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I need to show that $$F_{3}[x]/(x^{3}-x^{2}+1)\cong F_{3}[x]/(x^{3}-x^{2}+x+1)$$ Without using the classification of finite fields. It seems I can use some subsitution like $x\rightarrow x+1$, then we have $x^{3}\rightarrow x^{3}+1$, $x^{2}\rightarrow x^{2}-x+1$, together with $x\rightarrow x+1$ we may change the former field $F_{3}[x]/(x^{3}-x^{2}+1)$ into $F_{3}[u](u^{3}-u^{2}+u+1)$ with $u=x-1$. But is this really justified? I feel uncertain because obviously $$x\rightarrow x+1$$ is not a field isomorphism for $F_{3}[x]$. On the other hand all the other trivial field isomorphisms like Frobenius isomorphism looks quite complicated when write down.

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    There is a unique homomorphism $\mathbb{F}_3 [x] \to \mathbb{F}_3 [x]$ that sends $x$ to $x + 1$, but this is not the map that sends every element $p$ to the element $p + 1$!2012-11-04
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    @ZhenLin: what do you mean? we have $\phi(ab)=ab+1\not=(a+1)(b+1)$ for $x,y\in F_{3}[x]$. what is the unique homomorphism you are talking about?2012-11-04
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    Explicitly it maps a polynomial $p(x)$ to the polynomial $p(x + 1)$. Here $x$ is not a variable but rather the distinguished element of $\mathbb{F}_3 [x]$.2012-11-04
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    I see. So my gut intuition is not entirely wrong then. Thanks.2012-11-04

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