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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.

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  • $\begingroup$ EDIT: Nevermind. $\endgroup$ – Lolwat Apr 28 '13 at 22:28
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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $\rm\,\Bbb F_5.$

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  • $\begingroup$ 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

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