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


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.


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 $y^2+y+2=0$).

  • $\begingroup$ 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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