Suppose that $F$ is a field with $27$ elements. Show that for every element $a in F$, $5a = −a$
$begingroup$
Suppose that $F$ is a field with $27$ elements. Show that for every
element $a in F$, $5a = −a$.
I am not able to understand how to approach this problem.
abstract-algebra ring-theory field-theory
edited Jan 2 at 12:41
egreg
180k1485202
asked Jan 2 at 7:15
Sambhav KhuranaSambhav Khurana
163
$endgroup$
add a comment |
$begingroup$
Suppose that $F$ is a field with $27$ elements. Show that for every
element $a in F$, $5a = −a$.
I am not able to understand how to approach this problem.
abstract-algebra ring-theory field-theory
edited Jan 2 at 12:41
egreg
180k1485202
asked Jan 2 at 7:15
Sambhav KhuranaSambhav Khurana
163
$endgroup$
14
$begingroup$
Consider the characteristic.
$endgroup$
– Randall
Jan 2 at 7:16
2
$begingroup$
Sambhav Khurana Even $3a=0$
$endgroup$
– Michael Rozenberg
Jan 2 at 7:44
$begingroup$
Can I say that the characteristic must be prime since it's a field and it must divide 27. So 3a=0.?
$endgroup$
– Sambhav Khurana
Jan 2 at 7:56
$begingroup$
Yes, of course! Also, $27$ must be $p^n$, where $p$ is prime.
$endgroup$
– Michael Rozenberg
Jan 2 at 8:00
add a comment |
$begingroup$
Suppose that $F$ is a field with $27$ elements. Show that for every
element $a in F$, $5a = −a$.
I am not able to understand how to approach this problem.
abstract-algebra ring-theory field-theory
edited Jan 2 at 12:41
egreg
180k1485202
asked Jan 2 at 7:15
Sambhav KhuranaSambhav Khurana
163
$endgroup$
Suppose that $F$ is a field with $27$ elements. Show that for every
element $a in F$, $5a = −a$.
I am not able to understand how to approach this problem.
abstract-algebra ring-theory field-theory
abstract-algebra ring-theory field-theory
edited Jan 2 at 12:41
egreg
180k1485202
asked Jan 2 at 7:15
Sambhav KhuranaSambhav Khurana
163
edited Jan 2 at 12:41
egreg
180k1485202
asked Jan 2 at 7:15
Sambhav KhuranaSambhav Khurana
163
edited Jan 2 at 12:41
egreg
180k1485202
edited Jan 2 at 12:41
egreg
180k1485202
edited Jan 2 at 12:41
egreg
180k1485202
180k1485202
asked Jan 2 at 7:15
Sambhav KhuranaSambhav Khurana
163
asked Jan 2 at 7:15
Sambhav KhuranaSambhav Khurana
163
asked Jan 2 at 7:15
Sambhav KhuranaSambhav Khurana
163
163
14
$begingroup$
Consider the characteristic.
$endgroup$
– Randall
Jan 2 at 7:16
2
$begingroup$
Sambhav Khurana Even $3a=0$
$endgroup$
– Michael Rozenberg
Jan 2 at 7:44
$begingroup$
Can I say that the characteristic must be prime since it's a field and it must divide 27. So 3a=0.?
$endgroup$
– Sambhav Khurana
Jan 2 at 7:56
$begingroup$
Yes, of course! Also, $27$ must be $p^n$, where $p$ is prime.
$endgroup$
– Michael Rozenberg
Jan 2 at 8:00
add a comment |
14
$begingroup$
Consider the characteristic.
$endgroup$
– Randall
Jan 2 at 7:16
2
$begingroup$
Sambhav Khurana Even $3a=0$
$endgroup$
– Michael Rozenberg
Jan 2 at 7:44
$begingroup$
Can I say that the characteristic must be prime since it's a field and it must divide 27. So 3a=0.?
$endgroup$
– Sambhav Khurana
Jan 2 at 7:56
$begingroup$
Yes, of course! Also, $27$ must be $p^n$, where $p$ is prime.
$endgroup$
– Michael Rozenberg
Jan 2 at 8:00
14
14
$begingroup$
Consider the characteristic.
$endgroup$
– Randall
Jan 2 at 7:16
$begingroup$
Consider the characteristic.
$endgroup$
– Randall
Jan 2 at 7:16
2
2
$begingroup$
Sambhav Khurana Even $3a=0$
$endgroup$
– Michael Rozenberg
Jan 2 at 7:44
$begingroup$
Sambhav Khurana Even $3a=0$
$endgroup$
– Michael Rozenberg
Jan 2 at 7:44
$begingroup$
Can I say that the characteristic must be prime since it's a field and it must divide 27. So 3a=0.?
$endgroup$
– Sambhav Khurana
Jan 2 at 7:56
$begingroup$
Can I say that the characteristic must be prime since it's a field and it must divide 27. So 3a=0.?
$endgroup$
– Sambhav Khurana
Jan 2 at 7:56
$begingroup$
Yes, of course! Also, $27$ must be $p^n$, where $p$ is prime.
$endgroup$
– Michael Rozenberg
Jan 2 at 8:00
$begingroup$
Yes, of course! Also, $27$ must be $p^n$, where $p$ is prime.
$endgroup$
– Michael Rozenberg
Jan 2 at 8:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Recall that a field $F$ has a characteristic, which is defined as follows:
Define the homomorphism of rings
begin{align}
psi:mathbb{Z}&to F
end{align}
That sends $1$ to $1_F$. Its kernel is principal, say, generated (in particular) by a prime number $p$ or $0$ (if it is injective). That number is called the characteristic of the field $F$.
Now, you have to know that every finite field, has positive characteristic $p$ (i.e. the characteristic is not zero) and have $p^n$ elements. This is because as it is finite, via $psi$ we can not have infinitely many sums of $1_F$ that are different pairwise, and if $mathbb{F}_p$ is the field with $p$ elements, $F$ can be seen as a non-trivial $mathbb{F}_p$ vector space, so is isomorphic to $mathbb{F}^n$, for some $n>1$.
Because of the previous results, your field $F$ with $3^3$ elements, has characteristic $3$, so that $3a=0$ for all $ain F$, in particular $6a=0$, so that $5a=-a$.
answered Jan 2 at 12:19
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Recall that a field $F$ has a characteristic, which is defined as follows:
Define the homomorphism of rings
begin{align}
psi:mathbb{Z}&to F
end{align}
That sends $1$ to $1_F$. Its kernel is principal, say, generated (in particular) by a prime number $p$ or $0$ (if it is injective). That number is called the characteristic of the field $F$.
Now, you have to know that every finite field, has positive characteristic $p$ (i.e. the characteristic is not zero) and have $p^n$ elements. This is because as it is finite, via $psi$ we can not have infinitely many sums of $1_F$ that are different pairwise, and if $mathbb{F}_p$ is the field with $p$ elements, $F$ can be seen as a non-trivial $mathbb{F}_p$ vector space, so is isomorphic to $mathbb{F}^n$, for some $n>1$.
Because of the previous results, your field $F$ with $3^3$ elements, has characteristic $3$, so that $3a=0$ for all $ain F$, in particular $6a=0$, so that $5a=-a$.
answered Jan 2 at 12:19
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
$begingroup$
Recall that a field $F$ has a characteristic, which is defined as follows:
Define the homomorphism of rings
begin{align}
psi:mathbb{Z}&to F
end{align}
That sends $1$ to $1_F$. Its kernel is principal, say, generated (in particular) by a prime number $p$ or $0$ (if it is injective). That number is called the characteristic of the field $F$.
Now, you have to know that every finite field, has positive characteristic $p$ (i.e. the characteristic is not zero) and have $p^n$ elements. This is because as it is finite, via $psi$ we can not have infinitely many sums of $1_F$ that are different pairwise, and if $mathbb{F}_p$ is the field with $p$ elements, $F$ can be seen as a non-trivial $mathbb{F}_p$ vector space, so is isomorphic to $mathbb{F}^n$, for some $n>1$.
Because of the previous results, your field $F$ with $3^3$ elements, has characteristic $3$, so that $3a=0$ for all $ain F$, in particular $6a=0$, so that $5a=-a$.
answered Jan 2 at 12:19
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
$begingroup$
Recall that a field $F$ has a characteristic, which is defined as follows:
Define the homomorphism of rings
begin{align}
psi:mathbb{Z}&to F
end{align}
That sends $1$ to $1_F$. Its kernel is principal, say, generated (in particular) by a prime number $p$ or $0$ (if it is injective). That number is called the characteristic of the field $F$.
Now, you have to know that every finite field, has positive characteristic $p$ (i.e. the characteristic is not zero) and have $p^n$ elements. This is because as it is finite, via $psi$ we can not have infinitely many sums of $1_F$ that are different pairwise, and if $mathbb{F}_p$ is the field with $p$ elements, $F$ can be seen as a non-trivial $mathbb{F}_p$ vector space, so is isomorphic to $mathbb{F}^n$, for some $n>1$.
Because of the previous results, your field $F$ with $3^3$ elements, has characteristic $3$, so that $3a=0$ for all $ain F$, in particular $6a=0$, so that $5a=-a$.
answered Jan 2 at 12:19
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
Recall that a field $F$ has a characteristic, which is defined as follows:
Define the homomorphism of rings
begin{align}
psi:mathbb{Z}&to F
end{align}
That sends $1$ to $1_F$. Its kernel is principal, say, generated (in particular) by a prime number $p$ or $0$ (if it is injective). That number is called the characteristic of the field $F$.
Now, you have to know that every finite field, has positive characteristic $p$ (i.e. the characteristic is not zero) and have $p^n$ elements. This is because as it is finite, via $psi$ we can not have infinitely many sums of $1_F$ that are different pairwise, and if $mathbb{F}_p$ is the field with $p$ elements, $F$ can be seen as a non-trivial $mathbb{F}_p$ vector space, so is isomorphic to $mathbb{F}^n$, for some $n>1$.
Because of the previous results, your field $F$ with $3^3$ elements, has characteristic $3$, so that $3a=0$ for all $ain F$, in particular $6a=0$, so that $5a=-a$.
answered Jan 2 at 12:19
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
edited Jan 2 at 12:47
answered Jan 2 at 12:19
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Jan 2 at 12:19
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Jan 2 at 12:19
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
802110
add a comment |
add a comment |
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14
$begingroup$
Consider the characteristic.
$endgroup$
– Randall
Jan 2 at 7:16
2
$begingroup$
Sambhav Khurana Even $3a=0$
$endgroup$
– Michael Rozenberg
Jan 2 at 7:44
$begingroup$
Can I say that the characteristic must be prime since it's a field and it must divide 27. So 3a=0.?
$endgroup$
– Sambhav Khurana
Jan 2 at 7:56
$begingroup$
Yes, of course! Also, $27$ must be $p^n$, where $p$ is prime.
$endgroup$
– Michael Rozenberg
Jan 2 at 8:00