Showing Two Polynomials Have Isomorphic Splitting Fields over $mathbb{F}_p$












0












$begingroup$


Consider the polynomials, $f_1, f_2 in mathbb{F}_p[x]$, given by



$$
begin{align}
f_1 &= x^{2^n} + 1 \
f_2 &= x^{2^{n+1}} - 1
end{align}
$$



How can we show that $f_1$ and $f_2$ have isomorphic splitting fields over $mathbb{F}_p$?




I think I have a bit of a messy way of showing this using the fact that we can write



$$
begin{align}
f_2 &= x^{2^{n+1}} \
&= (x^{2^n} + 1)(x^{2^{n-1}} + 1) + cdots + (x^2 + 1)(x + 1)(x - 1)
end{align}
$$



However, my idea seems quite inelegant and doesn't use the fact that we are working over a finite field, so I feel like I am missing something.



Any ideas would be greatly appreciated. Thanks in advance!






galois-theory finite-fields cyclic-groups splitting-field







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Thinking about roots of unity looks promising...
    $endgroup$
    – Peter Taylor
    Dec 15 '18 at 15:37
















0












$begingroup$


Consider the polynomials, $f_1, f_2 in mathbb{F}_p[x]$, given by



$$
begin{align}
f_1 &= x^{2^n} + 1 \
f_2 &= x^{2^{n+1}} - 1
end{align}
$$



How can we show that $f_1$ and $f_2$ have isomorphic splitting fields over $mathbb{F}_p$?




I think I have a bit of a messy way of showing this using the fact that we can write



$$
begin{align}
f_2 &= x^{2^{n+1}} \
&= (x^{2^n} + 1)(x^{2^{n-1}} + 1) + cdots + (x^2 + 1)(x + 1)(x - 1)
end{align}
$$



However, my idea seems quite inelegant and doesn't use the fact that we are working over a finite field, so I feel like I am missing something.



Any ideas would be greatly appreciated. Thanks in advance!






galois-theory finite-fields cyclic-groups splitting-field







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Thinking about roots of unity looks promising...
    $endgroup$
    – Peter Taylor
    Dec 15 '18 at 15:37














0












0








0


1



$begingroup$


Consider the polynomials, $f_1, f_2 in mathbb{F}_p[x]$, given by



$$
begin{align}
f_1 &= x^{2^n} + 1 \
f_2 &= x^{2^{n+1}} - 1
end{align}
$$



How can we show that $f_1$ and $f_2$ have isomorphic splitting fields over $mathbb{F}_p$?




I think I have a bit of a messy way of showing this using the fact that we can write



$$
begin{align}
f_2 &= x^{2^{n+1}} \
&= (x^{2^n} + 1)(x^{2^{n-1}} + 1) + cdots + (x^2 + 1)(x + 1)(x - 1)
end{align}
$$



However, my idea seems quite inelegant and doesn't use the fact that we are working over a finite field, so I feel like I am missing something.



Any ideas would be greatly appreciated. Thanks in advance!






galois-theory finite-fields cyclic-groups splitting-field







share|cite|improve this question











$endgroup$




Consider the polynomials, $f_1, f_2 in mathbb{F}_p[x]$, given by



$$
begin{align}
f_1 &= x^{2^n} + 1 \
f_2 &= x^{2^{n+1}} - 1
end{align}
$$



How can we show that $f_1$ and $f_2$ have isomorphic splitting fields over $mathbb{F}_p$?




I think I have a bit of a messy way of showing this using the fact that we can write



$$
begin{align}
f_2 &= x^{2^{n+1}} \
&= (x^{2^n} + 1)(x^{2^{n-1}} + 1) + cdots + (x^2 + 1)(x + 1)(x - 1)
end{align}
$$



However, my idea seems quite inelegant and doesn't use the fact that we are working over a finite field, so I feel like I am missing something.



Any ideas would be greatly appreciated. Thanks in advance!






galois-theory finite-fields cyclic-groups splitting-field




galois-theory finite-fields cyclic-groups splitting-field






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 15:52







John Don

















asked Dec 15 '18 at 14:59









John DonJohn Don

337115




337115








  • 1




    $begingroup$
    Thinking about roots of unity looks promising...
    $endgroup$
    – Peter Taylor
    Dec 15 '18 at 15:37














  • 1




    $begingroup$
    Thinking about roots of unity looks promising...
    $endgroup$
    – Peter Taylor
    Dec 15 '18 at 15:37








1




1




$begingroup$
Thinking about roots of unity looks promising...
$endgroup$
– Peter Taylor
Dec 15 '18 at 15:37




$begingroup$
Thinking about roots of unity looks promising...
$endgroup$
– Peter Taylor
Dec 15 '18 at 15:37










1 Answer
1






active

oldest

votes


















2












$begingroup$

Indeed, this hasn't much to do with the fact that we're working over a finite field. If you want, consider the question over an arbitrary field $K$. We may suppose $2 neq 0$, or else there is nothing to prove. Call the corresponding splitting fields $L_1$ and $L_2$. Because $f_1$ divides $f_2$, there is a homomorphism $phi : L_1 to L_2$. To show that it is surjective, take a root $alpha$ of $f_1$ of order $2^{n+1}$. Take any root $beta$ of $f_2$. Then $beta$ has order dividing $2^{n+1}$, so it is a power of $phi(alpha)$. Hence $phi$ is surjective.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry if I'm being a bit dense here, but how do we know that $f_1$ has a root of order $2^{n+1}$? (And thanks for the answer btw!)
    $endgroup$
    – John Don
    Dec 15 '18 at 15:47










  • $begingroup$
    Never mind, I've got it! If $omega$ is a primitive $2^{n+1}$-th root of unity, then (by definition) it has order $2^{n+1}$, and is also a root of $f_1$ (for example, this follows from the definition of a primitive root of unity and the factorization of $f_2$ given in the question).
    $endgroup$
    – John Don
    Dec 15 '18 at 16:01










  • $begingroup$
    Indeed. Alternatively, as $omega^{2^{n+1}}=1$, we have that $(omega^{2^n})^2=1$, so $omega^{2^n} = pm 1$. It can't be $1$, because $omega$ has order $2^{n+1}$, so it must be $-1$.
    $endgroup$
    – Maximum
    Dec 15 '18 at 16:06













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1 Answer
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1 Answer
1






active

oldest

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active

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active

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votes









2












$begingroup$

Indeed, this hasn't much to do with the fact that we're working over a finite field. If you want, consider the question over an arbitrary field $K$. We may suppose $2 neq 0$, or else there is nothing to prove. Call the corresponding splitting fields $L_1$ and $L_2$. Because $f_1$ divides $f_2$, there is a homomorphism $phi : L_1 to L_2$. To show that it is surjective, take a root $alpha$ of $f_1$ of order $2^{n+1}$. Take any root $beta$ of $f_2$. Then $beta$ has order dividing $2^{n+1}$, so it is a power of $phi(alpha)$. Hence $phi$ is surjective.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry if I'm being a bit dense here, but how do we know that $f_1$ has a root of order $2^{n+1}$? (And thanks for the answer btw!)
    $endgroup$
    – John Don
    Dec 15 '18 at 15:47










  • $begingroup$
    Never mind, I've got it! If $omega$ is a primitive $2^{n+1}$-th root of unity, then (by definition) it has order $2^{n+1}$, and is also a root of $f_1$ (for example, this follows from the definition of a primitive root of unity and the factorization of $f_2$ given in the question).
    $endgroup$
    – John Don
    Dec 15 '18 at 16:01










  • $begingroup$
    Indeed. Alternatively, as $omega^{2^{n+1}}=1$, we have that $(omega^{2^n})^2=1$, so $omega^{2^n} = pm 1$. It can't be $1$, because $omega$ has order $2^{n+1}$, so it must be $-1$.
    $endgroup$
    – Maximum
    Dec 15 '18 at 16:06


















2












$begingroup$

Indeed, this hasn't much to do with the fact that we're working over a finite field. If you want, consider the question over an arbitrary field $K$. We may suppose $2 neq 0$, or else there is nothing to prove. Call the corresponding splitting fields $L_1$ and $L_2$. Because $f_1$ divides $f_2$, there is a homomorphism $phi : L_1 to L_2$. To show that it is surjective, take a root $alpha$ of $f_1$ of order $2^{n+1}$. Take any root $beta$ of $f_2$. Then $beta$ has order dividing $2^{n+1}$, so it is a power of $phi(alpha)$. Hence $phi$ is surjective.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry if I'm being a bit dense here, but how do we know that $f_1$ has a root of order $2^{n+1}$? (And thanks for the answer btw!)
    $endgroup$
    – John Don
    Dec 15 '18 at 15:47










  • $begingroup$
    Never mind, I've got it! If $omega$ is a primitive $2^{n+1}$-th root of unity, then (by definition) it has order $2^{n+1}$, and is also a root of $f_1$ (for example, this follows from the definition of a primitive root of unity and the factorization of $f_2$ given in the question).
    $endgroup$
    – John Don
    Dec 15 '18 at 16:01










  • $begingroup$
    Indeed. Alternatively, as $omega^{2^{n+1}}=1$, we have that $(omega^{2^n})^2=1$, so $omega^{2^n} = pm 1$. It can't be $1$, because $omega$ has order $2^{n+1}$, so it must be $-1$.
    $endgroup$
    – Maximum
    Dec 15 '18 at 16:06
















2












2








2





$begingroup$

Indeed, this hasn't much to do with the fact that we're working over a finite field. If you want, consider the question over an arbitrary field $K$. We may suppose $2 neq 0$, or else there is nothing to prove. Call the corresponding splitting fields $L_1$ and $L_2$. Because $f_1$ divides $f_2$, there is a homomorphism $phi : L_1 to L_2$. To show that it is surjective, take a root $alpha$ of $f_1$ of order $2^{n+1}$. Take any root $beta$ of $f_2$. Then $beta$ has order dividing $2^{n+1}$, so it is a power of $phi(alpha)$. Hence $phi$ is surjective.






share|cite|improve this answer









$endgroup$



Indeed, this hasn't much to do with the fact that we're working over a finite field. If you want, consider the question over an arbitrary field $K$. We may suppose $2 neq 0$, or else there is nothing to prove. Call the corresponding splitting fields $L_1$ and $L_2$. Because $f_1$ divides $f_2$, there is a homomorphism $phi : L_1 to L_2$. To show that it is surjective, take a root $alpha$ of $f_1$ of order $2^{n+1}$. Take any root $beta$ of $f_2$. Then $beta$ has order dividing $2^{n+1}$, so it is a power of $phi(alpha)$. Hence $phi$ is surjective.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 15 '18 at 15:34









MaximumMaximum

212




212












  • $begingroup$
    Sorry if I'm being a bit dense here, but how do we know that $f_1$ has a root of order $2^{n+1}$? (And thanks for the answer btw!)
    $endgroup$
    – John Don
    Dec 15 '18 at 15:47










  • $begingroup$
    Never mind, I've got it! If $omega$ is a primitive $2^{n+1}$-th root of unity, then (by definition) it has order $2^{n+1}$, and is also a root of $f_1$ (for example, this follows from the definition of a primitive root of unity and the factorization of $f_2$ given in the question).
    $endgroup$
    – John Don
    Dec 15 '18 at 16:01










  • $begingroup$
    Indeed. Alternatively, as $omega^{2^{n+1}}=1$, we have that $(omega^{2^n})^2=1$, so $omega^{2^n} = pm 1$. It can't be $1$, because $omega$ has order $2^{n+1}$, so it must be $-1$.
    $endgroup$
    – Maximum
    Dec 15 '18 at 16:06




















  • $begingroup$
    Sorry if I'm being a bit dense here, but how do we know that $f_1$ has a root of order $2^{n+1}$? (And thanks for the answer btw!)
    $endgroup$
    – John Don
    Dec 15 '18 at 15:47










  • $begingroup$
    Never mind, I've got it! If $omega$ is a primitive $2^{n+1}$-th root of unity, then (by definition) it has order $2^{n+1}$, and is also a root of $f_1$ (for example, this follows from the definition of a primitive root of unity and the factorization of $f_2$ given in the question).
    $endgroup$
    – John Don
    Dec 15 '18 at 16:01










  • $begingroup$
    Indeed. Alternatively, as $omega^{2^{n+1}}=1$, we have that $(omega^{2^n})^2=1$, so $omega^{2^n} = pm 1$. It can't be $1$, because $omega$ has order $2^{n+1}$, so it must be $-1$.
    $endgroup$
    – Maximum
    Dec 15 '18 at 16:06


















$begingroup$
Sorry if I'm being a bit dense here, but how do we know that $f_1$ has a root of order $2^{n+1}$? (And thanks for the answer btw!)
$endgroup$
– John Don
Dec 15 '18 at 15:47




$begingroup$
Sorry if I'm being a bit dense here, but how do we know that $f_1$ has a root of order $2^{n+1}$? (And thanks for the answer btw!)
$endgroup$
– John Don
Dec 15 '18 at 15:47












$begingroup$
Never mind, I've got it! If $omega$ is a primitive $2^{n+1}$-th root of unity, then (by definition) it has order $2^{n+1}$, and is also a root of $f_1$ (for example, this follows from the definition of a primitive root of unity and the factorization of $f_2$ given in the question).
$endgroup$
– John Don
Dec 15 '18 at 16:01




$begingroup$
Never mind, I've got it! If $omega$ is a primitive $2^{n+1}$-th root of unity, then (by definition) it has order $2^{n+1}$, and is also a root of $f_1$ (for example, this follows from the definition of a primitive root of unity and the factorization of $f_2$ given in the question).
$endgroup$
– John Don
Dec 15 '18 at 16:01












$begingroup$
Indeed. Alternatively, as $omega^{2^{n+1}}=1$, we have that $(omega^{2^n})^2=1$, so $omega^{2^n} = pm 1$. It can't be $1$, because $omega$ has order $2^{n+1}$, so it must be $-1$.
$endgroup$
– Maximum
Dec 15 '18 at 16:06






$begingroup$
Indeed. Alternatively, as $omega^{2^{n+1}}=1$, we have that $(omega^{2^n})^2=1$, so $omega^{2^n} = pm 1$. It can't be $1$, because $omega$ has order $2^{n+1}$, so it must be $-1$.
$endgroup$
– Maximum
Dec 15 '18 at 16:06




















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