_2,\ldots\}$, the conjecture asserts that the left ideal containing $b^{n_0},f_1^{n_1},f_2^{n_2},\ldots$ also contains a power of $a+b$.
For some sequences $\vec{n}$ (such as eventually constant sequences), it is not difficult to compute that the conjecture is true.  But in general the computations get extremely difficult.  Lam says in his "First Course in Noncommutative Rings" p. 171, that there has been a 

(long-held) suspicion that the Conjecture is false.

If so, then it is false for a sufficiently fast growing sequence $\vec{n}$.
However, one of the troubles here is that nobody knows how to control the behavior of $I$ for fast growing sequences $\vec{n}$, and there are also ways to modify $I$ only slightly and provably not produce a counter-example.
My current hunch is that the conjecture is false, but to see this we need tools to assert that (for fast growing vectors $\vec{n}$) the ideal $I$ stays complicated.  I'm not confident that the regular tools from ZFC are sufficient to the task; but have no idea where to go from here.
One last comment: There is always the danger of fearing that a problem one is working on is formally undecidable, just because you can't immediately solve it.  Because of this I haven't given up hope that there is such a solution.  However, there is also the chance that some stronger set theory may provide tools to look at the problem differently, and really that's what I'm after.

REPLY [4 votes]: Joel David Hamkins has given a good answer in the case of (2), where a certain notion when expressed formally in the language of a theory $T$ (and also the negation of this notion) may not be provable from $T$ using the appropriate proof system.  I would like to touch on the other case (1), where it may be possible to show not only that the conjecture is false, but that it can be false in a noncomputable way.  This would be a good resolution from the standpoint of the original poster, because it would shed light on the nature of algorithmic questions in general algebra.
(Disclaimer: In spite of my proximity to the subject, I am no expert on decidability.  I recommend Algorithmic Problems in Varieties by Sapir and Kharlampovich, and also section 6 and earlier of Ross Willard's "An Overview of Modern Universal Algebra", both available on the Web.  I had the privilege of being one of the first to see McKenzie present his work on Tarski's Finite Basis Problem: that work influences this post.)
Consider the possibility that Koethe's conjecture is false.  As Pace Nielsen points out in his question, one way that it could be false is that there exists a fast growing sequence $n_i$ and a (hopefully recursive) enumeration $f_1,f_2, \ldots$ of $Ra$ such that $a+b$ and all of its powers  avoids all the left-ideals generated by a finite subsequence of $b^{n_0}, {f_1}^{n_1}, {f_2}^{n_2}, \ldots$ .  (Unless I missed something, this should be equivalent to his formulation.)  It would be nice to be able to characterize those ideals which fail the conjecture, or given a particular ring, to characterize those sequences and enumerations which lead to failure.
As mentioned in the comments, to show that no recursive characterization is possible, one method would be to reduce an undecidable problem to this question.  In other words, construct a recursive map $\phi$ so that for every instance $I$ of a problem known to be undecidable, $\phi(I)$ is a sequence (say) of exponents $n_j$ such that a certain enumeration of $Ra$ raised to exponents $n_j$ produces an ideal which contains a power of $a+b$ iff $I$ satisfies the property in the problem known to be undecidable.  The Boone-Novikov result is well known (a finitely presented group with undecidable word problem), and the article of Sapir and Kharlampovich contains many more examples, and if the Koethe problem "looks" enough like one of the undecidable problems, one might then construct a recursive $\phi$ as suggested above.
Alternatively, one could have a mastery of the situation that one could try to build a sequence of ideals with such a property that "encoded" Turing machine computations.  In the case of Tarski's Finite Basis problem (is there a computer program that can take as input the function tables of a finite algebra A of finite type, and output "yes" if the equational theory of V(A) is finitely-based, and "no" otherwise), there was a series of related problems that McKenzie was working on that allowed him to express a Turing program as an algebra with certain operations corresponding to the state transition table of the program, such that if the halting state were reached, that would correspond to a certain algebraic property in an infinite power of the base algebra, and not correspond otherwise.  The properties involved had to do with residual smallness, and McKenzie told me that part of his thought process was to arrange all of the properties he needed in small groups, and make small changes in the algebraic construction to get the group of properties he needed (my wording based on a poor remembering of an event over twenty years in the past).
This is an attempt then to answer the question in a different way: if you suspect the answer is not just no, but no in a non-recursive way, try two approaches: reduction from something in Kharlampovich-Sapir (or other sources), or understand the properties of the ideals that satisfy Koethe's conjecture as well as those that don't, and see if those properties are recursively inseparable (can be related somehow to computations halting or not).
Gerhard "That's How I Would Start" Paseman, 2015.03.24