Mixing
The degree of a splitting field over $mathbb{F}_p$ of non-reducible monic polynomial of degree $n$
$begingroup$
Let $finmathbb{F}_p(x)$ be a monic irreducible polynomial, denoting $deg(f)=n$.
I wish to show (if it's true) that $f(x)$'s splitting field is $mathbb{F}_{p^n}$.
I did some manual test for some polynomials till degree $5$, using the expansion field $mathbb{F}_p[x]/langle f(x) rangle$ , and used Frobenius' automorphism to show that if $a$ is a root then $a^p , a^{p^2},a^{p^3}...a^{p^n}$ are unique roots. When $|mathbb{F}_p[x]/langle f(x) rangle| = p^n$.
Is there a way to show that this generally?
Other explanations will be appreciated, however I am not familiar with Galois theory.
abstract-algebra field-theory finite-fields splitting-field
asked Jan 30 at 16:42
dandan
621613
$endgroup$
add a comment |
$begingroup$
Let $finmathbb{F}_p(x)$ be a monic irreducible polynomial, denoting $deg(f)=n$.
I wish to show (if it's true) that $f(x)$'s splitting field is $mathbb{F}_{p^n}$.
I did some manual test for some polynomials till degree $5$, using the expansion field $mathbb{F}_p[x]/langle f(x) rangle$ , and used Frobenius' automorphism to show that if $a$ is a root then $a^p , a^{p^2},a^{p^3}...a^{p^n}$ are unique roots. When $|mathbb{F}_p[x]/langle f(x) rangle| = p^n$.
Is there a way to show that this generally?
Other explanations will be appreciated, however I am not familiar with Galois theory.
abstract-algebra field-theory finite-fields splitting-field
asked Jan 30 at 16:42
dandan
621613
$endgroup$
1
$begingroup$
Read about the finite fields of q=p^n elements, for example at Wikipedia. Here a link "en.wikipedia.org/wiki/Finite_field#Existence_and_uniqueness".
$endgroup$
– AlgebraicallyClosed
Jan 30 at 17:30
add a comment |
$begingroup$
Let $finmathbb{F}_p(x)$ be a monic irreducible polynomial, denoting $deg(f)=n$.
I wish to show (if it's true) that $f(x)$'s splitting field is $mathbb{F}_{p^n}$.
I did some manual test for some polynomials till degree $5$, using the expansion field $mathbb{F}_p[x]/langle f(x) rangle$ , and used Frobenius' automorphism to show that if $a$ is a root then $a^p , a^{p^2},a^{p^3}...a^{p^n}$ are unique roots. When $|mathbb{F}_p[x]/langle f(x) rangle| = p^n$.
Is there a way to show that this generally?
Other explanations will be appreciated, however I am not familiar with Galois theory.
abstract-algebra field-theory finite-fields splitting-field
asked Jan 30 at 16:42
dandan
621613
$endgroup$
Let $finmathbb{F}_p(x)$ be a monic irreducible polynomial, denoting $deg(f)=n$.
I wish to show (if it's true) that $f(x)$'s splitting field is $mathbb{F}_{p^n}$.
I did some manual test for some polynomials till degree $5$, using the expansion field $mathbb{F}_p[x]/langle f(x) rangle$ , and used Frobenius' automorphism to show that if $a$ is a root then $a^p , a^{p^2},a^{p^3}...a^{p^n}$ are unique roots. When $|mathbb{F}_p[x]/langle f(x) rangle| = p^n$.
Is there a way to show that this generally?
Other explanations will be appreciated, however I am not familiar with Galois theory.
abstract-algebra field-theory finite-fields splitting-field
abstract-algebra field-theory finite-fields splitting-field
asked Jan 30 at 16:42
dandan
621613
asked Jan 30 at 16:42
dandan
621613
asked Jan 30 at 16:42
dandan
621613
asked Jan 30 at 16:42
dandan
621613
asked Jan 30 at 16:42
dandan
621613
621613
1
$begingroup$
Read about the finite fields of q=p^n elements, for example at Wikipedia. Here a link "en.wikipedia.org/wiki/Finite_field#Existence_and_uniqueness".
$endgroup$
– AlgebraicallyClosed
Jan 30 at 17:30
add a comment |
1
$begingroup$
Read about the finite fields of q=p^n elements, for example at Wikipedia. Here a link "en.wikipedia.org/wiki/Finite_field#Existence_and_uniqueness".
$endgroup$
– AlgebraicallyClosed
Jan 30 at 17:30
1
1
$begingroup$
Read about the finite fields of q=p^n elements, for example at Wikipedia. Here a link "en.wikipedia.org/wiki/Finite_field#Existence_and_uniqueness".
$endgroup$
– AlgebraicallyClosed
Jan 30 at 17:30
$begingroup$
Read about the finite fields of q=p^n elements, for example at Wikipedia. Here a link "en.wikipedia.org/wiki/Finite_field#Existence_and_uniqueness".
$endgroup$
– AlgebraicallyClosed
Jan 30 at 17:30
add a comment |
1 Answer
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All right, after some reading as proposed in comments, I believe this is the answer, please let me know If I'm wrong:
let $a$ be a root of $f(x)$ in $mathbb{F}_p[x]/langle f rangle$ .
So $|mathbb{F}_p[x]/langle f rangle| = p^n$. Thus it is isomorphic to $mathbb{F}_{p^n}$ in which $x^{p^n}-x$ is splitting.
Thus $a$ is a root of $x^{p^n}-xin mathbb{F}_p[x]$.
$f$ is irreducible thus $f(x) | p(x)$ over $mathbb{F}_p[x]$ because $f$ divides any polynomial with $a$ as a root. $x^{p^n}-x$ is splitting in $mathbb{F}_p[x]/langle f ranglecongmathbb{F}_{p^n}$ thus $f$ must split in $mathbb{F}_p[x]/langle f rangle$.
There is not subfield in which $f$ splits in, because any field containing a roots of $f$ , $mathbb{F}_p(a)$ degree is $n$ and thus include $p^n$ elements.
answered Jan 30 at 16:58
dandan
621613
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All right, after some reading as proposed in comments, I believe this is the answer, please let me know If I'm wrong:
let $a$ be a root of $f(x)$ in $mathbb{F}_p[x]/langle f rangle$ .
So $|mathbb{F}_p[x]/langle f rangle| = p^n$. Thus it is isomorphic to $mathbb{F}_{p^n}$ in which $x^{p^n}-x$ is splitting.
Thus $a$ is a root of $x^{p^n}-xin mathbb{F}_p[x]$.
$f$ is irreducible thus $f(x) | p(x)$ over $mathbb{F}_p[x]$ because $f$ divides any polynomial with $a$ as a root. $x^{p^n}-x$ is splitting in $mathbb{F}_p[x]/langle f ranglecongmathbb{F}_{p^n}$ thus $f$ must split in $mathbb{F}_p[x]/langle f rangle$.
There is not subfield in which $f$ splits in, because any field containing a roots of $f$ , $mathbb{F}_p(a)$ degree is $n$ and thus include $p^n$ elements.
answered Jan 30 at 16:58
dandan
621613
$endgroup$
add a comment |
$begingroup$
All right, after some reading as proposed in comments, I believe this is the answer, please let me know If I'm wrong:
let $a$ be a root of $f(x)$ in $mathbb{F}_p[x]/langle f rangle$ .
So $|mathbb{F}_p[x]/langle f rangle| = p^n$. Thus it is isomorphic to $mathbb{F}_{p^n}$ in which $x^{p^n}-x$ is splitting.
Thus $a$ is a root of $x^{p^n}-xin mathbb{F}_p[x]$.
$f$ is irreducible thus $f(x) | p(x)$ over $mathbb{F}_p[x]$ because $f$ divides any polynomial with $a$ as a root. $x^{p^n}-x$ is splitting in $mathbb{F}_p[x]/langle f ranglecongmathbb{F}_{p^n}$ thus $f$ must split in $mathbb{F}_p[x]/langle f rangle$.
There is not subfield in which $f$ splits in, because any field containing a roots of $f$ , $mathbb{F}_p(a)$ degree is $n$ and thus include $p^n$ elements.
answered Jan 30 at 16:58
dandan
621613
$endgroup$
add a comment |
$begingroup$
All right, after some reading as proposed in comments, I believe this is the answer, please let me know If I'm wrong:
let $a$ be a root of $f(x)$ in $mathbb{F}_p[x]/langle f rangle$ .
So $|mathbb{F}_p[x]/langle f rangle| = p^n$. Thus it is isomorphic to $mathbb{F}_{p^n}$ in which $x^{p^n}-x$ is splitting.
Thus $a$ is a root of $x^{p^n}-xin mathbb{F}_p[x]$.
$f$ is irreducible thus $f(x) | p(x)$ over $mathbb{F}_p[x]$ because $f$ divides any polynomial with $a$ as a root. $x^{p^n}-x$ is splitting in $mathbb{F}_p[x]/langle f ranglecongmathbb{F}_{p^n}$ thus $f$ must split in $mathbb{F}_p[x]/langle f rangle$.
There is not subfield in which $f$ splits in, because any field containing a roots of $f$ , $mathbb{F}_p(a)$ degree is $n$ and thus include $p^n$ elements.
answered Jan 30 at 16:58
dandan
621613
$endgroup$
All right, after some reading as proposed in comments, I believe this is the answer, please let me know If I'm wrong:
let $a$ be a root of $f(x)$ in $mathbb{F}_p[x]/langle f rangle$ .
So $|mathbb{F}_p[x]/langle f rangle| = p^n$. Thus it is isomorphic to $mathbb{F}_{p^n}$ in which $x^{p^n}-x$ is splitting.
Thus $a$ is a root of $x^{p^n}-xin mathbb{F}_p[x]$.
$f$ is irreducible thus $f(x) | p(x)$ over $mathbb{F}_p[x]$ because $f$ divides any polynomial with $a$ as a root. $x^{p^n}-x$ is splitting in $mathbb{F}_p[x]/langle f ranglecongmathbb{F}_{p^n}$ thus $f$ must split in $mathbb{F}_p[x]/langle f rangle$.
There is not subfield in which $f$ splits in, because any field containing a roots of $f$ , $mathbb{F}_p(a)$ degree is $n$ and thus include $p^n$ elements.
answered Jan 30 at 16:58
dandan
621613
edited Jan 30 at 18:55
answered Jan 30 at 16:58
dandan
621613
answered Jan 30 at 16:58
dandan
621613
answered Jan 30 at 16:58
dandan
621613
621613
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Read about the finite fields of q=p^n elements, for example at Wikipedia. Here a link "en.wikipedia.org/wiki/Finite_field#Existence_and_uniqueness".
$endgroup$
– AlgebraicallyClosed
Jan 30 at 17:30