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Let $p$ be a prime number, and consider the mapping called the trace $$ Tr \quad : \quad \mathbb{F}_{p^n} \ \longrightarrow \ \mathbb{F}_{p^n} \quad : \quad x \ \longmapsto \ x + x^p + x^{p^2} + \cdots + x^{p^{n-1}}$$ My syllabus Abstract Algebra states the following:

  • Every element $x \in \mathbb{F}_{p^n}$, is mapped to $\mathbb{F}_p$.
  • The restricted mapping $Tr' \ : \ \mathbb{F}_{p^n} \ \rightarrow \mathbb{F}_p $ is surjective.

I failed to prove both these statements, and I ask you for some help.

Research effort for the first statement

We can see $\mathbb{F}_{p^n}$ as a vector space with the scalar field $\mathbb{F}_p$. I showed that $Tr$ is a linear mapping in this setting. I convinced myself that the mapping is not an ismorfism in general by looking at the case where $p=n=2$ and $x\neq 0 \neq y$. Assume that $Tr$ is multiplicative. then $$Tr(xy) \ = \ Tr(x)Tr(y) \quad \iff \quad xy + x^2y^2 \ = \ (x+x^2)(y+y^2) \ = \ x^2y^2 +xy^2+x^2y +xy$$ and by substracting we see $$x^2y+x^2y=0 \quad \iff \quad xy(x+y) = 0 \quad \iff \quad x = -y$$ And this is not generally true in a field with four elements... I had no inspiration to continue some other way.

Research effort for the second statement

I knew that for all elements $x \in \mathbb{F}_p, \ Tr(x) = \sum_{j=1}^n x^{p^j}=\sum_{j=1}^n x$. If this the mapping would be injective here, we would be done since the sum $x+ \cdots +x$ is an element of $\mathbb{F}_p$. If $p$ divides $n$ though, this won't work because every element $x \in \mathbb{F}_p$ would be mapped to 0.

I hope you can provide me hints to prove the statements. Thank you for your time.

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up vote 4 down vote accepted

First: The elements of $F_{p^n}$ that are in $F_p$ are exactly the elements $x$ that satisfy $x^p=x$ (if you know Galois theory, this is the generator of the Galois group of $F_{p^n}$ over $F_p$). Can you prove that for any image of the trace map?

By the way, the trace map is not multiplicative!

Second: The trace is a polynomial of degree $p^{n-1}$ so it can have at most so many zeros. This means that there is an element that has non-zero trace.

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Thank you. I will think about your answer and "accept" when I have found the solution. If not, I might ask you another question. – Koenraad van Duin Nov 29 '13 at 7:36

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