MHB Kernel of Linear Map $\theta$ in $\mathbb{F}_{q^n}$

[mathmari]

Hey!

Let $q$ be a power of a prime and $n\in \mathbb{N}$. We symbolize with $Tr$ the map of the trace from $\mathbb{F}_{q^n}$ to $\mathbb{F}_q$, i.e. $Tr:\mathbb{F}_{q^n}\rightarrow \mathbb{F}_q$, $\displaystyle{Tr(a)=\sum_{j=0}^{n-1}a^{q^j}}$. I want to calculate the dimension of the image of the linear map $\theta : \mathbb{F}_{q^n} \rightarrow \mathbb{F}_{q^n}$, $\theta (\beta)=\beta^q-\beta$.

We have that $\dim (\mathbb{F}_{q^n})=\dim (\ker (\theta))+\dim (\text{im} (\theta))$, right?

The dimension of $\mathbb{F}_{q^n}$ is $q^n$, isn't it?

The kernel of the linear map is $$\ker (\theta)=\{\beta \in \mathbb{F}_{q^n}: \theta (\beta )=0\}=\{\beta \in \mathbb{F}_{q^n}: \beta^q-\beta=0\}$$ How many elements does this set have? (Wondering)


After that I want to show that $\ker (Tr)=\{\beta^q-\beta :\beta \in \mathbb{F}_{q^n}\}$.

We have that $$\ker (Tr)=\{a\in \mathbb{F}_{q^n} : Tr(a)=0\}=\left \{a\in \mathbb{F}_{q^n} : \sum_{j=0}^{n-1}a^{q^j}=0\right \}$$ Could you give me a hint how we could continue? (Wondering)

 

[I like Serena]

Hey mathmari!

Let p be a prime such that $q=p^k$ for some k.
Then $\mathbb F_{q^n}$ is a Galois extension of $\mathbb F_p$, isn't it?
And it can be treated as a vector space.
As such $\mathbb F_{q^n}$ has a basis of $kn$ elements, which means that its dimension is $kn$, doesn't it? (Thinking)

And yes, it means that $\dim (\mathbb{F}_{q^n})=\dim(\ker (\theta))+\dim (\text{im} (\theta))$.

Did you already try to figure it out for an example?
Say $q=3$ and $n=2$? (Wondering)

 

[mathmari]

Let p be a prime such that $q=p^k$ for some k.
Then $\mathbb F_{q^n}$ is a Galois extension of $\mathbb F_p$, isn't it?
And it can be treated as a vector space.
As such $\mathbb F_{q^n}$ has a basis of $kn$ elements, which means that its dimension is $kn$, doesn't it? (Thinking)

So do we have that $\mathbb F_{q^n}=\{1, a, a^2, \ldots , a^{q^n-1}\}=\{1, a, a^2, \ldots , a^{(p^k)^n-1}\}=\{1, a, a^2, \ldots , a^{p^{k^n}-1}\}$ ? (Wondering)

Did you already try to figure it out for an example?
Say $q=3$ and $n=2$? (Wondering)

For an element $x$ of $\mathbb{F}_{3^2}$ it holds that $x^{3^2}=x$ and in general an element $x$ of $\mathbb{F}_{q^n}$ satisfies $x^{q^n}=x$, or not? (Wondering) For $\beta \in \mathbb{F}_{q^n}$ we have that $\beta^{q^n}=\beta\Rightarrow \left (\beta^q\right )^n=\beta$

So we get \begin{align*}\beta^q-\beta=0 &\Rightarrow \beta^q-\left (\beta^q\right )^n=0 \\ & \Rightarrow \beta^q\cdot \left (1-\left (\beta^q\right )^{n-1}\right )=0 \\ & \Rightarrow \beta^q=0 \text{ or } 1-\left (\beta^q\right )^{n-1}=0 \\ & \Rightarrow \beta^q=0 \text{ or } \left (\beta^q\right )^{n-1}=1\end{align*}

Is everything correct so far? (Wondering)

 

[I like Serena]

So do we have that $\mathbb F_{q^n}=\{1, a, a^2, \ldots , a^{q^n-1}\}=\{1, a, a^2, \ldots , a^{(p^k)^n-1}\}=\{1, a, a^2, \ldots , a^{p^{k^n}-1}\}$ ?

Shouldn't that be:
$$\text{basis }\mathbb F_{q^n}=\{1, a^p, a^{p^2}, \ldots , a^{q^{n-1}}\}=\{1, a^p, a^{p^2}, \ldots , a^{(p^k)^{n-1}}\}=\{1, a^p, a^{p^2}, \ldots , a^{p^{kn-1}}\}$$
for some element $a$ that is primitive? (Wondering)

For an element $x$ of $\mathbb{F}_{3^2}$ it holds that $x^{3^2}=x$ and in general an element $x$ of $\mathbb{F}_{q^n}$ satisfies $x^{q^n}=x$, or not?

Yes.

For $\beta \in \mathbb{F}_{q^n}$ we have that $\beta^{q^n}=\beta\Rightarrow \left (\beta^q\right )^n=\beta$

Isn't $(\beta^q)^n = \beta^{qn}$ instead? (Wondering)

 

[mathmari]

Shouldn't that be:
$$\text{basis }\mathbb F_{q^n}=\{1, a^p, a^{p^2}, \ldots , a^{q^{n-1}}\}=\{1, a^p, a^{p^2}, \ldots , a^{(p^k)^{n-1}}\}=\{1, a^p, a^{p^2}, \ldots , a^{p^{kn-1}}\}$$
for some element $a$ that is primitive? (Wondering)

Ahh ok!

Isn't $(\beta^q)^n = \beta^{qn}$ instead? (Wondering)

Ahh yes!

For $\beta\in \mathbb{F}_{q^n}$ we have $\beta^{q^n}=\beta\Rightarrow \left (\beta^q\right )^{q^{n-1}}=\beta$, right? (Wondering)

Then we get:
\begin{align*}\beta^q-\beta=0 &\Rightarrow \beta^q-\left (\beta^q\right )^{q^{n-1}}=0 \\ & \Rightarrow \beta^q\left (1-\left (\beta^q\right )^{q^{n-2}}\right )=0 \\ & \Rightarrow \beta^q=0 \ \text{ or } \ \left (\beta^q\right )^{q^{n-2}}=1\end{align*} Do we get from that $\beta=0$ or $\beta=1$ ? (Wondering)

 

[I like Serena]

For $\beta\in \mathbb{F}_{q^n}$ we have $\beta^{q^n}=\beta\Rightarrow \left (\beta^q\right )^{q^{n-1}}=\beta$, right?

Yes. (Nod)

Then we get:
\begin{align*}\beta^q-\beta=0 &\Rightarrow \beta^q-\left (\beta^q\right )^{q^{n-1}}=0 \\ & \Rightarrow \beta^q\left (1-\left (\beta^q\right )^{q^{n-2}}\right )=0 \\ & \Rightarrow \beta^q=0 \ \text{ or } \ \left (\beta^q\right )^{q^{n-2}}=1\end{align*}

Don't we have that:
$$\beta^q\cdot \left (\beta^q\right )^{q^{n-2}} = \left (\beta^q\right )^{q^{n-2}+1}
\ne \left (\beta^q\right )^{q^{n-1}}$$
(Worried)

Do we get from that $\beta=0$ or $\beta=1$ ?

I don't think so.
Let's pick $q=3$ and $n=2$, so that $q^n=9$.
Then $2^3-2\equiv 6 \equiv 0 \pmod 3$, so that $\theta(2)=0$.
Therefore it can't be that the kernel of $\theta$ is just $\{0,1\}$, can it? (Wondering)

 

[mathmari]

Don't we have that:
$$\beta^q\cdot \left (\beta^q\right )^{q^{n-2}} = \left (\beta^q\right )^{q^{n-2}+1}
\ne \left (\beta^q\right )^{q^{n-1}}$$
(Worried)

Ah yes (Worried)

But how can we continue from $\beta^q-\beta=0 \Rightarrow \beta^q-\left (\beta^q\right )^{q^{n-1}}=0$ ? Could you give me a hint? (Wondering)

 

[I like Serena]

But how can we continue from $\beta^q-\beta=0 \Rightarrow \beta^q-\left (\beta^q\right )^{q^{n-1}}=0$ ? Could you give me a hint?

Suppose $\alpha$ is a primitive element of $\mathbb F_{q^n}$.
Then $\mathbb F_{q^n} = \{0,\alpha,\alpha^2,\ldots,\alpha^{q^n-1}\}$, isn't it?
Furthermore, the order of $\alpha$ is $q^n-1$.
That is, $\alpha^{q^n-1}=1$, and $\alpha^i \ne 1$ for all $1\le i < q^n-1$.

We have $\beta^q-\beta=0 \iff \beta(\beta^{q-1}-1)=0 \iff \beta=0 \lor \beta^{q-1}=1$.
For which $\beta =\alpha^i$ will we have that $\beta^{q-1}=1$? (Wondering)

Alternatively, we have $\beta^q-\left (\beta^q\right )^{q^{n-1}}=0$.
And $q$ is co-prime with $q^n-1$ isn't it?
So $\beta\mapsto \beta^q$ is a bijection.
Therefore we can also count the $\gamma=\beta^q$ such that $\gamma-\gamma^{q^{n-1}}=0 \iff \gamma(1-\gamma^{q^{n-1}-1})=0
\iff \gamma=0 \lor \gamma^{q^{n-1}-1}=1$.
For which $\gamma=\alpha^i$ will that be the case? (Wondering)

 

[mathmari]

Suppose $\alpha$ is a primitive element of $\mathbb F_{q^n}$.
Then $\mathbb F_{q^n} = \{0,\alpha,\alpha^2,\ldots,\alpha^{q^n-1}\}$, isn't it?
Furthermore, the order of $\alpha$ is $q^n-1$.
That is, $\alpha^{q^n-1}=1$, and $\alpha^i \ne 1$ for all $1\le i < q^n-1$.

We have $\beta^q-\beta=0 \iff \beta(\beta^{q-1}-1)=0 \iff \beta=0 \lor \beta^{q-1}=1$.
For which $\beta =\alpha^i$ will we have that $\beta^{q-1}=1$? (Wondering)

We have that $\beta^{q-1} =1 \Rightarrow \left(\alpha^i\right )^{q-1}=1 \Rightarrow a^{i(q-1)}=1$. This holds for $1\leq i<q^n-1$ such that $i(q-1)$ is a multiple of $q^n-1$ :

$$i(q-1)=k\cdot (q^n-1)\Rightarrow i= \frac{k}{q-1}\cdot (q^n-1)$$
So we have to find all the values of $k$ such that $$1\leq \frac{k}{q-1}\cdot (q^n-1)<q^n-1$$ That is $1\leq k<q-1$, or not? (Wondering)

Does $i$ have to be an integer or can it be also a rational number? (Wondering)

 

[I like Serena]

We have that $\beta^{q-1} =1 \Rightarrow \left(\alpha^i\right )^{q-1}=1 \Rightarrow a^{i(q-1)}=1$. This holds for $1\leq i<q^n-1$ such that $i(q-1)$ is a multiple of $q^n-1$ :

$$i(q-1)=k\cdot (q^n-1)\Rightarrow i= \frac{k}{q-1}\cdot (q^n-1)$$
So we have to find all the values of $k$ such that $$1\leq \frac{k}{q-1}\cdot (q^n-1)<q^n-1$$ That is $1\leq k<q-1$, or not? (Wondering)

Does $i$ have to be an integer or can it be also a rational number? (Wondering)

We picked $i$ such that the $\alpha^i$ are exactly the elements of $\mathbb F_{q^n}\setminus \{0\}$.
That means $i\in \{1,\ldots,q^{n}-1\}$, so no, $i$ is not rational.
Btw, that is up to and including $i=q^{n}-1$.
(Alternativelly we could pick $i\in \{0,\ldots,q^{n}-2\}$.)

Will $i= \frac{k}{q-1}\cdot (q^n-1)$ be in that set for all values of $1\le k {\color{red}\le} q-1$? (Wondering)

 

[mathmari]

We picked $i$ such that the $\alpha^i$ are exactly the elements of $\mathbb F_{q^n}\setminus \{0\}$.
That means $i\in \{1,\ldots,q^{n}-1\}$, so no, $i$ is not rational.
Btw, that is up to and including $i=q^{n}-1$.
(Alternativelly we could pick $i\in \{0,\ldots,q^{n}-2\}$.)

Ah ok! I see! (Nerd)

Will $i= \frac{k}{q-1}\cdot (q^n-1)$ be in that set for all values of $1\le k {\color{red}\le} q-1$? (Wondering)

For any $k$ in that interval we get $k\cdot \frac{q^n-1}{q-1}=k\cdot \displaystyle{\sum_{i=0}^{n-1}q^i}$ which is an integer, right?

For $1\leq k\leq q-1$ we get $$\frac{q^n-1}{q-1}\leq k \cdot \frac{q^n-1}{q-1}\leq (q-1)\cdot \frac{q^n-1}{q-1} \Rightarrow \sum_{i=0}^{n-1}q^i\leq k \cdot \frac{q^n-1}{q-1}\leq q^n-1$$ So, it is in the set $\{1, \ldots , q^n-1\}$, isn't it?

 

[I like Serena]

For any $k$ in that interval we get $k\cdot \frac{q^n-1}{q-1}=k\cdot \displaystyle{\sum_{i=0}^{n-1}q^i}$ which is an integer, right?

For $1\leq k\leq q-1$ we get $$\frac{q^n-1}{q-1}\leq k \cdot \frac{q^n-1}{q-1}\leq (q-1)\cdot \frac{q^n-1}{q-1} \Rightarrow \sum_{i=0}^{n-1}q^i\leq k \cdot \frac{q^n-1}{q-1}\leq q^n-1$$ So, it is in the set $\{1, \ldots , q^n-1\}$, isn't it?

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

So, the kernel has the element $\beta=0 \Rightarrow \alpha=0$ (1 element) and the elements $\beta^{q-1} =1 \Rightarrow \left(\alpha^i\right )^{q-1}=1 \Rightarrow a^{i(q-1)}=1 \Rightarrow a^{k(q^n-1)}=1$, $1\leq k\leq q-1$ ($q-1$ elements).
So, in total the kernel has $q$ elements.

Is that correct? (Wondering) Therefore we get the following (where $q=p^k$): \begin{align*}\dim (\mathbb{F}_{q^n})=\dim (\ker (\theta))+\dim (\text{im} (\theta)) &\Rightarrow kn=q+\dim (\text{im} (\theta)) \\ & \Rightarrow kn=p^k+\dim (\text{im} (\theta)) \\ & \Rightarrow \dim (\text{im} (\theta))=kn-p^k\end{align*}
Is this correct? (Wondering)

 

[I like Serena]

So, the kernel has the element $\beta=0 \Rightarrow \alpha=0$ (1 element) and the elements $\beta^{q-1} =1 \Rightarrow \left(\alpha^i\right )^{q-1}=1 \Rightarrow a^{i(q-1)}=1 \Rightarrow a^{k(q^n-1)}=1$, $1\leq k\leq q-1$ ($q-1$ elements).
So, in total the kernel has $q$ elements.

Yep.
Although $\alpha\ne 0$. It's supposed to be a primitive element after all.
Note that we cannot write $0$ as $\alpha^i$. The element $0$ is separate. (Nerd)

Therefore we get the following (where $q=p^k$): \begin{align*}\dim (\mathbb{F}_{q^n})=\dim (\ker (\theta))+\dim (\text{im} (\theta)) &\Rightarrow kn=q+\dim (\text{im} (\theta)) \\ & \Rightarrow kn=p^k+\dim (\text{im} (\theta)) \\ & \Rightarrow \dim (\text{im} (\theta))=kn-p^k\end{align*}
Is this correct?

A dimension corresponds to $p$ elements.
So there are $k$ dimensions in $q=p^k$ elements, aren't there? (Thinking)

 

[mathmari]

Yep.
Although $\alpha\ne 0$. It's supposed to be a primitive element after all.
Note that we cannot write $0$ as $\alpha^i$. The element $0$ is separate. (Nerd)

So, the elements that are in $\ker (\theta)$ are the elements of $\mathbb{F}_q$, or not?

That means that $\dim (\ker (\theta ))=\dim (\mathbb{F}_q)$. The dimension of $\mathbb{F}_q$ over $\mathbb{F}_q$ is $1$, or not?

So, do we get that $\dim (\ker (\theta ))=1$ ?

Do we also have that $\dim (\mathbb{F}_{q^n})=n$ over $\mathbb{F}_q$ ?

(Wondering)

 

[I like Serena]

So, the elements that are in $\ker (\theta)$ are the elements of $\mathbb{F}_q$, or not?

That means that $\dim (\ker (\theta ))=\dim (\mathbb{F}_q)$. The dimension of $\mathbb{F}_q$ over $\mathbb{F}_q$ is $1$, or not?

So, do we get that $\dim (\ker (\theta ))=1$ ?

Do we also have that $\dim (\mathbb{F}_{q^n})=n$ over $\mathbb{F}_q$ ?

I think so yes. (Nod)

 

[mathmari]

I think so yes. (Nod)

That means that $$\dim (\mathbb{F}_{q^n})=\dim (\ker (\theta))+\dim (\text{im} (\theta)) \Rightarrow n=1+\dim (\text{im} (\theta)) \Rightarrow \dim (\text{im} (\theta))=n-1$$
Is that correct? (Wondering)
About the second part:
$$\ker (Tr)=\{a\in \mathbb{F}_{q^n} : Tr(a)=0\}=\left \{a\in \mathbb{F}_{q^n} : \sum_{j=0}^{n-1}a^{q^j}=0\right \}$$ How can we show that this is equal to $\{\beta^q-\beta:\beta \in \mathbb{F}_{q^n}\}$ ? (Wondering)

 

[I like Serena]

That means that $$\dim (\mathbb{F}_{q^n})=\dim (\ker (\theta))+\dim (\text{im} (\theta)) \Rightarrow n=1+\dim (\text{im} (\theta)) \Rightarrow \dim (\text{im} (\theta))=n-1$$
Is that correct?

Yep. (Nod)

About the second part:
$$\ker (Tr)=\{a\in \mathbb{F}_{q^n} : Tr(a)=0\}=\left \{a\in \mathbb{F}_{q^n} : \sum_{j=0}^{n-1}a^{q^j}=0\right \}$$ How can we show that this is equal to $\{\beta^q-\beta:\beta \in \mathbb{F}_{q^n}\}$ ?

Suppose we fill in $\beta^q-\beta$ in the formula for $Tr$? (Thinking)

 

[mathmari]

Suppose we fill in $\beta^q-\beta$ in the formula for $Tr$? (Thinking)

Is $Tr$ a linear map? If yes, then we have $Tr(\beta^q-\beta)=Tr(\beta^q)-Tr(\beta)$.

We have the following: \begin{align*}&Tr(\beta^q)=\sum_{j=0}^{n-1}\left (\beta^q\right )^{q^j}=\sum_{j=0}^{n-1}\beta^{q^{j+1}}=\beta^q+\beta^{q^2}+\ldots+\beta^{q^{n-1}}+\beta^{q^n}=\beta^q+\ldots+\beta^{q^{n-1}}+\beta \\ &Tr(\beta)=\sum_{j=0}^{n-1}\beta^{q^j}=\beta+\beta^q+\ldots +\beta^{q^{n-1}}\end{align*} That means that $Tr(\beta^q)=Tr(\beta)$.

So, we get $Tr(\beta^q-\beta)=0$ and so $\beta^q-q\in \ker(Tr)\Rightarrow \text{im}(\theta)\subseteq \ker (Tr)$, right? (Wondering)

Now it is left to get the equality. We know that $\dim (\text{im}(\theta))=n-1$. How can we calculate the dimension of $\ker (Tr)$ ? (Wondering)

 

[I like Serena]

Is $Tr$ a linear map? If yes, then we have $Tr(\beta^q-\beta)=Tr(\beta^q)-Tr(\beta)$.

We'll have to verify!
Do we have for all $a,b\in\mathbb F_{q^n}$ and $\lambda\in\mathbb F_q$ that $Tr(a+b)=Tr(a)+Tr(b)$ and $Tr(\lambda a)=\lambda Tr(a)$? (Wondering)

We have the following: \begin{align*}&Tr(\beta^q)=\sum_{j=0}^{n-1}\left (\beta^q\right )^{q^j}=\sum_{j=0}^{n-1}\beta^{q^{j+1}}=\beta^q+\beta^{q^2}+\ldots+\beta^{q^{n-1}}+\beta^{q^n}=\beta^q+\ldots+\beta^{q^{n-1}}+\beta \\ &Tr(\beta)=\sum_{j=0}^{n-1}\beta^{q^j}=\beta+\beta^q+\ldots +\beta^{q^{n-1}}\end{align*} That means that $Tr(\beta^q)=Tr(\beta)$.

So, we get $Tr(\beta^q-\beta)=0$ and so $\beta^q-q\in \ker(Tr)\Rightarrow \text{im}(\theta)\subseteq \ker (Tr)$, right? (Wondering)

Now it is left to get the equality. We know that $\dim (\text{im}(\theta))=n-1$. How can we calculate the dimension of $\ker (Tr)$ ? (Wondering)

If $Tr$ is a linear map, than $\dim(\text{im}(Tr))$ is either $0$ or $1$ isn't it?
Can we find an element $a$ such that $Tr(a)\ne 0$? (Wondering)

 

[mathmari]

We'll have to verify!
Do we have for all $a,b\in\mathbb F_{q^n}$ and $\lambda\in\mathbb F_q$ that $Tr(a+b)=Tr(a)+Tr(b)$ and $Tr(\lambda a)=\lambda Tr(a)$? (Wondering)

$$T(a+b)=\sum_{j=0}^{n-1}\left (a+b\right )^{q^j}$$ We apply here the binomial theorem and the intermediate terms are $0$ over $\mathbb{F}_q$ and then we get $$\sum_{j=0}^{n-1}\left (a^{q^j}+b^{q^j}\right )=\sum_{j=0}^{n-1}a^{q^j}+\sum_{j=0}^{n-1}b^{q^j}=Tr(a)+Tr(b)$$ right? (Wondering)

Furthermore, $$Tr(\lambda a)=\sum_{j=0}^{n-1}\left (\lambda a\right )^{q^j}=\sum_{j=0}^{n-1}\left (\lambda^{q^j} a^{q^j}\right )$$ Is this equal to $\lambda Tr(a)$ ? (Wondering)

If $Tr$ is a linear map, than $\dim(\text{im}(Tr))$ is either $0$ or $1$ isn't it?
Can we find an element $a$ such that $Tr(a)\ne 0$? (Wondering)

Why is the dimension $0$ or $1$ ? (Wondering)

For $a=1$ we get $Tr(1)=n\neq 0$, don't we? (Wondering)

 

[I like Serena]

$$T(a+b)=\sum_{j=0}^{n-1}\left (a+b\right )^{q^j}$$ We apply here the binomial theorem and the intermediate terms are $0$ over $\mathbb{F}_q$ and then we get $$\sum_{j=0}^{n-1}\left (a^{q^j}+b^{q^j}\right )=\sum_{j=0}^{n-1}a^{q^j}+\sum_{j=0}^{n-1}b^{q^j}=Tr(a)+Tr(b)$$ right? (Wondering)

Furthermore, $$Tr(\lambda a)=\sum_{j=0}^{n-1}\left (\lambda a\right )^{q^j}=\sum_{j=0}^{n-1}\left (\lambda^{q^j} a^{q^j}\right )$$ Is this equal to $\lambda Tr(a)$ ?

Since $\lambda\in \mathbb F_q$, don't we have that $\lambda^q=\lambda$? (Thinking)

Why is the dimension $0$ or $1$ ?

Since $\text{im}(\theta)\subseteq \ker(Tr)$, we have that $\dim(\ker(Tr))\ge \dim(\text{im}(\theta))=n-1$.
And $\dim(\mathbb F_{q^n})=\dim(\ker(Tr)) + \dim(\text{im}(Tr))$, isn't it? (Wondering)

For $a=1$ we get $Tr(1)=n\neq 0$, don't we?

Indeed.
But only if $p\not\mid n$, isn't it? (Wondering)

 

[mathmari]

Since $\lambda\in \mathbb F_q$, don't we have that $\lambda^q=\lambda$? (Thinking)

Oh yes! (Nerd)

Since $\text{im}(\theta)\subseteq \ker(Tr)$, we have that $\dim(\ker(Tr))\ge \dim(\text{im}(\theta))=n-1$.
And $\dim(\mathbb F_{q^n})=\dim(\ker(Tr)) + \dim(\text{im}(Tr))$, isn't it? (Wondering)

Ahh ok, I got it! (Yes)

Indeed.
But only if $p\not\mid n$, isn't it? (Wondering)

Do you mean $q$ instead of $p$, i.e. if $q\not\mid n$ then for $a=1$ we have $Tr(1)=n\neq 0$ ?

If $q\mid n$ what $a$ can we take then?

(Wondering)

 

[I like Serena]

Do you mean $q$ instead of $p$, i.e. if $q\not\mid n$ then for $a=1$ we have $Tr(1)=n\neq 0$ ?

No, I meant $p$, the prime that $q$ is a power of.
The characteristic of a finite field is the number of times that we must add $1$ together until we get $0$, which is this prime $p$.
So if $n$ is a multiple of $p$, then $Tr(1)=0$, regardless of what $q$ is. (Nerd)

If $q\mid n$ what $a$ can we take then?

I don't know yet what we can pick if $p \mid n$. (Sweating)

 

[I like Serena]

$Tr$ is a non-zero polynomial of degree at most $q^{n-1}$, isn't it?
So it has at most $q^{n-1}$ zeroes.
Since there are $q^n$ elements in total, there must be at least an element $a$ with $Tr(a)\ne 0$, mustn't it? (Wondering)