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I've tried to make it clear that this is a soft question. I wanted to express a few of my questions/concerns on what I call 'elementary questions,' hoping to find some answers.

I don't have a set definition of an 'elementary questions' but I characterize them as being mainly a) algebraic in nature, and b) requiring a prerequisite knowledge of only a few theorems and c) relatively simple in proof. I have been running into many that I characterize as this whilst self-teaching Abstract Algebra, currently with Gallian. I'll add a few examples of 'elementary problems' that I have encountered.

Prove: Every Boolean Ring ($R$ such that $x^2 = x$ for all $x \in R$) is Commutative

I eventually came up with the solution that for any $a, b \in R$, we have that

$$(a^2 + b^2) = (a + b)^2 = a^2 + ab + ba + b^2$$

$$-ab = ba \hspace{1cm} (1)$$

And for any $a \in R$, we can get

$$(a^2 + a^2) = (a + a)^2 = a^2 + 2a + a^2$$

$$-a = a \hspace{1cm} (2)$$

Then we can use (1) and (2) to get $$ab = ba$$ for all $a, b \in R$.

Prove: All non-trivial prime ideals are maximal ideals in a PID

The goal is to show that for some prime $\langle a \rangle$ in a PID, R, if there exists some $\langle a \rangle \subset \langle b \rangle$, $\langle b \rangle$ necessarily contains 1 (and so is equal to R).

If there exists some $\langle a \rangle \subset \langle b \rangle$, then we know that for some $r \in R$, we have that $br = a$. Since $\langle a \rangle $ is prime, then it must be that either $b$ or $r$ are in $\langle a \rangle$. Both result in a scenario $br^\prime a = a$ and by cancellation we have $br^\prime = 1$.

The characteristics of both of these elementary proofs is that once the 'trick' is realized, it is quite easy to do them. And although many 'higher-level' concepts are utilized by name (cancellation, knowledge of basic properties of rings/domains, etc...) the actual 'algebra' going on is no higher than that at a high-school level. In Abstract Algebra specifically, I have stumbled upon countless exercises similar to these, usually relating to proving some basic properties of elements in a specific type of ring / group / field.

The problem is, these 'elementary' problems are always quite difficult for me. The 'Every Boolean Ring' problem took me around four hours to solve the first time. For the second one, I spent 3.5 hours before peeking at the answer - with the intention of reading the first line and trying again - but with just the first line I was able to deduct the rest of the proof pretty easily.

Half of the problem seems to be the difficulty of 'not finding the trick' - yet the other half seems to be a problem of focus. Sometimes it feels like for these types of problems my already not-great attention span is further reduced, which makes it hard to keep relenting at an approach even when no results seem to occur.

I'm wondering two things:

  1. Is difficulty solving these 'elementary' problems a symptom of underlying weakness at problem solving? The fact that these problems require not much 'broadness' of knowledge and more require ingenuity with a small amount of tools imply, to me, that they are a good 'test' of 'pure' problem solving skill - whatever that may be.

  2. What is the best remedy for this? Of course, practice. Yet, I was wondering if specific types of mathematics are more suited for things like this... Elementary Number Theory, perhaps?

Thank you, and apologies for the soft question.

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    $\begingroup$ I don't think it's a sign of underlying weakness. You can't problem-solve effectively when you haven't built up the intuition about how to manipulate things like this yet. $\endgroup$
    – Zoe Allen
    Mar 12, 2023 at 8:03
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    $\begingroup$ The goal of these problems is to make you think really hard about what the axioms or definitions imply. It is natural to be stuck at these for a while before you develop the strategies to solve them. However, in my opinion if you are stuck on a problem for more than an hour without being able to come up with new ideas, you should ask your professor/TA for a hint or for help. Your professor will recognize if it is worthwhile for you to spend more time on it, or if you will gain more by them guiding you to the solution. $\endgroup$ Mar 12, 2023 at 8:19
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    $\begingroup$ One more element of reflexion : the domains you mention are in abstract algebra (I would say the same for the "discrete mathematics" domain) where intuition is hard to apply, contrasting with domains like geometry where intuition and deduction can be "coupled". An advice : vary the mathematical domains on which you work. Maybe you will discover that in one of them, your intuition is especially guiding you (for example a branch of applied mathematics like image processing, and/or in connection with computer science as well...) $\endgroup$
    – Jean Marie
    Mar 12, 2023 at 8:46
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    $\begingroup$ 1/2 As a HS student you probably can allow yourself to work on a problem for some time. I was talking from the perspective of giving advice to an undergrad student in maths: if a student has 20 hours of lectures per week and I give them 5 problems for HW, it would be unreasonable for me to expect of them to think about each individual problem for more than an hour, because they have other assignments to work on. At your stage peeking at solutions is fine, as long as you are making an effort to understand the solution. What is an exercise in Gallian is an explicit theorem in some other book. $\endgroup$ Mar 12, 2023 at 22:11
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    $\begingroup$ 2/2 For example your second problem is Proposition 4.13 in Aluffi's "Algebra: Chapter 0" for which a proof is given. So if you happened to work with Aluffi instead of Gallian, you wouldn't be peeking. You would be following the material in a way the author intended. $\endgroup$ Mar 12, 2023 at 22:16

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Rather than trying to give general suggestions (for this, I recommend Polya's How to Solve it), I'll go through your first example in a way that illustrates the kind of insights you (ideally) should be getting when working on it. Implicit in this is that the more work you put into solving the problem, at least up to a point (eventually there will be dimensioning returns, and in my opinion 4 hours is way past that point), the more you'll gain from solving it.

I think it's mostly a matter of getting the "right feel" for a topic as you begin studying it. In the case of your example, you know that $a^2 = a$ and $b^2 = b$ and $c^2 = c,$ etc. for the various elements $a$ and $b$ and $c,$ etc. in the ring. Now what? Well, squaring one of these again gives $(a^2)^2 = a^2,$ and since we know $a^2 = a,$ we get $a^4 = a.$ Similarly, $a^8 = a$ and $a^{16} = a,$ but this doesn't seem particularly relevant for what we want to show. However, these curious identities might be useful for other Boolean ring problems, or rings such that (say) $a^3 = a$ for every element $a,$ or such that $a^7 = a$ for every element $a,$ so maybe keep this successive-powering idea in our toolbox.

Looking back at the equations $a^2 = a$ and $b^2 = b$ and $c^2 = c,$ etc., what else can we do? We're dealing with a ring, so we can add and multiply. Multiplying (the first) two of these equations gives $(a^2)(b^2) = (ab)^2,$ so we get $aabb = abab.$ Now if the ring were commutative, then this last identity would be a consequence, but that's the wrong logical direction (i.e. it's the converse, namely the last identity implies the ring is commutative). On the other hand, this does show that Boolean rings have some "commutativity aspects" -- something which, if you were actually doing new research, would be very useful to notice if you were freshly investigating rings such that $a^2 = a$ for every element $a.$ Thus, the identity $aabb = abab$ suggests that such rings might be commutative, or at least share some properties of commutative rings (thereby suggesting possible avenues to then explore, such as are there any important consequences of commutativity that might also be true for Boolean rings).

Looking again at $aabb = abab$ and remembering that we have right and left multiplicative cancellability in rings, we get $abb = bab$ followed by $ab = ba,$ and we're done!

But maybe you didn't think of this. Maybe when staring at $a^2 = a$ and $b^2 = b$ and $c^2 = c,$ you thought to add a couple of equations instead of multiplying. This gives $a^2 + b^2 = a + b.$ Now what? I notice that the left side is the square of the right side if we forgot the 'O' and 'I' in "FOIL" (i.e. we squared via the Freshman's dream method), and since squares of elements are equal to the elements, we'd get $a^2 + b^2 = (a + b)^2$ if we ignore the cross-product terms. But since we know better, we know those cross-product terms must be included. At this point you would hopefully either be led to consider expanding the right side and then additively cancelling stuff (what you did); OR you would realize that the cross-product terms must add to zero since $a + b$ is equal to $(a + b)^2$ (Boolean ring property), which includes the cross-product terms, and this stuff is equal to $a^2 + b^2$ (because $a = a^2$ and $b = b^2),$ which doesn't include the cross-product terms.

So we have $ab = -ba$ (follows from the cross-product terms adding to zero), which while interesting (and something worth keeping note of for possible later use), seems to take us further from our goal. Indeed, if we were researching this from scratch (without knowing in advance that Boolean rings are commutative), then we likely would be thinking that Boolean rings, while having some commutativity aspects, are probably not "fully commutative". However, it follows that at least some of the elements are their own additive inverse, namely those elements that can be expressed as a product of two elements. Thinking a bit more about which elements of the ring can be expressed as the product of two elements, hopefully you'd realize that every element can be so expressed, since one of the elements we could use is the identity $e.$ That is, given any element $a$ in the ring, we have $ae = a,$ so any element in the ring can be expressed as the product of two elements in the ring. Thus, we have $a = -a$ for any element $a$ in the ring. After thinking through this, we'd probably be led (in retrospect) to shorten the argument, and argue this way: Let $a$ be any element of the ring. Then $ae = a$ and $ea = a$ (because $e$ is the identity) and $ae = -ea$ (identity we previously got, with $b = e),$ from which it follows that $a = -a.$

Of course, there are other ways of stumbling on the fact that $a = -a$ for any element $a,$ for example the way that you did it. In fact, coming up with more than one way, while overkill for a homework problem, is often very useful in research, because maybe the two ways individually use slightly less than all the ring properties, so one might be led to discover that Boolean lower-quasi rings and Boolean left-pseudo rings (made up terms) are each commutative with respect to multiplication.

At this point we know that each element is its additive inverse, so $ab = -ba$ (the anti-commutative law we got earlier) and $-ba = ba$ (each element is its additive inverse), which together give us $ab = ba.$

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    $\begingroup$ Your emphasis on 'feel' - i.e noticing a law that seems like 'pseudo-commutativity' and trying to expand on that to get a full answer - is especially brilliant, and I can quite easily see how this idea can be expanded to many different problems / research. Thank you for this insight. $\endgroup$ Mar 12, 2023 at 11:03

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