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I have tried get a version of the proof stating that a left ideals of a ring is not, in general, a right ideal, and viceversa. Is my formulation right? Comments and corrections are welcome. I have tried to understand some German papers on the topic I came across.......I hope it is fine......The proof takes the form of a reductio ad absurdum...... Thanks in advance:

The proof goes as follows: let us assume M to be a left module over the ring R with the external operation R × M → M. Then we write the product with elements from R, proceeding from right to left. For that purpose, the symbol ◦ is used. Let then α·x := x◦α, ∀α ∈ R, x ∈ M. Now, generally speaking, the left module M with the external product ◦ : M × R → M is no right module whatsoever. This contradicts the Associative Law. According to the initial assumption M is a left module, that is, it is precisely the Associative Law that comes into play for scalar multiplication of a ring element from the left side. We then suppose that M with the operation ◦ is also a right module. Assuming, without loss of generality, that all other conditions apply as usual, it can be seen that instead of the Associative Law (M3) the following holds: (αβ) · x = x ◦ (αβ) = (x ◦ α) ◦ β = (α · x) ◦ β = β · (α · x)⊥, which constitutes a contradiction to the Associativity of the left module.

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  • $\begingroup$ Are you trying to prove that left modules can never be right modules? This is, of course, not true. Maybe you should try to find an explicit example of a left module that is also a right module, and a left module that is not a right module. $\endgroup$ – Max Jul 30 '15 at 8:33
  • $\begingroup$ No, I am trying to prove that they are not automatically right modules. Of course I know they are sometimes so. The point is, they are not always such a thing. I provide examples in some other place. Here I am just concerned with the proof. $\endgroup$ – Javier Arias Jul 30 '15 at 8:37
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    $\begingroup$ IMO a counterexample is the only proof you can have. Why would you not call the existence of a counterexample a proof that the fact is not always true? An example is never a proof, but a single counterexample is always a disproof (=proof that the claim is wrong). $\endgroup$ – Jyrki Lahtonen Jul 30 '15 at 8:48
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It is IMHO never a good idea to try and prove that some structure is not some other structure using the axioms only. After all, in many cases the said structure also IS that other structure, so a proof "generally speaking" is doomed to fail. Here you cannot rule out the possibility that, in addition to what you describe as a substitute to the associativity, the usual associativity may still hold also. For example, if $R$ happens to be commutative every left ideal is also a right ideal. Therefore any argument that works in any ring is automatically wrong.

Furthermore, you cannot just add the assumption that $R$ is not commutative to your proof, and expect the readers to be convinced. This is because there are non-commutative rings were every left ideal is also a right ideal. The ring of Hamilton's quaternions is an example of such a ring.

No. When you are given the task to show that something is not always true, the first reaction should be to look for a counterexample. Here your task is to find a ring $R$ and its left ideal $L$ such that $L$ is not a right ideal of $R$. Hint: Look at the ring of $2\times2$ matrices.

Students often want to answer questions like this by starting out with one of the proofs from the textbook/lecture notes, and then point out a step that does not work. This gives me the teacher pimples. Such an answer simply allows us to conclude that this line of reasoning did not work. It leaves open the possibility that another argument still might.

Sorry about the rantish answer. I just feel that learning to appreciate the role of counterexamples is a very important part of the education of a mathematical mind. Hence the emphasis.

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  • $\begingroup$ I was trying to formulate in English what I found in German on oages 12-13 of this paper....mathematik-netz.de/pdf/Moduln.pdf . I do not know whether I succeeded $\endgroup$ – Javier Arias Jul 30 '15 at 8:50
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    $\begingroup$ Why is that not a formal proof? AFAICT the point of that German paper is pedagogical and it only shows that the process described there does not allows one to turn a left module into a right module unless that identity happens to hold for all the elements of the module and the ring. I think it is more than a bit misleading to call it "Beweis." $\endgroup$ – Jyrki Lahtonen Jul 30 '15 at 9:10
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    $\begingroup$ I don't understand. To make things clearer to me can you (for example) give a formal proof that a continuous function is not always differentiable. Or a formal proof that a group is not always abelian. Then I would better understand what you mean. A counterexample is a proof by contradiction: We claim that a left ideal is not always a right ideal. Assume contrariwise that a left ideal is always a right ideal. Look, here's a left ideal that is not a right ideal. A contradiction. Q.E.D. $\endgroup$ – Jyrki Lahtonen Jul 30 '15 at 9:44
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    $\begingroup$ @JavierArias I think someone has already pointed out to you in your proof that if the $R$ in the proof is commutative, it will be proving a commutative ring has a left ideal that isn't a right ideal, which is absurd. The strategy of beginning with "Let $M$ be a module over a ring $R$..." and then proving not all left ideals are right ideals is doomed to failure for that reason. $\endgroup$ – rschwieb Jul 30 '15 at 14:06
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    $\begingroup$ yes, you should use reasoning along those lines. I might say "Note that while over a commutative ring, left modules can be thought of as right modules (and vice versa), in general this is not the case. A counterexample is provided in section..." $\endgroup$ – Max Jul 31 '15 at 16:42

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