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Prove that $ f(z)=z^p-x $ is an irreducible polynomial over the fraction field $ mathbbZ_p(x) $ where $ p $ is a prime number.
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Prove that $ f(z)=z^p-x $ is an irreducible polynomial over the fraction field $ mathbbZ_p(x) $ where $ p $ is a prime number and $ z $ is an indeterminate over the field $ mathbbZ_p(x) $.
A little help here? I get stuck in this.
abstract-algebra
asked Jul 14 at 16:08
Yuchen
751114
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Prove that $ f(z)=z^p-x $ is an irreducible polynomial over the fraction field $ mathbbZ_p(x) $ where $ p $ is a prime number and $ z $ is an indeterminate over the field $ mathbbZ_p(x) $.
A little help here? I get stuck in this.
abstract-algebra
asked Jul 14 at 16:08
Yuchen
751114
1
Is $z$ an element of $mathbbZ_p$? If this is the case, then $z^pequiv_p z$, so you should consider $mathbbZ_p(x)/(z-x)$.
– cansomeonehelpmeout
Jul 14 at 16:14
@cansomeonehelpmeout No, $ z $ is not an element of $ mathbbZ_p $. $ z $ is an indeterminate over the fraction field $ mathbbZ_p(x) $.
– Yuchen
Jul 14 at 16:27
1
If your $BbbZ_p=BbbZ/pBbbZ$, then you can use Eisenstein's criterion with the irreducible element $x$ of the polynomial ring $BbbZ_p[x]$ in the role of a prime. In other words, the same proof that $z^p-2$ is irreducible in $BbbZ[z]$.
– Jyrki Lahtonen
Jul 14 at 19:37
1
Eisenstein does work. You have the wrong domain. Use $D=BbbZ_p[x]$. Certainly $x$ is an irreducible polynomial, and the coefficients of $z^p-xin D[z]$ meet the Eisenstein's criteria. Then you need the counterpart of Gauss' Lemma to conclude that irreducibility over $D$ implies irreducibility over its field of fractions $Q_D$. Here $Q_D=BbbZ_p(x)$, just what the doctor orderer!
– Jyrki Lahtonen
Jul 15 at 4:58
1
Generalized Eisenstein's criterion.
– Jyrki Lahtonen
Jul 15 at 4:59
 |Â
show 4 more comments
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0
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favorite
up vote
0
down vote
favorite
Prove that $ f(z)=z^p-x $ is an irreducible polynomial over the fraction field $ mathbbZ_p(x) $ where $ p $ is a prime number and $ z $ is an indeterminate over the field $ mathbbZ_p(x) $.
A little help here? I get stuck in this.
abstract-algebra
asked Jul 14 at 16:08
Yuchen
751114
Prove that $ f(z)=z^p-x $ is an irreducible polynomial over the fraction field $ mathbbZ_p(x) $ where $ p $ is a prime number and $ z $ is an indeterminate over the field $ mathbbZ_p(x) $.
A little help here? I get stuck in this.
abstract-algebra
asked Jul 14 at 16:08
Yuchen
751114
edited Jul 14 at 16:31
asked Jul 14 at 16:08
Yuchen
751114
asked Jul 14 at 16:08
Yuchen
751114
asked Jul 14 at 16:08
Yuchen
751114
751114
1
Is $z$ an element of $mathbbZ_p$? If this is the case, then $z^pequiv_p z$, so you should consider $mathbbZ_p(x)/(z-x)$.
– cansomeonehelpmeout
Jul 14 at 16:14
@cansomeonehelpmeout No, $ z $ is not an element of $ mathbbZ_p $. $ z $ is an indeterminate over the fraction field $ mathbbZ_p(x) $.
– Yuchen
Jul 14 at 16:27
1
If your $BbbZ_p=BbbZ/pBbbZ$, then you can use Eisenstein's criterion with the irreducible element $x$ of the polynomial ring $BbbZ_p[x]$ in the role of a prime. In other words, the same proof that $z^p-2$ is irreducible in $BbbZ[z]$.
– Jyrki Lahtonen
Jul 14 at 19:37
1
Eisenstein does work. You have the wrong domain. Use $D=BbbZ_p[x]$. Certainly $x$ is an irreducible polynomial, and the coefficients of $z^p-xin D[z]$ meet the Eisenstein's criteria. Then you need the counterpart of Gauss' Lemma to conclude that irreducibility over $D$ implies irreducibility over its field of fractions $Q_D$. Here $Q_D=BbbZ_p(x)$, just what the doctor orderer!
– Jyrki Lahtonen
Jul 15 at 4:58
1
Generalized Eisenstein's criterion.
– Jyrki Lahtonen
Jul 15 at 4:59
 |Â
show 4 more comments
1
Is $z$ an element of $mathbbZ_p$? If this is the case, then $z^pequiv_p z$, so you should consider $mathbbZ_p(x)/(z-x)$.
– cansomeonehelpmeout
Jul 14 at 16:14
@cansomeonehelpmeout No, $ z $ is not an element of $ mathbbZ_p $. $ z $ is an indeterminate over the fraction field $ mathbbZ_p(x) $.
– Yuchen
Jul 14 at 16:27
1
If your $BbbZ_p=BbbZ/pBbbZ$, then you can use Eisenstein's criterion with the irreducible element $x$ of the polynomial ring $BbbZ_p[x]$ in the role of a prime. In other words, the same proof that $z^p-2$ is irreducible in $BbbZ[z]$.
– Jyrki Lahtonen
Jul 14 at 19:37
1
Eisenstein does work. You have the wrong domain. Use $D=BbbZ_p[x]$. Certainly $x$ is an irreducible polynomial, and the coefficients of $z^p-xin D[z]$ meet the Eisenstein's criteria. Then you need the counterpart of Gauss' Lemma to conclude that irreducibility over $D$ implies irreducibility over its field of fractions $Q_D$. Here $Q_D=BbbZ_p(x)$, just what the doctor orderer!
– Jyrki Lahtonen
Jul 15 at 4:58
1
Generalized Eisenstein's criterion.
– Jyrki Lahtonen
Jul 15 at 4:59
1
1
Is $z$ an element of $mathbbZ_p$? If this is the case, then $z^pequiv_p z$, so you should consider $mathbbZ_p(x)/(z-x)$.
– cansomeonehelpmeout
Jul 14 at 16:14
Is $z$ an element of $mathbbZ_p$? If this is the case, then $z^pequiv_p z$, so you should consider $mathbbZ_p(x)/(z-x)$.
– cansomeonehelpmeout
Jul 14 at 16:14
@cansomeonehelpmeout No, $ z $ is not an element of $ mathbbZ_p $. $ z $ is an indeterminate over the fraction field $ mathbbZ_p(x) $.
– Yuchen
Jul 14 at 16:27
@cansomeonehelpmeout No, $ z $ is not an element of $ mathbbZ_p $. $ z $ is an indeterminate over the fraction field $ mathbbZ_p(x) $.
– Yuchen
Jul 14 at 16:27
1
1
If your $BbbZ_p=BbbZ/pBbbZ$, then you can use Eisenstein's criterion with the irreducible element $x$ of the polynomial ring $BbbZ_p[x]$ in the role of a prime. In other words, the same proof that $z^p-2$ is irreducible in $BbbZ[z]$.
– Jyrki Lahtonen
Jul 14 at 19:37
If your $BbbZ_p=BbbZ/pBbbZ$, then you can use Eisenstein's criterion with the irreducible element $x$ of the polynomial ring $BbbZ_p[x]$ in the role of a prime. In other words, the same proof that $z^p-2$ is irreducible in $BbbZ[z]$.
– Jyrki Lahtonen
Jul 14 at 19:37
1
1
Eisenstein does work. You have the wrong domain. Use $D=BbbZ_p[x]$. Certainly $x$ is an irreducible polynomial, and the coefficients of $z^p-xin D[z]$ meet the Eisenstein's criteria. Then you need the counterpart of Gauss' Lemma to conclude that irreducibility over $D$ implies irreducibility over its field of fractions $Q_D$. Here $Q_D=BbbZ_p(x)$, just what the doctor orderer!
– Jyrki Lahtonen
Jul 15 at 4:58
Eisenstein does work. You have the wrong domain. Use $D=BbbZ_p[x]$. Certainly $x$ is an irreducible polynomial, and the coefficients of $z^p-xin D[z]$ meet the Eisenstein's criteria. Then you need the counterpart of Gauss' Lemma to conclude that irreducibility over $D$ implies irreducibility over its field of fractions $Q_D$. Here $Q_D=BbbZ_p(x)$, just what the doctor orderer!
– Jyrki Lahtonen
Jul 15 at 4:58
1
1
Generalized Eisenstein's criterion.
– Jyrki Lahtonen
Jul 15 at 4:59
Generalized Eisenstein's criterion.
– Jyrki Lahtonen
Jul 15 at 4:59
 |Â
show 4 more comments
2 Answers
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up vote
2
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accepted
As Jyrki Lahtonen showed, consider the polynomial $ z^p-x $ in $ (mathbbZ_p[x])[z] $. Since $ x $ is a prime element in the integral domain $ mathbbZ_p[x] $, then by Eisenstein's criterion, $ z^p-x $ is an irreducible polynomial over $ mathbbZ_p[x] $. Furthermore, by Gauss' Lemma, $ z^p-x $ is also an irreducible polynomial over the fraction field $ mathbbZ_p(x) $.
answered Jul 15 at 5:38
Yuchen
751114
add a comment |Â
up vote
0
down vote
By $mathbfZ_p$ I hope you mean the integers mod $p$, commonly written $mathbfZ/p$ or $mathbfF_p$ nowadays, and not the $p$-adic integers.
Unfortunately, I cannot yet comment to figure out what you mean. Regardless, we can do this with simple casework. (The argument is exactly the same, but I named the variables differently just because.)
Case 1: You meant the integers mod $p$
Let $k=mathbfF_p(x)$. Consider the splitting field $L$ of $f(z)=z^p-x$ over $k$. Let $g$ be a primitive $p$th root of unity for $mathbfF_p$. Then the roots of $f(z)$ here are $g^k x^1/p$ for each $0le k < p$, in the field $L=k(g, x^1/p)$.
In a certain sense, the reason why the integers mod $p$ are so nice to work in is the fact that $g$ exists in $mathbfF_p$. That is, $L=k(x^1/p)=mathbfF_p(x^1/p)$. Note that this field is isomorphic to $k$ via the map $x^1/pmapsto x$. Thus, any ring homomorphism $varphi:Lto M$ is completely determined by the image of $x^1/p$, and the maps $varphi(x^1/p)=alpha$ for $alphane 0$ are all ring homomorphisms.
To show $f$ is irreducible, it suffices to show that the Galois group $G(L/k)$ acts transitively on the roots of $f(z)$, meaning that for any roots $alpha, alpha'$, we can always find an automorphism $sigma$ of $L$ fixing $K$ and sending $sigma(alpha)=alpha'$.
(If you don't know Galois theory, the intuition here is that invertible homomorphisms $sigma:Lto L$ that fix $k$ will fix the coefficients of polynomials $q(z)$ in $k[z]$. Hence, if $sigmacirc q = qcirc sigma$, so if $alpha$ is a root of $q$, then so is $sigma(alpha)$. In particular, we know that $f$ has no double roots, so if we could break up $f(z)=q(z)r(z)$, then we would not be able to send any of the roots of $q$ to any of the roots of $r$.)
So pick two roots of $f(z)$, say, $g^jx^1/p$ and $g^k x^1/p$. Let $sigma$ be the homomorphism determined by $sigma(x^1/p)=g^k-jx^1/p$. Then $sigma$ is clearly injective and surjective, so it is an automorphism. Furthermore, since $sigma(x)=x$, $sigma$ fixes $k$, so it is an automorphism of $L/k$. Finally, $sigma(g^jx^1/p) = g^kx^1/p$, as desired.
So $G(L/k)$ indeed acts transitively, and thus $f(z)$ is irreducible.
Case 2: You meant the $p$-adic integers for $p>2$
Let $k=mathbfZ_p(x)$. Consider the splitting field $L$ of $f(z)=z^p-x$ over $k$. Let $zeta$ be a primitive $p$th root of unity for $mathbfZ_p$. Then the roots of $f(z)$ here are $zeta^k x^1/p$ for each $0le k < p$, in the field $L=k(zeta, x^1/p)$.
In a certain sense, the reason why the $p$-adic integers so terrible to work in is the fact that $zeta$ doesn't exist in $mathbfZ_p$. That is, $L=mathbfZ_p(x^1/p, zeta)$. Note that any ring homomorphism $varphi:Lto M$ is completely determined by the images $varphi(x^1/p)$ and $varphi(zeta)$. Such maps exist with the provision that $varphi(x^1/p)ne 0$ and $varphi(zeta)$ is be a primitve $p$th root of unity.
To show $f$ is irreducible, it suffices to show that the Galois group $G(L/k)$ acts transitively on the roots of $f(z)$, meaning that for any roots $alpha, alpha'$, we can always find an automorphism $sigma$ of $L$ fixing $K$ and sending $sigma(alpha)=alpha'$.
(If you don't know Galois theory, the intuition here is that invertible homomorphisms $sigma:Lto L$ that fix $k$ will fix the coefficients of polynomials $q(z)$ in $k[z]$. Hence, if $sigmacirc q = qcirc sigma$, so if $alpha$ is a root of $q$, then so is $sigma(alpha)$. In particular, we know that $f$ has no double roots, so if we could break up $f(z)=q(z)r(z)$, then we would not be able to send any of the roots of $q$ to any of the roots of $r$.)
So pick two roots of $f(z)$, say, $zeta^jx^1/p$ and $zeta^k x^1/p$. Let $sigma$ be the homomorphism determined by $sigma(x^1/p)=zeta^k-jx^1/p$, $sigma(zeta)=zeta$. Then $sigma$ is clearly injective and surjective, so it is an automorphism. Furthermore, since $sigma(x)=x$, $sigma$ fixes $k$, so it is an automorphism of $L/k$. Finally, $sigma(zeta^jx^1/p) = zeta^kx^1/p$, as desired.
So $G(L/k)$ indeed acts transitively, and thus $f(z)$ is irreducible.
Case 3: You meant the $2$-adics
Exact same thing again, but $zeta in k$ just like in case 1. Not that there was actually any difference at all between the cases.
answered Jul 14 at 19:26
Roshan Padaki
111
Thank you! I mean $ mathbbZ_p $ the field $ mathbbF_p $.
– Yuchen
Jul 15 at 0:53
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
As Jyrki Lahtonen showed, consider the polynomial $ z^p-x $ in $ (mathbbZ_p[x])[z] $. Since $ x $ is a prime element in the integral domain $ mathbbZ_p[x] $, then by Eisenstein's criterion, $ z^p-x $ is an irreducible polynomial over $ mathbbZ_p[x] $. Furthermore, by Gauss' Lemma, $ z^p-x $ is also an irreducible polynomial over the fraction field $ mathbbZ_p(x) $.
answered Jul 15 at 5:38
Yuchen
751114
add a comment |Â
up vote
2
down vote
accepted
As Jyrki Lahtonen showed, consider the polynomial $ z^p-x $ in $ (mathbbZ_p[x])[z] $. Since $ x $ is a prime element in the integral domain $ mathbbZ_p[x] $, then by Eisenstein's criterion, $ z^p-x $ is an irreducible polynomial over $ mathbbZ_p[x] $. Furthermore, by Gauss' Lemma, $ z^p-x $ is also an irreducible polynomial over the fraction field $ mathbbZ_p(x) $.
answered Jul 15 at 5:38
Yuchen
751114
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
As Jyrki Lahtonen showed, consider the polynomial $ z^p-x $ in $ (mathbbZ_p[x])[z] $. Since $ x $ is a prime element in the integral domain $ mathbbZ_p[x] $, then by Eisenstein's criterion, $ z^p-x $ is an irreducible polynomial over $ mathbbZ_p[x] $. Furthermore, by Gauss' Lemma, $ z^p-x $ is also an irreducible polynomial over the fraction field $ mathbbZ_p(x) $.
answered Jul 15 at 5:38
Yuchen
751114
As Jyrki Lahtonen showed, consider the polynomial $ z^p-x $ in $ (mathbbZ_p[x])[z] $. Since $ x $ is a prime element in the integral domain $ mathbbZ_p[x] $, then by Eisenstein's criterion, $ z^p-x $ is an irreducible polynomial over $ mathbbZ_p[x] $. Furthermore, by Gauss' Lemma, $ z^p-x $ is also an irreducible polynomial over the fraction field $ mathbbZ_p(x) $.
answered Jul 15 at 5:38
Yuchen
751114
answered Jul 15 at 5:38
Yuchen
751114
answered Jul 15 at 5:38
Yuchen
751114
answered Jul 15 at 5:38
Yuchen
751114
751114
add a comment |Â
add a comment |Â
up vote
0
down vote
By $mathbfZ_p$ I hope you mean the integers mod $p$, commonly written $mathbfZ/p$ or $mathbfF_p$ nowadays, and not the $p$-adic integers.
Unfortunately, I cannot yet comment to figure out what you mean. Regardless, we can do this with simple casework. (The argument is exactly the same, but I named the variables differently just because.)
Case 1: You meant the integers mod $p$
Let $k=mathbfF_p(x)$. Consider the splitting field $L$ of $f(z)=z^p-x$ over $k$. Let $g$ be a primitive $p$th root of unity for $mathbfF_p$. Then the roots of $f(z)$ here are $g^k x^1/p$ for each $0le k < p$, in the field $L=k(g, x^1/p)$.
In a certain sense, the reason why the integers mod $p$ are so nice to work in is the fact that $g$ exists in $mathbfF_p$. That is, $L=k(x^1/p)=mathbfF_p(x^1/p)$. Note that this field is isomorphic to $k$ via the map $x^1/pmapsto x$. Thus, any ring homomorphism $varphi:Lto M$ is completely determined by the image of $x^1/p$, and the maps $varphi(x^1/p)=alpha$ for $alphane 0$ are all ring homomorphisms.
To show $f$ is irreducible, it suffices to show that the Galois group $G(L/k)$ acts transitively on the roots of $f(z)$, meaning that for any roots $alpha, alpha'$, we can always find an automorphism $sigma$ of $L$ fixing $K$ and sending $sigma(alpha)=alpha'$.
(If you don't know Galois theory, the intuition here is that invertible homomorphisms $sigma:Lto L$ that fix $k$ will fix the coefficients of polynomials $q(z)$ in $k[z]$. Hence, if $sigmacirc q = qcirc sigma$, so if $alpha$ is a root of $q$, then so is $sigma(alpha)$. In particular, we know that $f$ has no double roots, so if we could break up $f(z)=q(z)r(z)$, then we would not be able to send any of the roots of $q$ to any of the roots of $r$.)
So pick two roots of $f(z)$, say, $g^jx^1/p$ and $g^k x^1/p$. Let $sigma$ be the homomorphism determined by $sigma(x^1/p)=g^k-jx^1/p$. Then $sigma$ is clearly injective and surjective, so it is an automorphism. Furthermore, since $sigma(x)=x$, $sigma$ fixes $k$, so it is an automorphism of $L/k$. Finally, $sigma(g^jx^1/p) = g^kx^1/p$, as desired.
So $G(L/k)$ indeed acts transitively, and thus $f(z)$ is irreducible.
Case 2: You meant the $p$-adic integers for $p>2$
Let $k=mathbfZ_p(x)$. Consider the splitting field $L$ of $f(z)=z^p-x$ over $k$. Let $zeta$ be a primitive $p$th root of unity for $mathbfZ_p$. Then the roots of $f(z)$ here are $zeta^k x^1/p$ for each $0le k < p$, in the field $L=k(zeta, x^1/p)$.
In a certain sense, the reason why the $p$-adic integers so terrible to work in is the fact that $zeta$ doesn't exist in $mathbfZ_p$. That is, $L=mathbfZ_p(x^1/p, zeta)$. Note that any ring homomorphism $varphi:Lto M$ is completely determined by the images $varphi(x^1/p)$ and $varphi(zeta)$. Such maps exist with the provision that $varphi(x^1/p)ne 0$ and $varphi(zeta)$ is be a primitve $p$th root of unity.
To show $f$ is irreducible, it suffices to show that the Galois group $G(L/k)$ acts transitively on the roots of $f(z)$, meaning that for any roots $alpha, alpha'$, we can always find an automorphism $sigma$ of $L$ fixing $K$ and sending $sigma(alpha)=alpha'$.
(If you don't know Galois theory, the intuition here is that invertible homomorphisms $sigma:Lto L$ that fix $k$ will fix the coefficients of polynomials $q(z)$ in $k[z]$. Hence, if $sigmacirc q = qcirc sigma$, so if $alpha$ is a root of $q$, then so is $sigma(alpha)$. In particular, we know that $f$ has no double roots, so if we could break up $f(z)=q(z)r(z)$, then we would not be able to send any of the roots of $q$ to any of the roots of $r$.)
So pick two roots of $f(z)$, say, $zeta^jx^1/p$ and $zeta^k x^1/p$. Let $sigma$ be the homomorphism determined by $sigma(x^1/p)=zeta^k-jx^1/p$, $sigma(zeta)=zeta$. Then $sigma$ is clearly injective and surjective, so it is an automorphism. Furthermore, since $sigma(x)=x$, $sigma$ fixes $k$, so it is an automorphism of $L/k$. Finally, $sigma(zeta^jx^1/p) = zeta^kx^1/p$, as desired.
So $G(L/k)$ indeed acts transitively, and thus $f(z)$ is irreducible.
Case 3: You meant the $2$-adics
Exact same thing again, but $zeta in k$ just like in case 1. Not that there was actually any difference at all between the cases.
answered Jul 14 at 19:26
Roshan Padaki
111
Thank you! I mean $ mathbbZ_p $ the field $ mathbbF_p $.
– Yuchen
Jul 15 at 0:53
add a comment |Â
up vote
0
down vote
By $mathbfZ_p$ I hope you mean the integers mod $p$, commonly written $mathbfZ/p$ or $mathbfF_p$ nowadays, and not the $p$-adic integers.
Unfortunately, I cannot yet comment to figure out what you mean. Regardless, we can do this with simple casework. (The argument is exactly the same, but I named the variables differently just because.)
Case 1: You meant the integers mod $p$
Let $k=mathbfF_p(x)$. Consider the splitting field $L$ of $f(z)=z^p-x$ over $k$. Let $g$ be a primitive $p$th root of unity for $mathbfF_p$. Then the roots of $f(z)$ here are $g^k x^1/p$ for each $0le k < p$, in the field $L=k(g, x^1/p)$.
In a certain sense, the reason why the integers mod $p$ are so nice to work in is the fact that $g$ exists in $mathbfF_p$. That is, $L=k(x^1/p)=mathbfF_p(x^1/p)$. Note that this field is isomorphic to $k$ via the map $x^1/pmapsto x$. Thus, any ring homomorphism $varphi:Lto M$ is completely determined by the image of $x^1/p$, and the maps $varphi(x^1/p)=alpha$ for $alphane 0$ are all ring homomorphisms.
To show $f$ is irreducible, it suffices to show that the Galois group $G(L/k)$ acts transitively on the roots of $f(z)$, meaning that for any roots $alpha, alpha'$, we can always find an automorphism $sigma$ of $L$ fixing $K$ and sending $sigma(alpha)=alpha'$.
(If you don't know Galois theory, the intuition here is that invertible homomorphisms $sigma:Lto L$ that fix $k$ will fix the coefficients of polynomials $q(z)$ in $k[z]$. Hence, if $sigmacirc q = qcirc sigma$, so if $alpha$ is a root of $q$, then so is $sigma(alpha)$. In particular, we know that $f$ has no double roots, so if we could break up $f(z)=q(z)r(z)$, then we would not be able to send any of the roots of $q$ to any of the roots of $r$.)
So pick two roots of $f(z)$, say, $g^jx^1/p$ and $g^k x^1/p$. Let $sigma$ be the homomorphism determined by $sigma(x^1/p)=g^k-jx^1/p$. Then $sigma$ is clearly injective and surjective, so it is an automorphism. Furthermore, since $sigma(x)=x$, $sigma$ fixes $k$, so it is an automorphism of $L/k$. Finally, $sigma(g^jx^1/p) = g^kx^1/p$, as desired.
So $G(L/k)$ indeed acts transitively, and thus $f(z)$ is irreducible.
Case 2: You meant the $p$-adic integers for $p>2$
Let $k=mathbfZ_p(x)$. Consider the splitting field $L$ of $f(z)=z^p-x$ over $k$. Let $zeta$ be a primitive $p$th root of unity for $mathbfZ_p$. Then the roots of $f(z)$ here are $zeta^k x^1/p$ for each $0le k < p$, in the field $L=k(zeta, x^1/p)$.
In a certain sense, the reason why the $p$-adic integers so terrible to work in is the fact that $zeta$ doesn't exist in $mathbfZ_p$. That is, $L=mathbfZ_p(x^1/p, zeta)$. Note that any ring homomorphism $varphi:Lto M$ is completely determined by the images $varphi(x^1/p)$ and $varphi(zeta)$. Such maps exist with the provision that $varphi(x^1/p)ne 0$ and $varphi(zeta)$ is be a primitve $p$th root of unity.
To show $f$ is irreducible, it suffices to show that the Galois group $G(L/k)$ acts transitively on the roots of $f(z)$, meaning that for any roots $alpha, alpha'$, we can always find an automorphism $sigma$ of $L$ fixing $K$ and sending $sigma(alpha)=alpha'$.
(If you don't know Galois theory, the intuition here is that invertible homomorphisms $sigma:Lto L$ that fix $k$ will fix the coefficients of polynomials $q(z)$ in $k[z]$. Hence, if $sigmacirc q = qcirc sigma$, so if $alpha$ is a root of $q$, then so is $sigma(alpha)$. In particular, we know that $f$ has no double roots, so if we could break up $f(z)=q(z)r(z)$, then we would not be able to send any of the roots of $q$ to any of the roots of $r$.)
So pick two roots of $f(z)$, say, $zeta^jx^1/p$ and $zeta^k x^1/p$. Let $sigma$ be the homomorphism determined by $sigma(x^1/p)=zeta^k-jx^1/p$, $sigma(zeta)=zeta$. Then $sigma$ is clearly injective and surjective, so it is an automorphism. Furthermore, since $sigma(x)=x$, $sigma$ fixes $k$, so it is an automorphism of $L/k$. Finally, $sigma(zeta^jx^1/p) = zeta^kx^1/p$, as desired.
So $G(L/k)$ indeed acts transitively, and thus $f(z)$ is irreducible.
Case 3: You meant the $2$-adics
Exact same thing again, but $zeta in k$ just like in case 1. Not that there was actually any difference at all between the cases.
answered Jul 14 at 19:26
Roshan Padaki
111
Thank you! I mean $ mathbbZ_p $ the field $ mathbbF_p $.
– Yuchen
Jul 15 at 0:53
add a comment |Â
up vote
0
down vote
up vote
0
down vote
By $mathbfZ_p$ I hope you mean the integers mod $p$, commonly written $mathbfZ/p$ or $mathbfF_p$ nowadays, and not the $p$-adic integers.
Unfortunately, I cannot yet comment to figure out what you mean. Regardless, we can do this with simple casework. (The argument is exactly the same, but I named the variables differently just because.)
Case 1: You meant the integers mod $p$
Let $k=mathbfF_p(x)$. Consider the splitting field $L$ of $f(z)=z^p-x$ over $k$. Let $g$ be a primitive $p$th root of unity for $mathbfF_p$. Then the roots of $f(z)$ here are $g^k x^1/p$ for each $0le k < p$, in the field $L=k(g, x^1/p)$.
In a certain sense, the reason why the integers mod $p$ are so nice to work in is the fact that $g$ exists in $mathbfF_p$. That is, $L=k(x^1/p)=mathbfF_p(x^1/p)$. Note that this field is isomorphic to $k$ via the map $x^1/pmapsto x$. Thus, any ring homomorphism $varphi:Lto M$ is completely determined by the image of $x^1/p$, and the maps $varphi(x^1/p)=alpha$ for $alphane 0$ are all ring homomorphisms.
To show $f$ is irreducible, it suffices to show that the Galois group $G(L/k)$ acts transitively on the roots of $f(z)$, meaning that for any roots $alpha, alpha'$, we can always find an automorphism $sigma$ of $L$ fixing $K$ and sending $sigma(alpha)=alpha'$.
(If you don't know Galois theory, the intuition here is that invertible homomorphisms $sigma:Lto L$ that fix $k$ will fix the coefficients of polynomials $q(z)$ in $k[z]$. Hence, if $sigmacirc q = qcirc sigma$, so if $alpha$ is a root of $q$, then so is $sigma(alpha)$. In particular, we know that $f$ has no double roots, so if we could break up $f(z)=q(z)r(z)$, then we would not be able to send any of the roots of $q$ to any of the roots of $r$.)
So pick two roots of $f(z)$, say, $g^jx^1/p$ and $g^k x^1/p$. Let $sigma$ be the homomorphism determined by $sigma(x^1/p)=g^k-jx^1/p$. Then $sigma$ is clearly injective and surjective, so it is an automorphism. Furthermore, since $sigma(x)=x$, $sigma$ fixes $k$, so it is an automorphism of $L/k$. Finally, $sigma(g^jx^1/p) = g^kx^1/p$, as desired.
So $G(L/k)$ indeed acts transitively, and thus $f(z)$ is irreducible.
Case 2: You meant the $p$-adic integers for $p>2$
Let $k=mathbfZ_p(x)$. Consider the splitting field $L$ of $f(z)=z^p-x$ over $k$. Let $zeta$ be a primitive $p$th root of unity for $mathbfZ_p$. Then the roots of $f(z)$ here are $zeta^k x^1/p$ for each $0le k < p$, in the field $L=k(zeta, x^1/p)$.
In a certain sense, the reason why the $p$-adic integers so terrible to work in is the fact that $zeta$ doesn't exist in $mathbfZ_p$. That is, $L=mathbfZ_p(x^1/p, zeta)$. Note that any ring homomorphism $varphi:Lto M$ is completely determined by the images $varphi(x^1/p)$ and $varphi(zeta)$. Such maps exist with the provision that $varphi(x^1/p)ne 0$ and $varphi(zeta)$ is be a primitve $p$th root of unity.
To show $f$ is irreducible, it suffices to show that the Galois group $G(L/k)$ acts transitively on the roots of $f(z)$, meaning that for any roots $alpha, alpha'$, we can always find an automorphism $sigma$ of $L$ fixing $K$ and sending $sigma(alpha)=alpha'$.
(If you don't know Galois theory, the intuition here is that invertible homomorphisms $sigma:Lto L$ that fix $k$ will fix the coefficients of polynomials $q(z)$ in $k[z]$. Hence, if $sigmacirc q = qcirc sigma$, so if $alpha$ is a root of $q$, then so is $sigma(alpha)$. In particular, we know that $f$ has no double roots, so if we could break up $f(z)=q(z)r(z)$, then we would not be able to send any of the roots of $q$ to any of the roots of $r$.)
So pick two roots of $f(z)$, say, $zeta^jx^1/p$ and $zeta^k x^1/p$. Let $sigma$ be the homomorphism determined by $sigma(x^1/p)=zeta^k-jx^1/p$, $sigma(zeta)=zeta$. Then $sigma$ is clearly injective and surjective, so it is an automorphism. Furthermore, since $sigma(x)=x$, $sigma$ fixes $k$, so it is an automorphism of $L/k$. Finally, $sigma(zeta^jx^1/p) = zeta^kx^1/p$, as desired.
So $G(L/k)$ indeed acts transitively, and thus $f(z)$ is irreducible.
Case 3: You meant the $2$-adics
Exact same thing again, but $zeta in k$ just like in case 1. Not that there was actually any difference at all between the cases.
answered Jul 14 at 19:26
Roshan Padaki
111
By $mathbfZ_p$ I hope you mean the integers mod $p$, commonly written $mathbfZ/p$ or $mathbfF_p$ nowadays, and not the $p$-adic integers.
Unfortunately, I cannot yet comment to figure out what you mean. Regardless, we can do this with simple casework. (The argument is exactly the same, but I named the variables differently just because.)
Case 1: You meant the integers mod $p$
Let $k=mathbfF_p(x)$. Consider the splitting field $L$ of $f(z)=z^p-x$ over $k$. Let $g$ be a primitive $p$th root of unity for $mathbfF_p$. Then the roots of $f(z)$ here are $g^k x^1/p$ for each $0le k < p$, in the field $L=k(g, x^1/p)$.
In a certain sense, the reason why the integers mod $p$ are so nice to work in is the fact that $g$ exists in $mathbfF_p$. That is, $L=k(x^1/p)=mathbfF_p(x^1/p)$. Note that this field is isomorphic to $k$ via the map $x^1/pmapsto x$. Thus, any ring homomorphism $varphi:Lto M$ is completely determined by the image of $x^1/p$, and the maps $varphi(x^1/p)=alpha$ for $alphane 0$ are all ring homomorphisms.
To show $f$ is irreducible, it suffices to show that the Galois group $G(L/k)$ acts transitively on the roots of $f(z)$, meaning that for any roots $alpha, alpha'$, we can always find an automorphism $sigma$ of $L$ fixing $K$ and sending $sigma(alpha)=alpha'$.
(If you don't know Galois theory, the intuition here is that invertible homomorphisms $sigma:Lto L$ that fix $k$ will fix the coefficients of polynomials $q(z)$ in $k[z]$. Hence, if $sigmacirc q = qcirc sigma$, so if $alpha$ is a root of $q$, then so is $sigma(alpha)$. In particular, we know that $f$ has no double roots, so if we could break up $f(z)=q(z)r(z)$, then we would not be able to send any of the roots of $q$ to any of the roots of $r$.)
So pick two roots of $f(z)$, say, $g^jx^1/p$ and $g^k x^1/p$. Let $sigma$ be the homomorphism determined by $sigma(x^1/p)=g^k-jx^1/p$. Then $sigma$ is clearly injective and surjective, so it is an automorphism. Furthermore, since $sigma(x)=x$, $sigma$ fixes $k$, so it is an automorphism of $L/k$. Finally, $sigma(g^jx^1/p) = g^kx^1/p$, as desired.
So $G(L/k)$ indeed acts transitively, and thus $f(z)$ is irreducible.
Case 2: You meant the $p$-adic integers for $p>2$
Let $k=mathbfZ_p(x)$. Consider the splitting field $L$ of $f(z)=z^p-x$ over $k$. Let $zeta$ be a primitive $p$th root of unity for $mathbfZ_p$. Then the roots of $f(z)$ here are $zeta^k x^1/p$ for each $0le k < p$, in the field $L=k(zeta, x^1/p)$.
In a certain sense, the reason why the $p$-adic integers so terrible to work in is the fact that $zeta$ doesn't exist in $mathbfZ_p$. That is, $L=mathbfZ_p(x^1/p, zeta)$. Note that any ring homomorphism $varphi:Lto M$ is completely determined by the images $varphi(x^1/p)$ and $varphi(zeta)$. Such maps exist with the provision that $varphi(x^1/p)ne 0$ and $varphi(zeta)$ is be a primitve $p$th root of unity.
To show $f$ is irreducible, it suffices to show that the Galois group $G(L/k)$ acts transitively on the roots of $f(z)$, meaning that for any roots $alpha, alpha'$, we can always find an automorphism $sigma$ of $L$ fixing $K$ and sending $sigma(alpha)=alpha'$.
(If you don't know Galois theory, the intuition here is that invertible homomorphisms $sigma:Lto L$ that fix $k$ will fix the coefficients of polynomials $q(z)$ in $k[z]$. Hence, if $sigmacirc q = qcirc sigma$, so if $alpha$ is a root of $q$, then so is $sigma(alpha)$. In particular, we know that $f$ has no double roots, so if we could break up $f(z)=q(z)r(z)$, then we would not be able to send any of the roots of $q$ to any of the roots of $r$.)
So pick two roots of $f(z)$, say, $zeta^jx^1/p$ and $zeta^k x^1/p$. Let $sigma$ be the homomorphism determined by $sigma(x^1/p)=zeta^k-jx^1/p$, $sigma(zeta)=zeta$. Then $sigma$ is clearly injective and surjective, so it is an automorphism. Furthermore, since $sigma(x)=x$, $sigma$ fixes $k$, so it is an automorphism of $L/k$. Finally, $sigma(zeta^jx^1/p) = zeta^kx^1/p$, as desired.
So $G(L/k)$ indeed acts transitively, and thus $f(z)$ is irreducible.
Case 3: You meant the $2$-adics
Exact same thing again, but $zeta in k$ just like in case 1. Not that there was actually any difference at all between the cases.
answered Jul 14 at 19:26
Roshan Padaki
111
answered Jul 14 at 19:26
Roshan Padaki
111
answered Jul 14 at 19:26
Roshan Padaki
111
answered Jul 14 at 19:26
Roshan Padaki
111
111
Thank you! I mean $ mathbbZ_p $ the field $ mathbbF_p $.
– Yuchen
Jul 15 at 0:53
add a comment |Â
Thank you! I mean $ mathbbZ_p $ the field $ mathbbF_p $.
– Yuchen
Jul 15 at 0:53
Thank you! I mean $ mathbbZ_p $ the field $ mathbbF_p $.
– Yuchen
Jul 15 at 0:53
Thank you! I mean $ mathbbZ_p $ the field $ mathbbF_p $.
– Yuchen
Jul 15 at 0:53
add a comment |Â
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1
Is $z$ an element of $mathbbZ_p$? If this is the case, then $z^pequiv_p z$, so you should consider $mathbbZ_p(x)/(z-x)$.
– cansomeonehelpmeout
Jul 14 at 16:14
@cansomeonehelpmeout No, $ z $ is not an element of $ mathbbZ_p $. $ z $ is an indeterminate over the fraction field $ mathbbZ_p(x) $.
– Yuchen
Jul 14 at 16:27
1
If your $BbbZ_p=BbbZ/pBbbZ$, then you can use Eisenstein's criterion with the irreducible element $x$ of the polynomial ring $BbbZ_p[x]$ in the role of a prime. In other words, the same proof that $z^p-2$ is irreducible in $BbbZ[z]$.
– Jyrki Lahtonen
Jul 14 at 19:37
1
Eisenstein does work. You have the wrong domain. Use $D=BbbZ_p[x]$. Certainly $x$ is an irreducible polynomial, and the coefficients of $z^p-xin D[z]$ meet the Eisenstein's criteria. Then you need the counterpart of Gauss' Lemma to conclude that irreducibility over $D$ implies irreducibility over its field of fractions $Q_D$. Here $Q_D=BbbZ_p(x)$, just what the doctor orderer!
– Jyrki Lahtonen
Jul 15 at 4:58
1
Generalized Eisenstein's criterion.
– Jyrki Lahtonen
Jul 15 at 4:59