In the second edition of Artin's algebra book, page 472, the following exercise is given:

Let $\alpha$ be a complex root of $x^3-3x+4$. Find the inverse of $\alpha^2+\alpha+1$ in the form $a\alpha^2+b\alpha+c$, with $a$, $b$, $c$ in $\mathbb{Q}$.

The exercise itself is not difficult. One can use the extended Euclidean Algorithm or brute force to find $a$, $b$ and $c$. I do not understand how the fact that $\alpha$ is a complex root of $x^3-3x+4$ relevant. As far as I can see, both the approaches do not make use of the fact that $\alpha$ is a complex root. There is possibly a simple explanation, but it eludes me. Can someone explain why this condition is there? Is it possible that the above condition gives a shorter way of solving the exercise?

  • $\begingroup$ After a moment’s reflection, I too can not see any significance in the complexness of $\alpha$. $\endgroup$ – Lubin Mar 25 '18 at 4:48
  • 1
    $\begingroup$ One can only suppose that the exercise is intended to work not only for real roots but also for nonreal roots. $\endgroup$ – MPW Mar 25 '18 at 4:49

Well, if you just said "a root of $x^3-3x+4$", then it would be unclear what field this root is supposed to be living in. So "complex" specifies that we are looking at a root in $\mathbb{C}$, as opposed to (say) a root in $\mathbb{F}_{37}$. As it turns out, it doesn't really matter that the field is specifically $\mathbb{C}$, but it does matter that it is some field of characteristic $0$ as opposed to a field of positive characteristic (and requiring $a,b,c\in\mathbb{Q}$ would not make sense in the latter case anyways).

| cite | improve this answer | |
  • $\begingroup$ Saying complex is not necessary to say that we are not looking at a root in $\mathbb{F}_{37}$ because the question clearly means that we are looking at the root in $\mathbb{Q}[\alpha]$ by mentioning that $a$, $b$, $c\in\mathbb{Q}$. $\endgroup$ – S. Venkataraman Mar 25 '18 at 15:01
  • $\begingroup$ I agree that the intent of the problem is clear without "complex", but it's simply wrong as a matter of mathematical terminology without it. It is meaningless to talk about a root of a polynomial unless you specify an algebraic structure in which the root lives. $\endgroup$ – Eric Wofsey Mar 25 '18 at 15:39
  • $\begingroup$ I suppose the confusion is because the equation has real root as well as two complex roots. So, it makes one suspect that the root being complex is important. Any real number is also a complex number, so in that sense, the question refers to all the roots, not just the complex roots which are not real. My doubt is, is it all there is to it or are we missing a subtle point in the proof where the fact that the root is not real matters. $\endgroup$ – S. Venkataraman Mar 25 '18 at 15:53
  • $\begingroup$ It does not matter whether the root is real, nor would "complex" be the correct way to indicate that the root is not real. As I said, the only purpose of "complex" here is because the problem statement would not make sense without some indication of where the root lives. $\endgroup$ – Eric Wofsey Mar 25 '18 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.