Express $a^5$ in terms of $c_0+c_1a+c_2a^2.$

Question

Let $F=\mathbb Z_2,f(x)=x^3+x+1\in F[x].$ Suppose $a$ is a zero of $f(x)$ in some extension of $F.$ Then $F(a)\simeq F[x]/\langle f(x)\rangle$ and there is an isomorphism $$\phi:F[x]/\langle f(x)\rangle\to F(a):g(x)+\langle f(x)\rangle\mapsto g(a)$$ and $F(a)=\{c_0+c_1a+c_2a^2:c_i\in F\}.$ I would like to express $a^5$ in terms of $c_0+c_1a+c_2a^2.$

My Attempt: I've first consider the element corresponding to $a^5$ in $F[x]/\langle f(x)$ viz $x^5+\langle f(x)\rangle.$ Now $x^5+\langle f(x)\rangle\\=x^2(x^3+x+1)-(x^3+x^2)+\langle f(x)\rangle\\=-(x^3+x^2)+\langle f(x)\rangle\\=-(x^3+x+1)+x-x^2+1+\langle f(x)\rangle\\=x-x^2+1+\langle f(x)\rangle$

So the element corresponding to $x-x^2+1+\langle f(x)\rangle$ is $a-a^2+1.$ But the answer is $a^2+a+1.$

I don't understand where did I go wrong!

Answer 1

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