Let R be a p-ring (chR=p,a prime and a^p=a for all a in R). If R is field then it is isomorphic to Zp....
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I want to prove that the ring R has exactly p elements. I am able to show that a Boolean ring(2-ring) is isomorphic to Z2 provided it is field by showing it has exactly 2 elements.
Please help me to prove the equation x^p=x has exactly p roots in R.
field-theory
asked Jan 23 at 11:38
Afzal AnsariAfzal Ansari
856
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closed as off-topic by amWhy, metamorphy, Gibbs, Paul Frost, Cesareo Jan 24 at 0:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
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$begingroup$
I want to prove that the ring R has exactly p elements. I am able to show that a Boolean ring(2-ring) is isomorphic to Z2 provided it is field by showing it has exactly 2 elements.
Please help me to prove the equation x^p=x has exactly p roots in R.
field-theory
asked Jan 23 at 11:38
Afzal AnsariAfzal Ansari
856
$endgroup$
closed as off-topic by amWhy, metamorphy, Gibbs, Paul Frost, Cesareo Jan 24 at 0:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, metamorphy, Gibbs, Paul Frost, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I want to prove that the ring R has exactly p elements. I am able to show that a Boolean ring(2-ring) is isomorphic to Z2 provided it is field by showing it has exactly 2 elements.
Please help me to prove the equation x^p=x has exactly p roots in R.
field-theory
asked Jan 23 at 11:38
Afzal AnsariAfzal Ansari
856
$endgroup$
I want to prove that the ring R has exactly p elements. I am able to show that a Boolean ring(2-ring) is isomorphic to Z2 provided it is field by showing it has exactly 2 elements.
Please help me to prove the equation x^p=x has exactly p roots in R.
field-theory
field-theory
asked Jan 23 at 11:38
Afzal AnsariAfzal Ansari
856
asked Jan 23 at 11:38
Afzal AnsariAfzal Ansari
856
asked Jan 23 at 11:38
Afzal AnsariAfzal Ansari
856
asked Jan 23 at 11:38
Afzal AnsariAfzal Ansari
856
asked Jan 23 at 11:38
Afzal AnsariAfzal Ansari
856
856
closed as off-topic by amWhy, metamorphy, Gibbs, Paul Frost, Cesareo Jan 24 at 0:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, metamorphy, Gibbs, Paul Frost, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, metamorphy, Gibbs, Paul Frost, Cesareo Jan 24 at 0:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, metamorphy, Gibbs, Paul Frost, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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Let $F$ be the prime field of $R$. Then $F$ has $p$ elements.
A polynomial of degree $p$ has at most $p$ zeros in a field.
Therefore, $F$ is exactly the set of zeros of $x^p-x$, which by hypothesis is $R$. So $R=F$.
answered Jan 23 at 11:44
lhflhf
166k10171396
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $F$ be the prime field of $R$. Then $F$ has $p$ elements.
A polynomial of degree $p$ has at most $p$ zeros in a field.
Therefore, $F$ is exactly the set of zeros of $x^p-x$, which by hypothesis is $R$. So $R=F$.
answered Jan 23 at 11:44
lhflhf
166k10171396
$endgroup$
add a comment |
$begingroup$
Let $F$ be the prime field of $R$. Then $F$ has $p$ elements.
A polynomial of degree $p$ has at most $p$ zeros in a field.
Therefore, $F$ is exactly the set of zeros of $x^p-x$, which by hypothesis is $R$. So $R=F$.
answered Jan 23 at 11:44
lhflhf
166k10171396
$endgroup$
add a comment |
$begingroup$
Let $F$ be the prime field of $R$. Then $F$ has $p$ elements.
A polynomial of degree $p$ has at most $p$ zeros in a field.
Therefore, $F$ is exactly the set of zeros of $x^p-x$, which by hypothesis is $R$. So $R=F$.
answered Jan 23 at 11:44
lhflhf
166k10171396
$endgroup$
Let $F$ be the prime field of $R$. Then $F$ has $p$ elements.
A polynomial of degree $p$ has at most $p$ zeros in a field.
Therefore, $F$ is exactly the set of zeros of $x^p-x$, which by hypothesis is $R$. So $R=F$.
answered Jan 23 at 11:44
lhflhf
166k10171396
answered Jan 23 at 11:44
lhflhf
166k10171396
answered Jan 23 at 11:44
lhflhf
166k10171396
answered Jan 23 at 11:44
lhflhf
166k10171396
166k10171396
add a comment |
add a comment |
