Characteristic of a field is not equal to zero
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Let $F$ be a field such that for every $x in F$ there exists a $k >0$ such that $x^k=1$. Does this imply that the characteristic of $F$ is strictly greater than zero?
field-theory
asked 16 hours ago
J.Bosser
314210
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up vote
3
down vote
favorite
Let $F$ be a field such that for every $x in F$ there exists a $k >0$ such that $x^k=1$. Does this imply that the characteristic of $F$ is strictly greater than zero?
field-theory
asked 16 hours ago
J.Bosser
314210
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $F$ be a field such that for every $x in F$ there exists a $k >0$ such that $x^k=1$. Does this imply that the characteristic of $F$ is strictly greater than zero?
field-theory
asked 16 hours ago
J.Bosser
314210
Let $F$ be a field such that for every $x in F$ there exists a $k >0$ such that $x^k=1$. Does this imply that the characteristic of $F$ is strictly greater than zero?
field-theory
field-theory
asked 16 hours ago
J.Bosser
314210
asked 16 hours ago
J.Bosser
314210
asked 16 hours ago
J.Bosser
314210
asked 16 hours ago
J.Bosser
314210
asked 16 hours ago
J.Bosser
314210
314210
add a comment |
add a comment |
2 Answers
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4
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Yes. If the characteristic is zero then the prime subfield is isomorphic to $mathbb{Q}$, and this contains elements - e.g. 2 - whose nonzero powers are never equal to 1.
answered 16 hours ago
Matthew Towers
7,42922244
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up vote
3
down vote
Every field $F$ of characteristic $0$ contains (up to isomorphism) the field of rational numbers. If you take the rational number $frac{1}{2}$, say, then there is no integer $k>0$ such that $left(frac{1}{2}right)^k = 1$.
For a finite field with $q$ (prime power) elements, one has $x^{q-1}=1$ for each element $xne 0$ and $x^q=x$ for each element $x$.
answered 16 hours ago
Wuestenfux
3,0351410
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Yes. If the characteristic is zero then the prime subfield is isomorphic to $mathbb{Q}$, and this contains elements - e.g. 2 - whose nonzero powers are never equal to 1.
answered 16 hours ago
Matthew Towers
7,42922244
add a comment |
up vote
4
down vote
accepted
Yes. If the characteristic is zero then the prime subfield is isomorphic to $mathbb{Q}$, and this contains elements - e.g. 2 - whose nonzero powers are never equal to 1.
answered 16 hours ago
Matthew Towers
7,42922244
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Yes. If the characteristic is zero then the prime subfield is isomorphic to $mathbb{Q}$, and this contains elements - e.g. 2 - whose nonzero powers are never equal to 1.
answered 16 hours ago
Matthew Towers
7,42922244
Yes. If the characteristic is zero then the prime subfield is isomorphic to $mathbb{Q}$, and this contains elements - e.g. 2 - whose nonzero powers are never equal to 1.
answered 16 hours ago
Matthew Towers
7,42922244
answered 16 hours ago
Matthew Towers
7,42922244
answered 16 hours ago
Matthew Towers
7,42922244
answered 16 hours ago
Matthew Towers
7,42922244
7,42922244
add a comment |
add a comment |
up vote
3
down vote
Every field $F$ of characteristic $0$ contains (up to isomorphism) the field of rational numbers. If you take the rational number $frac{1}{2}$, say, then there is no integer $k>0$ such that $left(frac{1}{2}right)^k = 1$.
For a finite field with $q$ (prime power) elements, one has $x^{q-1}=1$ for each element $xne 0$ and $x^q=x$ for each element $x$.
answered 16 hours ago
Wuestenfux
3,0351410
add a comment |
up vote
3
down vote
Every field $F$ of characteristic $0$ contains (up to isomorphism) the field of rational numbers. If you take the rational number $frac{1}{2}$, say, then there is no integer $k>0$ such that $left(frac{1}{2}right)^k = 1$.
For a finite field with $q$ (prime power) elements, one has $x^{q-1}=1$ for each element $xne 0$ and $x^q=x$ for each element $x$.
answered 16 hours ago
Wuestenfux
3,0351410
add a comment |
up vote
3
down vote
up vote
3
down vote
Every field $F$ of characteristic $0$ contains (up to isomorphism) the field of rational numbers. If you take the rational number $frac{1}{2}$, say, then there is no integer $k>0$ such that $left(frac{1}{2}right)^k = 1$.
For a finite field with $q$ (prime power) elements, one has $x^{q-1}=1$ for each element $xne 0$ and $x^q=x$ for each element $x$.
answered 16 hours ago
Wuestenfux
3,0351410
Every field $F$ of characteristic $0$ contains (up to isomorphism) the field of rational numbers. If you take the rational number $frac{1}{2}$, say, then there is no integer $k>0$ such that $left(frac{1}{2}right)^k = 1$.
For a finite field with $q$ (prime power) elements, one has $x^{q-1}=1$ for each element $xne 0$ and $x^q=x$ for each element $x$.
answered 16 hours ago
Wuestenfux
3,0351410
edited 16 hours ago
answered 16 hours ago
Wuestenfux
3,0351410
answered 16 hours ago
Wuestenfux
3,0351410
answered 16 hours ago
Wuestenfux
3,0351410
3,0351410
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