$f(X):= X^n- xin K[X]$ is irreducible and $Gal(K(y)/K)cong mathbb{Z}/nmathbb{Z}$ if $y$ is a root of $f$
$begingroup$
Hello I need help with solving the following task
Let $K$ be a field with algebraic closure $overline{K}$ and $K^{times}:=Kbackslash {0}$. Let $2le n in mathbb{Z}$ such that $K^{times}$ contains a primitive nth root of unity. If $charK=p>0$ assume that $pnmid n$ holds and $K^{times n}:={y^{n} mid yin K^{times}}$.
i) Let $xin K^{times}$ be an element with the property that if $0<iin mathbb{Z}$ and $yin K^{times}$ exists such that $x^{i} =y^n$ then $nmid i $ holds. Show that the polynomial $f(X):= X^n- xin K[X]$ is irreducible.
ii) Now let $yin overline{K}$ be a root of $f$. Show that $Ksubset K(y)$ is a galois extension with $Gal(K(y)/K)cong mathbb{Z}/nmathbb{Z}$.
Thanks in advance for any help.
abstract-algebra galois-theory irreducible-polynomials
$endgroup$
add a comment |
$begingroup$
Hello I need help with solving the following task
Let $K$ be a field with algebraic closure $overline{K}$ and $K^{times}:=Kbackslash {0}$. Let $2le n in mathbb{Z}$ such that $K^{times}$ contains a primitive nth root of unity. If $charK=p>0$ assume that $pnmid n$ holds and $K^{times n}:={y^{n} mid yin K^{times}}$.
i) Let $xin K^{times}$ be an element with the property that if $0<iin mathbb{Z}$ and $yin K^{times}$ exists such that $x^{i} =y^n$ then $nmid i $ holds. Show that the polynomial $f(X):= X^n- xin K[X]$ is irreducible.
ii) Now let $yin overline{K}$ be a root of $f$. Show that $Ksubset K(y)$ is a galois extension with $Gal(K(y)/K)cong mathbb{Z}/nmathbb{Z}$.
Thanks in advance for any help.
abstract-algebra galois-theory irreducible-polynomials
$endgroup$
$begingroup$
The minimal polynomial of $y=x^{1/n}$ is $F(X) = prod_{j=1}^l (X-zeta_n^{m_j} y)$ where the $m_j$ are a subset of $1ldots n$. $F(0) = ?$
$endgroup$
– reuns
Jan 16 at 3:05
$begingroup$
$F(0) = prod_{j=1}^l (-zeta_n^{m_j} y)$? Is the aim to show that that minimal polynomial is the same as the function $f$ already?
$endgroup$
– Anzu
Jan 16 at 11:45
$begingroup$
$y = x^{1/n}$ and product of roots of unity stays a root of unity so $F(0) = ?$
$endgroup$
– reuns
Jan 16 at 12:06
add a comment |
$begingroup$
Hello I need help with solving the following task
Let $K$ be a field with algebraic closure $overline{K}$ and $K^{times}:=Kbackslash {0}$. Let $2le n in mathbb{Z}$ such that $K^{times}$ contains a primitive nth root of unity. If $charK=p>0$ assume that $pnmid n$ holds and $K^{times n}:={y^{n} mid yin K^{times}}$.
i) Let $xin K^{times}$ be an element with the property that if $0<iin mathbb{Z}$ and $yin K^{times}$ exists such that $x^{i} =y^n$ then $nmid i $ holds. Show that the polynomial $f(X):= X^n- xin K[X]$ is irreducible.
ii) Now let $yin overline{K}$ be a root of $f$. Show that $Ksubset K(y)$ is a galois extension with $Gal(K(y)/K)cong mathbb{Z}/nmathbb{Z}$.
Thanks in advance for any help.
abstract-algebra galois-theory irreducible-polynomials
$endgroup$
Hello I need help with solving the following task
Let $K$ be a field with algebraic closure $overline{K}$ and $K^{times}:=Kbackslash {0}$. Let $2le n in mathbb{Z}$ such that $K^{times}$ contains a primitive nth root of unity. If $charK=p>0$ assume that $pnmid n$ holds and $K^{times n}:={y^{n} mid yin K^{times}}$.
i) Let $xin K^{times}$ be an element with the property that if $0<iin mathbb{Z}$ and $yin K^{times}$ exists such that $x^{i} =y^n$ then $nmid i $ holds. Show that the polynomial $f(X):= X^n- xin K[X]$ is irreducible.
ii) Now let $yin overline{K}$ be a root of $f$. Show that $Ksubset K(y)$ is a galois extension with $Gal(K(y)/K)cong mathbb{Z}/nmathbb{Z}$.
Thanks in advance for any help.
abstract-algebra galois-theory irreducible-polynomials
abstract-algebra galois-theory irreducible-polynomials
edited Mar 16 at 11:20
Sil
5,71621746
5,71621746
asked Jan 15 at 22:17
AnzuAnzu
246
246
$begingroup$
The minimal polynomial of $y=x^{1/n}$ is $F(X) = prod_{j=1}^l (X-zeta_n^{m_j} y)$ where the $m_j$ are a subset of $1ldots n$. $F(0) = ?$
$endgroup$
– reuns
Jan 16 at 3:05
$begingroup$
$F(0) = prod_{j=1}^l (-zeta_n^{m_j} y)$? Is the aim to show that that minimal polynomial is the same as the function $f$ already?
$endgroup$
– Anzu
Jan 16 at 11:45
$begingroup$
$y = x^{1/n}$ and product of roots of unity stays a root of unity so $F(0) = ?$
$endgroup$
– reuns
Jan 16 at 12:06
add a comment |
$begingroup$
The minimal polynomial of $y=x^{1/n}$ is $F(X) = prod_{j=1}^l (X-zeta_n^{m_j} y)$ where the $m_j$ are a subset of $1ldots n$. $F(0) = ?$
$endgroup$
– reuns
Jan 16 at 3:05
$begingroup$
$F(0) = prod_{j=1}^l (-zeta_n^{m_j} y)$? Is the aim to show that that minimal polynomial is the same as the function $f$ already?
$endgroup$
– Anzu
Jan 16 at 11:45
$begingroup$
$y = x^{1/n}$ and product of roots of unity stays a root of unity so $F(0) = ?$
$endgroup$
– reuns
Jan 16 at 12:06
$begingroup$
The minimal polynomial of $y=x^{1/n}$ is $F(X) = prod_{j=1}^l (X-zeta_n^{m_j} y)$ where the $m_j$ are a subset of $1ldots n$. $F(0) = ?$
$endgroup$
– reuns
Jan 16 at 3:05
$begingroup$
The minimal polynomial of $y=x^{1/n}$ is $F(X) = prod_{j=1}^l (X-zeta_n^{m_j} y)$ where the $m_j$ are a subset of $1ldots n$. $F(0) = ?$
$endgroup$
– reuns
Jan 16 at 3:05
$begingroup$
$F(0) = prod_{j=1}^l (-zeta_n^{m_j} y)$? Is the aim to show that that minimal polynomial is the same as the function $f$ already?
$endgroup$
– Anzu
Jan 16 at 11:45
$begingroup$
$F(0) = prod_{j=1}^l (-zeta_n^{m_j} y)$? Is the aim to show that that minimal polynomial is the same as the function $f$ already?
$endgroup$
– Anzu
Jan 16 at 11:45
$begingroup$
$y = x^{1/n}$ and product of roots of unity stays a root of unity so $F(0) = ?$
$endgroup$
– reuns
Jan 16 at 12:06
$begingroup$
$y = x^{1/n}$ and product of roots of unity stays a root of unity so $F(0) = ?$
$endgroup$
– reuns
Jan 16 at 12:06
add a comment |
1 Answer
1
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$begingroup$
Your problem constitutes the preliminaries of the so called Kummer theory, for which you can consult any textbook on Galois theory. The particular case considered here is simple enough to admit a direct approach. To avoid overlapping with classical proofs, let us follow more or less the hint given by @Reuns.
1) I keep your notations and hypotheses for the base field $K$, except that I prefer to write $ain K^*$ instead of $x$. Let $alpha$ be a fixed $n$-th root of $a$ in $bar K$. All the $n$-th roots of $a$ are of the form $zeta alpha$, where $zeta$ runs through the group $mu_n$ of $n$-th roots of $1$, which is contained in $K^*$, hence $K(alpha)$ is the splitting field of $X^n-a$ over $K$. These roots are distinct because of the hypotheses on char($K$), so finally the extension $L/K$ is galois. To compute its degree, let $f(X)$ be the minimal polynomial of $alpha$ over $K$, of degree $d$ say. As $f(X)$ divides $X^n-a$, it decomposes in $bar K[X]$ into a product of $d$ distinct linear factors of the form $(X-zeta alpha)$. The computation of the constant term $f(0)$ shows immediately that ${alpha}^d in K^*$. From $n=dq+r$, where $r<d$ or $r=0$ (euclidian division), it follows that $alpha^r in K^*$, hence $a^rin {{K}^*}^n$, and the third hypothesis (not used yet) implies that $r=0, n=dq$. But then ${alpha}^d =bin K^*$ implies $a^d={alpha}^{nd}=({alpha}^d)^{dq}=b^n$, and the third hypothesis again implies $nmid d$, hence $n=d$. NB: This is purely an academic exercise. In Kummer theory, it's much easier to show first the property 2), from which 1) follows.
2) We have seen in 1) that $K(alpha)/K$ is galois, and its degree is $n$ if $a$ mod ${K^*}^n$ has multiplicative order $n$ (this is the meaning of the third hypothesis). Let $G$ denote the Galois group. Any $sin G$ permutes the roots of $X^n-a$, hence $s(alpha)=zeta_salpha$, with $zeta_sin mu_n$, and the map $s to zeta_s$ is obviously an injective homomorphism of groups $Gtomu_n$. Since $G$ has order $n$, this is actually an isomorphism. NB: The third hypothesis is necessary. In general, one has only that the splitting field of $X^n-a$ is cyclic of degree $d$ over $K$, where $d$ is the multiplicative order of $a$ mod ${K^*}^n$ .
$endgroup$
add a comment |
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$begingroup$
Your problem constitutes the preliminaries of the so called Kummer theory, for which you can consult any textbook on Galois theory. The particular case considered here is simple enough to admit a direct approach. To avoid overlapping with classical proofs, let us follow more or less the hint given by @Reuns.
1) I keep your notations and hypotheses for the base field $K$, except that I prefer to write $ain K^*$ instead of $x$. Let $alpha$ be a fixed $n$-th root of $a$ in $bar K$. All the $n$-th roots of $a$ are of the form $zeta alpha$, where $zeta$ runs through the group $mu_n$ of $n$-th roots of $1$, which is contained in $K^*$, hence $K(alpha)$ is the splitting field of $X^n-a$ over $K$. These roots are distinct because of the hypotheses on char($K$), so finally the extension $L/K$ is galois. To compute its degree, let $f(X)$ be the minimal polynomial of $alpha$ over $K$, of degree $d$ say. As $f(X)$ divides $X^n-a$, it decomposes in $bar K[X]$ into a product of $d$ distinct linear factors of the form $(X-zeta alpha)$. The computation of the constant term $f(0)$ shows immediately that ${alpha}^d in K^*$. From $n=dq+r$, where $r<d$ or $r=0$ (euclidian division), it follows that $alpha^r in K^*$, hence $a^rin {{K}^*}^n$, and the third hypothesis (not used yet) implies that $r=0, n=dq$. But then ${alpha}^d =bin K^*$ implies $a^d={alpha}^{nd}=({alpha}^d)^{dq}=b^n$, and the third hypothesis again implies $nmid d$, hence $n=d$. NB: This is purely an academic exercise. In Kummer theory, it's much easier to show first the property 2), from which 1) follows.
2) We have seen in 1) that $K(alpha)/K$ is galois, and its degree is $n$ if $a$ mod ${K^*}^n$ has multiplicative order $n$ (this is the meaning of the third hypothesis). Let $G$ denote the Galois group. Any $sin G$ permutes the roots of $X^n-a$, hence $s(alpha)=zeta_salpha$, with $zeta_sin mu_n$, and the map $s to zeta_s$ is obviously an injective homomorphism of groups $Gtomu_n$. Since $G$ has order $n$, this is actually an isomorphism. NB: The third hypothesis is necessary. In general, one has only that the splitting field of $X^n-a$ is cyclic of degree $d$ over $K$, where $d$ is the multiplicative order of $a$ mod ${K^*}^n$ .
$endgroup$
add a comment |
$begingroup$
Your problem constitutes the preliminaries of the so called Kummer theory, for which you can consult any textbook on Galois theory. The particular case considered here is simple enough to admit a direct approach. To avoid overlapping with classical proofs, let us follow more or less the hint given by @Reuns.
1) I keep your notations and hypotheses for the base field $K$, except that I prefer to write $ain K^*$ instead of $x$. Let $alpha$ be a fixed $n$-th root of $a$ in $bar K$. All the $n$-th roots of $a$ are of the form $zeta alpha$, where $zeta$ runs through the group $mu_n$ of $n$-th roots of $1$, which is contained in $K^*$, hence $K(alpha)$ is the splitting field of $X^n-a$ over $K$. These roots are distinct because of the hypotheses on char($K$), so finally the extension $L/K$ is galois. To compute its degree, let $f(X)$ be the minimal polynomial of $alpha$ over $K$, of degree $d$ say. As $f(X)$ divides $X^n-a$, it decomposes in $bar K[X]$ into a product of $d$ distinct linear factors of the form $(X-zeta alpha)$. The computation of the constant term $f(0)$ shows immediately that ${alpha}^d in K^*$. From $n=dq+r$, where $r<d$ or $r=0$ (euclidian division), it follows that $alpha^r in K^*$, hence $a^rin {{K}^*}^n$, and the third hypothesis (not used yet) implies that $r=0, n=dq$. But then ${alpha}^d =bin K^*$ implies $a^d={alpha}^{nd}=({alpha}^d)^{dq}=b^n$, and the third hypothesis again implies $nmid d$, hence $n=d$. NB: This is purely an academic exercise. In Kummer theory, it's much easier to show first the property 2), from which 1) follows.
2) We have seen in 1) that $K(alpha)/K$ is galois, and its degree is $n$ if $a$ mod ${K^*}^n$ has multiplicative order $n$ (this is the meaning of the third hypothesis). Let $G$ denote the Galois group. Any $sin G$ permutes the roots of $X^n-a$, hence $s(alpha)=zeta_salpha$, with $zeta_sin mu_n$, and the map $s to zeta_s$ is obviously an injective homomorphism of groups $Gtomu_n$. Since $G$ has order $n$, this is actually an isomorphism. NB: The third hypothesis is necessary. In general, one has only that the splitting field of $X^n-a$ is cyclic of degree $d$ over $K$, where $d$ is the multiplicative order of $a$ mod ${K^*}^n$ .
$endgroup$
add a comment |
$begingroup$
Your problem constitutes the preliminaries of the so called Kummer theory, for which you can consult any textbook on Galois theory. The particular case considered here is simple enough to admit a direct approach. To avoid overlapping with classical proofs, let us follow more or less the hint given by @Reuns.
1) I keep your notations and hypotheses for the base field $K$, except that I prefer to write $ain K^*$ instead of $x$. Let $alpha$ be a fixed $n$-th root of $a$ in $bar K$. All the $n$-th roots of $a$ are of the form $zeta alpha$, where $zeta$ runs through the group $mu_n$ of $n$-th roots of $1$, which is contained in $K^*$, hence $K(alpha)$ is the splitting field of $X^n-a$ over $K$. These roots are distinct because of the hypotheses on char($K$), so finally the extension $L/K$ is galois. To compute its degree, let $f(X)$ be the minimal polynomial of $alpha$ over $K$, of degree $d$ say. As $f(X)$ divides $X^n-a$, it decomposes in $bar K[X]$ into a product of $d$ distinct linear factors of the form $(X-zeta alpha)$. The computation of the constant term $f(0)$ shows immediately that ${alpha}^d in K^*$. From $n=dq+r$, where $r<d$ or $r=0$ (euclidian division), it follows that $alpha^r in K^*$, hence $a^rin {{K}^*}^n$, and the third hypothesis (not used yet) implies that $r=0, n=dq$. But then ${alpha}^d =bin K^*$ implies $a^d={alpha}^{nd}=({alpha}^d)^{dq}=b^n$, and the third hypothesis again implies $nmid d$, hence $n=d$. NB: This is purely an academic exercise. In Kummer theory, it's much easier to show first the property 2), from which 1) follows.
2) We have seen in 1) that $K(alpha)/K$ is galois, and its degree is $n$ if $a$ mod ${K^*}^n$ has multiplicative order $n$ (this is the meaning of the third hypothesis). Let $G$ denote the Galois group. Any $sin G$ permutes the roots of $X^n-a$, hence $s(alpha)=zeta_salpha$, with $zeta_sin mu_n$, and the map $s to zeta_s$ is obviously an injective homomorphism of groups $Gtomu_n$. Since $G$ has order $n$, this is actually an isomorphism. NB: The third hypothesis is necessary. In general, one has only that the splitting field of $X^n-a$ is cyclic of degree $d$ over $K$, where $d$ is the multiplicative order of $a$ mod ${K^*}^n$ .
$endgroup$
Your problem constitutes the preliminaries of the so called Kummer theory, for which you can consult any textbook on Galois theory. The particular case considered here is simple enough to admit a direct approach. To avoid overlapping with classical proofs, let us follow more or less the hint given by @Reuns.
1) I keep your notations and hypotheses for the base field $K$, except that I prefer to write $ain K^*$ instead of $x$. Let $alpha$ be a fixed $n$-th root of $a$ in $bar K$. All the $n$-th roots of $a$ are of the form $zeta alpha$, where $zeta$ runs through the group $mu_n$ of $n$-th roots of $1$, which is contained in $K^*$, hence $K(alpha)$ is the splitting field of $X^n-a$ over $K$. These roots are distinct because of the hypotheses on char($K$), so finally the extension $L/K$ is galois. To compute its degree, let $f(X)$ be the minimal polynomial of $alpha$ over $K$, of degree $d$ say. As $f(X)$ divides $X^n-a$, it decomposes in $bar K[X]$ into a product of $d$ distinct linear factors of the form $(X-zeta alpha)$. The computation of the constant term $f(0)$ shows immediately that ${alpha}^d in K^*$. From $n=dq+r$, where $r<d$ or $r=0$ (euclidian division), it follows that $alpha^r in K^*$, hence $a^rin {{K}^*}^n$, and the third hypothesis (not used yet) implies that $r=0, n=dq$. But then ${alpha}^d =bin K^*$ implies $a^d={alpha}^{nd}=({alpha}^d)^{dq}=b^n$, and the third hypothesis again implies $nmid d$, hence $n=d$. NB: This is purely an academic exercise. In Kummer theory, it's much easier to show first the property 2), from which 1) follows.
2) We have seen in 1) that $K(alpha)/K$ is galois, and its degree is $n$ if $a$ mod ${K^*}^n$ has multiplicative order $n$ (this is the meaning of the third hypothesis). Let $G$ denote the Galois group. Any $sin G$ permutes the roots of $X^n-a$, hence $s(alpha)=zeta_salpha$, with $zeta_sin mu_n$, and the map $s to zeta_s$ is obviously an injective homomorphism of groups $Gtomu_n$. Since $G$ has order $n$, this is actually an isomorphism. NB: The third hypothesis is necessary. In general, one has only that the splitting field of $X^n-a$ is cyclic of degree $d$ over $K$, where $d$ is the multiplicative order of $a$ mod ${K^*}^n$ .
answered Jan 17 at 21:27
nguyen quang donguyen quang do
9,2141724
9,2141724
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$begingroup$
The minimal polynomial of $y=x^{1/n}$ is $F(X) = prod_{j=1}^l (X-zeta_n^{m_j} y)$ where the $m_j$ are a subset of $1ldots n$. $F(0) = ?$
$endgroup$
– reuns
Jan 16 at 3:05
$begingroup$
$F(0) = prod_{j=1}^l (-zeta_n^{m_j} y)$? Is the aim to show that that minimal polynomial is the same as the function $f$ already?
$endgroup$
– Anzu
Jan 16 at 11:45
$begingroup$
$y = x^{1/n}$ and product of roots of unity stays a root of unity so $F(0) = ?$
$endgroup$
– reuns
Jan 16 at 12:06