Mixing
The trace map in a finite field.
$begingroup$
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 $xneq 0 neq y$. 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.
abstract-algebra finite-fields
edited Nov 22 '15 at 19:39
darij grinberg
11.2k33167
asked Nov 28 '13 at 21:56
Koenraad van DuinKoenraad van Duin
1,393629
$endgroup$
add a comment |
$begingroup$
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 $xneq 0 neq y$. 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.
abstract-algebra finite-fields
edited Nov 22 '15 at 19:39
darij grinberg
11.2k33167
asked Nov 28 '13 at 21:56
Koenraad van DuinKoenraad van Duin
1,393629
$endgroup$
add a comment |
$begingroup$
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 $xneq 0 neq y$. 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.
abstract-algebra finite-fields
edited Nov 22 '15 at 19:39
darij grinberg
11.2k33167
asked Nov 28 '13 at 21:56
Koenraad van DuinKoenraad van Duin
1,393629
$endgroup$
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 $xneq 0 neq y$. 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.
abstract-algebra finite-fields
abstract-algebra finite-fields
edited Nov 22 '15 at 19:39
darij grinberg
11.2k33167
asked Nov 28 '13 at 21:56
Koenraad van DuinKoenraad van Duin
1,393629
edited Nov 22 '15 at 19:39
darij grinberg
11.2k33167
asked Nov 28 '13 at 21:56
Koenraad van DuinKoenraad van Duin
1,393629
edited Nov 22 '15 at 19:39
darij grinberg
11.2k33167
edited Nov 22 '15 at 19:39
darij grinberg
11.2k33167
edited Nov 22 '15 at 19:39
darij grinberg
11.2k33167
11.2k33167
asked Nov 28 '13 at 21:56
Koenraad van DuinKoenraad van Duin
1,393629
asked Nov 28 '13 at 21:56
Koenraad van DuinKoenraad van Duin
1,393629
asked Nov 28 '13 at 21:56
Koenraad van DuinKoenraad van Duin
1,393629
1,393629
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
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.
answered Nov 28 '13 at 22:59
MichalisNMichalisN
4,5971327
$endgroup$
1
$begingroup$
Thank you. I will think about your answer and "accept" when I have found the solution. If not, I might ask you another question.
$endgroup$
– Koenraad van Duin
Nov 29 '13 at 7:36
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
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votes
$begingroup$
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.
answered Nov 28 '13 at 22:59
MichalisNMichalisN
4,5971327
$endgroup$
1
$begingroup$
Thank you. I will think about your answer and "accept" when I have found the solution. If not, I might ask you another question.
$endgroup$
– Koenraad van Duin
Nov 29 '13 at 7:36
add a comment |
$begingroup$
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.
answered Nov 28 '13 at 22:59
MichalisNMichalisN
4,5971327
$endgroup$
1
$begingroup$
Thank you. I will think about your answer and "accept" when I have found the solution. If not, I might ask you another question.
$endgroup$
– Koenraad van Duin
Nov 29 '13 at 7:36
add a comment |
$begingroup$
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.
answered Nov 28 '13 at 22:59
MichalisNMichalisN
4,5971327
$endgroup$
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.
answered Nov 28 '13 at 22:59
MichalisNMichalisN
4,5971327
answered Nov 28 '13 at 22:59
MichalisNMichalisN
4,5971327
answered Nov 28 '13 at 22:59
MichalisNMichalisN
4,5971327
answered Nov 28 '13 at 22:59
MichalisNMichalisN
4,5971327
4,5971327
1
$begingroup$
Thank you. I will think about your answer and "accept" when I have found the solution. If not, I might ask you another question.
$endgroup$
– Koenraad van Duin
Nov 29 '13 at 7:36
add a comment |
1
$begingroup$
Thank you. I will think about your answer and "accept" when I have found the solution. If not, I might ask you another question.
$endgroup$
– Koenraad van Duin
Nov 29 '13 at 7:36
1
1
$begingroup$
Thank you. I will think about your answer and "accept" when I have found the solution. If not, I might ask you another question.
$endgroup$
– Koenraad van Duin
Nov 29 '13 at 7:36
$begingroup$
Thank you. I will think about your answer and "accept" when I have found the solution. If not, I might ask you another question.
$endgroup$
– Koenraad van Duin
Nov 29 '13 at 7:36
add a comment |
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