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.
What is a good way to show $F_{3}[x]/(x^{3}-x^{2}+1)\cong F_{3}[x]/(x^{3}-x^{2}+x+1)$?
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galois-theory
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0I see. So my gut intuition is not entirely wrong then. Thanks. – 2012-11-04
1 Answers
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To prove $\mathbb F_{3}[x]/(x^{3}-x^{2}+1) \simeq \mathbb F_{3}[y]/(y^{3}-y^{2}+y+1)$ we would like to find a field isomorphism, clearly $-1,0,1$ is fixed by it, so we must relate $x$ and $y$.
Let $\bar x = -y^2 + y$, then notice that $\bar x^3 - \bar x^2 + 1 = (y^2) - (-y^2 + 1) + 1 = 0$ in $\mathbb F_{3}[y]/(y^{3}-y^{2}+y+1).$
Therefore the "evaluation" map $a + b x + c x^2 \mapsto a + b \bar x + c \bar x^2$ (where $a,b,c \in \mathbb F_{3}$) is probably a field isomorphism, but we better check.
- Injective: Suppose $a + b \bar x + c \bar x^2 = a' + b' \bar x + c' \bar x^2$ then $(a+c) + (-c)x + (b-c)x^2 = (a'+c') + (-c')x + (b'-c')x^2$ so equating coefficients $a+c = a'+c'$, $-c = -c'$, $b-c = b'-c'$ which implies $a=a'$, $b=b'$, $c=c'$ hence we have injectivity.
- Surjectivity: Injective functions on a finite set are Surjective.
- Homomorphism: Immediate as it is an evaluation map.
Therefore we have an isomorphism.