Normal form of elements of $\mathbb{F}_{25}$?

Question

I have a question that says using the normal form of elements of $\mathbb{F}_{25}$ find explicit expressions for addition, multiplication and inversion of element of $\mathbb{F}_{25}$, but I don't even know what the normal form is. Help!

My attempt at the actual question now: Let $x = c + d\alpha, y = e + f\alpha \in \mathbb{F}_{25}$. Then we can define addition of x and y to be $x+y=(c+e) + (d+f)\alpha$ and multiplication to be $xy = (c + d\alpha)(e + f\alpha) = ce + df\alpha + df\alpha^2 = ce + df\alpha + -adf\alpha - bdf.$

For inverse we want to find $x^{-1}$ such that $xx^{-1} =1$ but I'm not sure where to go from here

Answer 1

$\Bbb F_{25}$ is created from $\Bbb F_5$ by adjoining a root $\alpha$ of some irreducible second-degree polynomial $x^2+ax+b$. This means that $\alpha^2=-a\alpha-b$, so any expression involving $\alpha$ to some power greater than $1$ may be reduced to the form $m\alpha+n$ without changing its value. This is (probably) the normal form.

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Comment on attempt (and solution): I would pick a second-degree polynomial, like $x^2-x+1$, and stick with it. This makes $\alpha^2 = \alpha - 1$. That way we actually get concrete answers. Your addition is fine. Your multiplication suffers from some minor mistakes. We have $$ xy = (c+d\alpha)(e+f\alpha) = ce + (cf + de)\alpha + df\alpha^2\\ = ce + (cf+de)\alpha + df(-1+\alpha)\\ = ce-df + (cf+de+df)\alpha $$ Now, let's use the above to figure out which $e, f$ makes $y = x^{-1}$. First, if $d = 0$, then it's easy: $f = 0$, and $e = \frac1c$ is the inverse of $c$ as elements of $\Bbb F_5$. So let's assume $d\neq 0$. We get $$ 1 = ce-df + (cf+de+df)\alpha\\ ce-df = 1\quad\text{and}\quad cf+de+df = 0 $$ The first equation, using $d\neq 0$, can be rewritten as $f = \frac{ce}{d} - \frac1d$. We insert that in the second equation and get $$ \frac{c^2e}{d} - \frac {c}d + de + ce - 1 = 0\\ e\left(\frac{c^2}{d} + d + c\right) = \frac cd + 1\\ e = \frac{c + d}{c^2 + d^2 + cd} $$ And once we have our $e$ (which is always defined, as $c^2+d^2+cd \neq 0$ for all $c$ and all $d\neq 0$), we can calculate our $f$ as $$ f = \frac{c^2 + cd}{c^2d + d^3 + cd^2} - \frac1d\\ = \frac{c^2 + cd - (c^2 + d^2 + cd)}{c^2d + d^3+cd^2}\\ = \frac{d^2}{c^2d + d^3+ cd^2}\\ = \frac{d}{c^2 + d^2 + cd} $$

Side note: These expressions for $e$ and $f$ work even when $d = 0$, as long as not both $c$ and $d$ are $0$.

Answer 2

There is no canonical normal form of elements of ${\mathbb F}_{25}$. To construct ${\mathbb F}_{25}$, you have adjoin a root $\alpha$ of some monic irreducible polynomial $x^2 + ax + b$ of degree $2$ to ${\mathbb F}_{5}$. Then ${\mathbb F}_{25} = {\mathbb F}_5[\alpha]$ and every element of ${\mathbb F}_{25}$ can be written in a unique way as $c \alpha + d$ for some $c, d \in {\mathbb F}_5$. Note that this 'unique way' depends on $\alpha$, i.e., on the specific monic irreducible polynomial of degree $2$ over ${\mathbb F}_5$ that you picked. In this representation, you can compute the representation of the product and/or sum of two elements using the fact that $\alpha^2 = -a\alpha - b$.