Express a function $g$ in the vector space $\mathbb{Z}/3$ I have a question I can't seem to solve. It is stated as follows:
Recall that $\mathbb{Z}$ denotes the ring of integers, and $\mathbb{Q}$ is the rationals. We let $F[x]$ denote the polynomial ring in the variable $x$ over a field $F$, and we let $f(x)=x^2+1\in F[x]$
Let $E={F}[x]/(f(x))$ with $F=\mathbb{Z}/(3)$, and let $\alpha \in E$ be a root of $f(x)$. Express $g(1+\alpha)$ when $g(x)=2x^4+x^2+1$ in the basis $\{ 1, \alpha \}$ of the $F-$ vector space $E$.
Could anyone help me with this question? Thanks beforehand
 A: This is one of the cases in math in which you don't have to make extraordinary arguments or being excessively smart. The ideas will come later, when you will have confidence with the formalism of polynomials and fields.
At the present moment, just stick to the definitions. Use the rules as you were a computer: substitute $1+\alpha$ into $g$, expand the powers, and use the fact that $\alpha^2= -1$.
You will end up with what you are looking for. Maybe it's not clear to you why, but let yourself be driven by the automatic calculations, and you will find yourself understanding better along the way!!
A: The defining property of $\alpha$ is that $\alpha^2 + 1 = 0$, or put in more 'human readable' form: $\alpha^2 = -1$.
Knowing this you can just write out $g(1 + \alpha)$ as some expression in $\alpha^4$, $\alpha^3$, $\alpha^2$, $\alpha$ and a bunch of numbers from $F$ in the standard high school way and next eliminate all the higher powers of $\alpha$ using $\alpha^2 = -1$ until an expression of the desired form remains.
What exactly is the ground field here is not terribly important, we could have $F = \mathbb{R}$ and $E = \mathbb{C}$ and the procedure would not be different in very important ways. (Depending on what you find important of course.)
A: $\mathbb{Z}$ is the ring of integers, $\mathbb{Q}$ the set of rational numbers. $F[x]$ denotes the polynomial ring in the variable and we let $f(x) = x^2+1$ in $F[x]$.
Let $E =\frac{F[x]}{(f(x))}$ with $F=\mathbb{Z}  /(3)$. Evaluating $\bar{f}(n)\;$ for all $n\in \mathbb{Z_3} $ :
\begin{align}
    \bar{f}(0) &= 0^2+1 = 1\nonumber \\
    \bar{f}(1) &= 1^2+1 = 2 \label{eq:fbar2}\\
    \bar{f}(2) &=2^2+1 = 2 \nonumber
\end{align}
$f(x)$ has no roots in $\mathbb{Z_3}$ , hence $\mathbb{Z }/ (3)$  is irreducible. Note that $(f(x))$ is a maximal ideal in $F[x]$ and $E =\frac{F[x]}{(f(x))}$ is a field. Let $\alpha $ be a root of $f(x)$, then we have $\frac{F[x]}{(f(x))}\simeq {F[\alpha]}$ with $F[\alpha]=\{ a+b\alpha | a,b \in \mathbb{Z}/(3) \}$. Clearly $\{1,\alpha \}$ is a basis of $F[\alpha]$. Since $\alpha$ is a root of $f(x)$ then $\alpha^2+1=0$. Evaluating $g(1+\alpha )$ :
\begin{align}
    \bar{g}(1+\alpha) &= 2(1+\alpha)^4+(1+\alpha)^2+1 \nonumber \\
     &= 2(1+\alpha)^2(1+\alpha)^2+(1+\alpha)^2+1 \label{eq:g}\\
     &= (1+\alpha)^2[2(1+\alpha)^2+1]+1\nonumber \\
     &= (1+\alpha^2+2\alpha)[2(1+\alpha^2+2\alpha)+1]+1\nonumber \quad \text{Use $1+\alpha^2=0$ in $\mathbb{Z}/(3)$}\\ 
     &= (0+2\alpha)[2(0+2\alpha)+1]+1\nonumber \\
     &= 2\alpha[4\alpha+1]+2 \nonumber  \\
     &= 8\alpha^2+2\alpha+2 \nonumber  \\
     &= 2\alpha^2+2\alpha+1 \nonumber  \\
     &= \alpha^2 +2\alpha +\alpha^2+1 \nonumber  \\
     &= \alpha^2+2\alpha+0 \nonumber  
\end{align}
Rewriting $g(1+\alpha)=\alpha^2+2\alpha=\alpha^2+1+2\alpha- 1=2\alpha-1$. So $g(1+\alpha) = 2\alpha - 1$ in the basis $\{1, \alpha\}$.
Answer : $g(1+\alpha)=2\alpha-1$ expressed in the basis $\{1, \alpha\}$.
This is my attempt but I feel like the stuff I state in the beginning isn't really necessary? I also wonder If what I've done is correct? Thanks beforehand
