If a field $K$ (of characteristic 0) has no proper extensions of the form $K[sqrt[n]{x}]$, is it...
$begingroup$
Let $K$ be a field, assume characteristic 0 if this simplifies things. Suppose we know that for any positive integer $n$ and any $x in K$ the polynomial $X^n - x$ has a root in $K$. Is $K$ algebraically closed?
Working on the assumption that the answer is probably "no", how would one construct a counterexample?
abstract-algebra field-theory radicals
edited Jan 4 at 21:29
user26857
39.3k124183
asked Jan 4 at 9:40
Bib-lostBib-lost
2,040629
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add a comment |
$begingroup$
Let $K$ be a field, assume characteristic 0 if this simplifies things. Suppose we know that for any positive integer $n$ and any $x in K$ the polynomial $X^n - x$ has a root in $K$. Is $K$ algebraically closed?
Working on the assumption that the answer is probably "no", how would one construct a counterexample?
abstract-algebra field-theory radicals
edited Jan 4 at 21:29
user26857
39.3k124183
asked Jan 4 at 9:40
Bib-lostBib-lost
2,040629
$endgroup$
add a comment |
$begingroup$
Let $K$ be a field, assume characteristic 0 if this simplifies things. Suppose we know that for any positive integer $n$ and any $x in K$ the polynomial $X^n - x$ has a root in $K$. Is $K$ algebraically closed?
Working on the assumption that the answer is probably "no", how would one construct a counterexample?
abstract-algebra field-theory radicals
edited Jan 4 at 21:29
user26857
39.3k124183
asked Jan 4 at 9:40
Bib-lostBib-lost
2,040629
$endgroup$
Let $K$ be a field, assume characteristic 0 if this simplifies things. Suppose we know that for any positive integer $n$ and any $x in K$ the polynomial $X^n - x$ has a root in $K$. Is $K$ algebraically closed?
Working on the assumption that the answer is probably "no", how would one construct a counterexample?
abstract-algebra field-theory radicals
abstract-algebra field-theory radicals
edited Jan 4 at 21:29
user26857
39.3k124183
asked Jan 4 at 9:40
Bib-lostBib-lost
2,040629
edited Jan 4 at 21:29
user26857
39.3k124183
asked Jan 4 at 9:40
Bib-lostBib-lost
2,040629
edited Jan 4 at 21:29
user26857
39.3k124183
edited Jan 4 at 21:29
user26857
39.3k124183
edited Jan 4 at 21:29
user26857
39.3k124183
39.3k124183
asked Jan 4 at 9:40
Bib-lostBib-lost
2,040629
asked Jan 4 at 9:40
Bib-lostBib-lost
2,040629
asked Jan 4 at 9:40
Bib-lostBib-lost
2,040629
2,040629
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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The answer in characteristic $0$ is no. For instance, the "root closure"* of $Bbb Q$ still wouldn't let you solve, say, an unsolvable quintic like $x^5-x-1 = 0$, because we know its solutions can't be written using roots in $Bbb Q$, so it can't be written using roots in the root closure either.
*What I mean by "root closure" is this: Start with $Bbb Q$, and consider it a subfield of $Bbb C$ (or of $overline{Bbb Q}$). To $Bbb Q$, add all $n$th roots of all elements of $Bbb Q$. Then add all $n$'th roots of all elements in that new field, and so on. The root closure is the final result, i.e. the union of these fields. Any element in that field may be written with a finite expression involving only addition, subtraction, multiplication, division, roots and rational numbers.
answered Jan 4 at 10:12
ArthurArthur
116k7116199
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$begingroup$
In positive characteristic the situation is even worse. For instance, the field $K$ with two elements trivially satisfies your property, but the polynomial $x^{2} + x + 1$ has no roots in $K$.
edited Jan 4 at 10:50
Arthur
116k7116199
answered Jan 4 at 10:15
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
$begingroup$
@DietrichBurde That's what happens when I only read titles. I need to stop doing that.
$endgroup$
– Arthur
Jan 4 at 10:49
2
$begingroup$
Part of the conversation is gone, but I suppose it is related to the fact that the title appears to restrict the question to characteristic $0$, whereas in the question body it is only stated assume characteristic $0$ if this simplifies things.
$endgroup$
– Andreas Caranti
Jan 4 at 11:07
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@DietrichBurde, thanks. And a Happy New Year to you!
$endgroup$
– Andreas Caranti
Jan 4 at 11:57
$begingroup$
Thank you, you too!!!
$endgroup$
– Dietrich Burde
Jan 4 at 11:59
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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oldest
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$begingroup$
The answer in characteristic $0$ is no. For instance, the "root closure"* of $Bbb Q$ still wouldn't let you solve, say, an unsolvable quintic like $x^5-x-1 = 0$, because we know its solutions can't be written using roots in $Bbb Q$, so it can't be written using roots in the root closure either.
*What I mean by "root closure" is this: Start with $Bbb Q$, and consider it a subfield of $Bbb C$ (or of $overline{Bbb Q}$). To $Bbb Q$, add all $n$th roots of all elements of $Bbb Q$. Then add all $n$'th roots of all elements in that new field, and so on. The root closure is the final result, i.e. the union of these fields. Any element in that field may be written with a finite expression involving only addition, subtraction, multiplication, division, roots and rational numbers.
answered Jan 4 at 10:12
ArthurArthur
116k7116199
$endgroup$
add a comment |
$begingroup$
The answer in characteristic $0$ is no. For instance, the "root closure"* of $Bbb Q$ still wouldn't let you solve, say, an unsolvable quintic like $x^5-x-1 = 0$, because we know its solutions can't be written using roots in $Bbb Q$, so it can't be written using roots in the root closure either.
*What I mean by "root closure" is this: Start with $Bbb Q$, and consider it a subfield of $Bbb C$ (or of $overline{Bbb Q}$). To $Bbb Q$, add all $n$th roots of all elements of $Bbb Q$. Then add all $n$'th roots of all elements in that new field, and so on. The root closure is the final result, i.e. the union of these fields. Any element in that field may be written with a finite expression involving only addition, subtraction, multiplication, division, roots and rational numbers.
answered Jan 4 at 10:12
ArthurArthur
116k7116199
$endgroup$
add a comment |
$begingroup$
The answer in characteristic $0$ is no. For instance, the "root closure"* of $Bbb Q$ still wouldn't let you solve, say, an unsolvable quintic like $x^5-x-1 = 0$, because we know its solutions can't be written using roots in $Bbb Q$, so it can't be written using roots in the root closure either.
*What I mean by "root closure" is this: Start with $Bbb Q$, and consider it a subfield of $Bbb C$ (or of $overline{Bbb Q}$). To $Bbb Q$, add all $n$th roots of all elements of $Bbb Q$. Then add all $n$'th roots of all elements in that new field, and so on. The root closure is the final result, i.e. the union of these fields. Any element in that field may be written with a finite expression involving only addition, subtraction, multiplication, division, roots and rational numbers.
answered Jan 4 at 10:12
ArthurArthur
116k7116199
$endgroup$
The answer in characteristic $0$ is no. For instance, the "root closure"* of $Bbb Q$ still wouldn't let you solve, say, an unsolvable quintic like $x^5-x-1 = 0$, because we know its solutions can't be written using roots in $Bbb Q$, so it can't be written using roots in the root closure either.
*What I mean by "root closure" is this: Start with $Bbb Q$, and consider it a subfield of $Bbb C$ (or of $overline{Bbb Q}$). To $Bbb Q$, add all $n$th roots of all elements of $Bbb Q$. Then add all $n$'th roots of all elements in that new field, and so on. The root closure is the final result, i.e. the union of these fields. Any element in that field may be written with a finite expression involving only addition, subtraction, multiplication, division, roots and rational numbers.
answered Jan 4 at 10:12
ArthurArthur
116k7116199
edited Jan 4 at 11:41
answered Jan 4 at 10:12
ArthurArthur
116k7116199
answered Jan 4 at 10:12
ArthurArthur
116k7116199
answered Jan 4 at 10:12
ArthurArthur
116k7116199
116k7116199
add a comment |
add a comment |
$begingroup$
In positive characteristic the situation is even worse. For instance, the field $K$ with two elements trivially satisfies your property, but the polynomial $x^{2} + x + 1$ has no roots in $K$.
edited Jan 4 at 10:50
Arthur
116k7116199
answered Jan 4 at 10:15
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
$begingroup$
@DietrichBurde That's what happens when I only read titles. I need to stop doing that.
$endgroup$
– Arthur
Jan 4 at 10:49
2
$begingroup$
Part of the conversation is gone, but I suppose it is related to the fact that the title appears to restrict the question to characteristic $0$, whereas in the question body it is only stated assume characteristic $0$ if this simplifies things.
$endgroup$
– Andreas Caranti
Jan 4 at 11:07
$begingroup$
@DietrichBurde, thanks. And a Happy New Year to you!
$endgroup$
– Andreas Caranti
Jan 4 at 11:57
$begingroup$
Thank you, you too!!!
$endgroup$
– Dietrich Burde
Jan 4 at 11:59
add a comment |
$begingroup$
In positive characteristic the situation is even worse. For instance, the field $K$ with two elements trivially satisfies your property, but the polynomial $x^{2} + x + 1$ has no roots in $K$.
edited Jan 4 at 10:50
Arthur
116k7116199
answered Jan 4 at 10:15
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
$begingroup$
@DietrichBurde That's what happens when I only read titles. I need to stop doing that.
$endgroup$
– Arthur
Jan 4 at 10:49
2
$begingroup$
Part of the conversation is gone, but I suppose it is related to the fact that the title appears to restrict the question to characteristic $0$, whereas in the question body it is only stated assume characteristic $0$ if this simplifies things.
$endgroup$
– Andreas Caranti
Jan 4 at 11:07
$begingroup$
@DietrichBurde, thanks. And a Happy New Year to you!
$endgroup$
– Andreas Caranti
Jan 4 at 11:57
$begingroup$
Thank you, you too!!!
$endgroup$
– Dietrich Burde
Jan 4 at 11:59
add a comment |
$begingroup$
In positive characteristic the situation is even worse. For instance, the field $K$ with two elements trivially satisfies your property, but the polynomial $x^{2} + x + 1$ has no roots in $K$.
edited Jan 4 at 10:50
Arthur
116k7116199
answered Jan 4 at 10:15
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
In positive characteristic the situation is even worse. For instance, the field $K$ with two elements trivially satisfies your property, but the polynomial $x^{2} + x + 1$ has no roots in $K$.
edited Jan 4 at 10:50
Arthur
116k7116199
answered Jan 4 at 10:15
Andreas CarantiAndreas Caranti
56.6k34395
edited Jan 4 at 10:50
Arthur
116k7116199
edited Jan 4 at 10:50
Arthur
116k7116199
edited Jan 4 at 10:50
Arthur
116k7116199
116k7116199
answered Jan 4 at 10:15
Andreas CarantiAndreas Caranti
56.6k34395
answered Jan 4 at 10:15
Andreas CarantiAndreas Caranti
56.6k34395
answered Jan 4 at 10:15
Andreas CarantiAndreas Caranti
56.6k34395
56.6k34395
$begingroup$
@DietrichBurde That's what happens when I only read titles. I need to stop doing that.
$endgroup$
– Arthur
Jan 4 at 10:49
2
$begingroup$
Part of the conversation is gone, but I suppose it is related to the fact that the title appears to restrict the question to characteristic $0$, whereas in the question body it is only stated assume characteristic $0$ if this simplifies things.
$endgroup$
– Andreas Caranti
Jan 4 at 11:07
$begingroup$
@DietrichBurde, thanks. And a Happy New Year to you!
$endgroup$
– Andreas Caranti
Jan 4 at 11:57
$begingroup$
Thank you, you too!!!
$endgroup$
– Dietrich Burde
Jan 4 at 11:59
add a comment |
$begingroup$
@DietrichBurde That's what happens when I only read titles. I need to stop doing that.
$endgroup$
– Arthur
Jan 4 at 10:49
2
$begingroup$
Part of the conversation is gone, but I suppose it is related to the fact that the title appears to restrict the question to characteristic $0$, whereas in the question body it is only stated assume characteristic $0$ if this simplifies things.
$endgroup$
– Andreas Caranti
Jan 4 at 11:07
$begingroup$
@DietrichBurde, thanks. And a Happy New Year to you!
$endgroup$
– Andreas Caranti
Jan 4 at 11:57
$begingroup$
Thank you, you too!!!
$endgroup$
– Dietrich Burde
Jan 4 at 11:59
$begingroup$
@DietrichBurde That's what happens when I only read titles. I need to stop doing that.
$endgroup$
– Arthur
Jan 4 at 10:49
$begingroup$
@DietrichBurde That's what happens when I only read titles. I need to stop doing that.
$endgroup$
– Arthur
Jan 4 at 10:49
2
2
$begingroup$
Part of the conversation is gone, but I suppose it is related to the fact that the title appears to restrict the question to characteristic $0$, whereas in the question body it is only stated assume characteristic $0$ if this simplifies things.
$endgroup$
– Andreas Caranti
Jan 4 at 11:07
$begingroup$
Part of the conversation is gone, but I suppose it is related to the fact that the title appears to restrict the question to characteristic $0$, whereas in the question body it is only stated assume characteristic $0$ if this simplifies things.
$endgroup$
– Andreas Caranti
Jan 4 at 11:07
$begingroup$
@DietrichBurde, thanks. And a Happy New Year to you!
$endgroup$
– Andreas Caranti
Jan 4 at 11:57
$begingroup$
@DietrichBurde, thanks. And a Happy New Year to you!
$endgroup$
– Andreas Caranti
Jan 4 at 11:57
$begingroup$
Thank you, you too!!!
$endgroup$
– Dietrich Burde
Jan 4 at 11:59
$begingroup$
Thank you, you too!!!
$endgroup$
– Dietrich Burde
Jan 4 at 11:59
add a comment |
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