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I think this problem is from Gallian, prof couldn't solve it. Notice that both polynomials have no roots. I tried to construct an onto homomorphism $\varphi:\mathbb{Z}_5[x]\to\mathbb{Z}_5/(x^2+x+2)$ whose kernel is $(x^2+x+1)$. The most obvious attempt is $\varphi(f(x))=a(x)f(x)$, but since the quadratics are relatively prime there's no natural way to do this. Any ideas?

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    $\begingroup$ Did you have finite fields? They are unique. And both things you mention are a field of order 25 $\endgroup$
    – Sorfosh
    Sep 18, 2018 at 22:30
  • $\begingroup$ Does $\equiv$ mean isomorphic? $\endgroup$
    – lhf
    Sep 18, 2018 at 23:05
  • $\begingroup$ @lhf yes, I was on the phone so couldn't check the tex character $\endgroup$ Sep 19, 2018 at 0:30
  • $\begingroup$ @Sorfosh prof mentioned that we result but we didn't see it in class $\endgroup$ Sep 19, 2018 at 0:31

2 Answers 2

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Well, $x^2+x+1$ has for its roots the primitive cube roots of unity. So to get an isomorphism from the field $\Bbb F_5[x]/(x^2+x+1)=k$ to the field $\Bbb F_5[t]/(t^2+t+2)=\ell$, you have to find a cube root of unity in $\ell$, call it $\beta$. Then your isomorphism takes $\bar x$ (in the quotient ring $k$) to $\beta\in\ell$.

On a relatively philosophical level, this interesting problem shows why there is no “the” field with $25$ elements: any two are isomorphic, yes, but you have to construct the isomorphism, especially, as here, where there is no preferred isomorphism staring you in the face.

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We wish to show that $\mathbb{Z}_5[x]/(x^2+x+1) \equiv \mathbb{Z}_5[y]/(y^2+y+1)$. Suppose that $x \rightarrow ay+b$ , consider where $x^2$ is mapped to gives \begin{eqnarray*} 4a+2b&=&4 \\ b^2+3a^2&=&4b+4 \end{eqnarray*} and by inspection this has solution $a=3,b=1$. One can verify that $x \rightarrow 3y+1$ maps $x^2+x+1$ to zero (in light of $y^2+y+2=0$).

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  • $\begingroup$ But would this map be a homomorphism? It doesn't seem that f(x+z)=f(x)+f(z). $\endgroup$ Sep 25, 2018 at 16:25

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