DiscMat assignment 10

exercise

10.3 Characteristic of a field

1)

By definition of characteristic, $q\times 1_F=0$ $q\times a=\underset{q\text{ times}}{\underbrace{a+a+\dots a}}$ $=\underset{q\text{ times}}{\underbrace{(1_F\star a)+(1_F\star a)+\cdots +(1_F\star a)}}$ $=(\underset{q\text{ times}}{\underbrace{1_F+1_F+\cdots +1_F}})\star a$ $=(q\times 1_F)\star a$ $=0_F\star a$ $=0_F$

2)

By using the binomial theorem $(a+b{)}^q={\sum }_{k=0}^q\left[\begin{array}{c}\left(\begin{array}{c}q\\ k\end{array}\right)\times (a^k\star b^{q-k})\end{array}\right]$ So $(a+b{)}^q=\left(\begin{array}{c}q\\ 0\end{array}\right)a^q{b}^0+\left(\begin{array}{c}q\\ q\end{array}\right)a^0{b}^q+{\sum }_{k=1}^{q-1}\left[\begin{array}{c}\left(\begin{array}{c}q\\ k\end{array}\right)\times (a^k\star b^{q-k})\end{array}\right]$ $=a^q+b^q+{\sum }_{k=1}^{q-1}\left[\begin{array}{c}\left(\begin{array}{c}q\\ k\end{array}\right)\times (a^k\star b^{q-k})\end{array}\right]$

We prove that the equation $(a+b{)}^q=a^q+b^q$ holds by showing ${\sum }_{k=1}^{q-1}\left[\begin{array}{c}\left(\begin{array}{c}q\\ k\end{array}\right)\times (a^k\star b^{q-k})\end{array}\right]=0_F$ Consider $\left(\begin{array}{c}q\\ k\end{array}\right)$ for all $1\le k\le q-1$

$\left(\begin{array}{c}q\\ k\end{array}\right)=\frac{q(q-1)\cdots (q-k+1)}{k(k-1)\cdots 2\cdot 1}$ For all $l\in \{1,\dots ,k\}$, $gcd(l,q)=1$, because $k$ is smaller than $q$ (so $l<q$), and $q$ is prime. $⟹$ so the factor $q$ remains, making $\left(\begin{array}{c}q\\ k\end{array}\right)$ a multiple of $q$

If an integer is divisible by $q$ in $\mathbb{Z}$, its image in $F$ is $0_F$ (the kernel of the ring homomorphism $\mathbb{Z}\to F$ is $(q)$) $⟹\left(\begin{array}{c}q\\ k\end{array}\right)=0_F$ in $F$ ${\sum }_{k=1}^{q-1}\left[\begin{array}{c}\left(\begin{array}{c}q\\ k\end{array}\right)\times (a^k\star b^{q-k})\end{array}\right]={\sum }_{k=1}^{q-1}\left[\begin{array}{c}{0}_F\star (a^k\star b^{q-k})\end{array}\right]$ $={\sum }_{k=1}^{q-1}{0}_F$ $=0_F$ Therefore, the $(a+b{)}^q=a^q+b^q+0_F=a^q+b^q$ The claim is proved.

3)

Let $S_k={\sum }_{i=1}^k{a}_i$ where $k\in \mathbb{N}$, and for all $i\in [k]$, $a_i\in F$, Claim (statement P): For all $k\ge 1$ and for all $a_1,\dots ,a_k\in F$

$$S_k^q=\sum _{i=1}^k{a}_i^q$$

Base: $k=1$ $S_1^q={\left({\sum }_{i=1}^1{a}_i\right)}^q=(a_1{)}^q=a_1^q$

Induction Steps: $k\to k+1$ $S_{k+1}^q={\left({\sum }_{i=1}^{k+1}{a}_i\right)}^q={\left({\sum }_{i=1}^k{a}_i+a_{k+1}\right)}^q={\left(S_k+a_{k+1}\right)}^q$ $=S_k^q+a_{k+1}^q$ (using the result from question 2) $=\left({\sum }_{i=1}^k{a}_i^q\right)+a_{k+1}^q$ $={\sum }_{i=1}^{k+1}{a}_i^q$

Thus, by induction, the claim holds for all $k\in \mathbb{N}$

4)

Let $a$ be $m\times 1_q$ where $m\in \mathbb{N}$ $a^q=(m\times 1_q{)}^q={\left({\sum }_{i=1}^q{1}_q\right)}^q={\sum }_{i=1}^q{1}_q^q={\sum }_{i=1}^q{1}_q=m\times 1_q=a$ The claim is proved.