OK so basically the question asks me to find an explicit ring isomorphism $ \alpha: x^2+x+2 \mapsto x^2+2x+2$

Where both are under the field of five elements ie (0,1,2,3,4). I don't know how to get the F[5] symbol up, sorry.

So how do I find this?

I've got another example question which is to show that there is an isomorphism of fields

$x^2+2\cong x^2+x+2$

Again worded a bit differently but obviously asking the same thing and again over the field of five elements. Can someone guide me on how to do the first one so that I can do it myself on the second question and post here if it is correct? Thanks.

  • $\begingroup$ EDIT: Nevermind. $\endgroup$ – Lolwat Apr 28 '13 at 22:28
  • 2
    $\begingroup$ Is it standard to refer to polynomials as if they were fields? As I understand it, you want an isomorphism ${\bf F}_5[x]/(x^2+x+2)\cong{\bf F}_5[x]/(x^2+2x+2)$, right? The second is not a field, though, whereas the first is a field, so there is no isomorphism between them. The splitting fields are also non-isomorphic, if that's what you intended. Is there something I'm not seeing? Also, it appears your question is tagged as (abstract-algebra), like you wanted, right? $\endgroup$ – anon Apr 28 '13 at 22:34
  • 1
    $\begingroup$ What "nevermind"? Your question or your comment? $\endgroup$ – DonAntonio Apr 28 '13 at 22:35
  • $\begingroup$ The comment, most likely. When you say "nevermind" to a previous comment but then delete the comment you are referring to, you will tend to confuse others, OP. $\endgroup$ – anon Apr 28 '13 at 22:36
  • $\begingroup$ To the OP, you can right click existing notational symbols here and bring up their TeX commands. $\endgroup$ – user41442 Apr 29 '13 at 1:32

Hint $\rm\ X^2\!+\!2,\ x^2\!+\!x\!+\!2\ \:$ both have discriminant $\rm = -2 \in\Bbb F_5.\:$ Since $\rm\:-2\:$ isn't a square in $\,\Bbb F_5,\:$ and $\rm\:1/2 \in \Bbb F_5,\:$ we can use the quadratic formula to deduce that both extension fields are generated by adjoining $\rm\:\sqrt{-2}\:$ to $\,\Bbb F_5,\:$ i.e. $\rm\: R = \Bbb F_5[X]/(X^2\!+2)\, \cong\, \Bbb F_5[\sqrt{-2}]\,\cong\, \Bbb F_5[x]/(x^2\!+\!x\!+\!2) = R'.\:$ To construct an explicit isomorphism, use that, in $\rm\:R,\ \ X =\, \pm\sqrt{-2},\,$ and $\rm\:\pm\sqrt{-2}\,=\, 2x\!+\!1\,$ in $\rm\,R',\:$ since $\rm\:x=(-1\pm\sqrt{-2})/2\:$ by the quadratic formula (or, completing square $\rm\,\Rightarrow\, (2x\!+\!1)^2\! = -2),\:$ and use: $\rm\:h\,$ a ring hom and $\rm\:- 2 = X^2\Rightarrow\,-2 = h(X^2) = h(X)^2,\:$ i.e. $\rm\:h(\pm\sqrt{-2})\, =\, \pm\sqrt{-2}.$

If the first is a typo for $\rm\:x^2\!+\!2x\!\color{#c00}{-\!2},\:$ with discriminant $\rm\:2,\:$ utilize $\rm\:X^2\!= -2\:\Rightarrow\,(2X)^2\! = 2\,$ in $\rm\,\Bbb F_5.$

  • $\begingroup$ Thanks for the solution but my mark scheme does it in a colmpletely different way? $\endgroup$ – Lolwat Apr 29 '13 at 19:25
  • $\begingroup$ @Lolwat Does it tell you how to discover the isomorphism, or does it simply pull it out of a hat, like magic? Above I try to give you some intuition about how to discover the correct isomorphism. If you post the solution in your answer then I can explain further. $\endgroup$ – Math Gems Apr 29 '13 at 19:50

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.