show that if $K$ is a field then $K[x]$ is principal [duplicate]
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This question already has an answer here:
$F$ is a field iff $F[x]$ is a Principal Ideal Domain
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I want to show that if $K$ is a field then $K[x]$ is principal
Here is what I did:
I took 2 polynomial p(x) and q(x). Since we're on a field we can do a euclidean division of polynomials:
$q(x) = p(x)a(x) + r(x)$ with $deg(r) < deg(p)$
Now if $r= 0$ we obviously have $p in (q)$
But I don't know how to show that it is principal if $r neq 0$
Also I would like to show that the assertion is false if we replace "$K$ is a field" by "$K$ is a principal domain and an integral domain". I can't figure out how to do that
abstract-algebra ring-theory field-theory
asked Mar 12 '17 at 15:56
Sylvester StalloneSylvester Stallone
515311
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marked as duplicate by rschwieb abstract-algebra
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Mar 12 '17 at 16:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
$F$ is a field iff $F[x]$ is a Principal Ideal Domain
1 answer
I want to show that if $K$ is a field then $K[x]$ is principal
Here is what I did:
I took 2 polynomial p(x) and q(x). Since we're on a field we can do a euclidean division of polynomials:
$q(x) = p(x)a(x) + r(x)$ with $deg(r) < deg(p)$
Now if $r= 0$ we obviously have $p in (q)$
But I don't know how to show that it is principal if $r neq 0$
Also I would like to show that the assertion is false if we replace "$K$ is a field" by "$K$ is a principal domain and an integral domain". I can't figure out how to do that
abstract-algebra ring-theory field-theory
asked Mar 12 '17 at 15:56
Sylvester StalloneSylvester Stallone
515311
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marked as duplicate by rschwieb abstract-algebra
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Mar 12 '17 at 16:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This and variants have been asked before. Check for example these: math.stackexchange.com/q/1214667/133781 math.stackexchange.com/q/2014146/133781 and math.stackexchange.com/q/873088/133781
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– Xam
Mar 12 '17 at 16:02
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$begingroup$
This question already has an answer here:
$F$ is a field iff $F[x]$ is a Principal Ideal Domain
1 answer
I want to show that if $K$ is a field then $K[x]$ is principal
Here is what I did:
I took 2 polynomial p(x) and q(x). Since we're on a field we can do a euclidean division of polynomials:
$q(x) = p(x)a(x) + r(x)$ with $deg(r) < deg(p)$
Now if $r= 0$ we obviously have $p in (q)$
But I don't know how to show that it is principal if $r neq 0$
Also I would like to show that the assertion is false if we replace "$K$ is a field" by "$K$ is a principal domain and an integral domain". I can't figure out how to do that
abstract-algebra ring-theory field-theory
asked Mar 12 '17 at 15:56
Sylvester StalloneSylvester Stallone
515311
$endgroup$
This question already has an answer here:
$F$ is a field iff $F[x]$ is a Principal Ideal Domain
1 answer
I want to show that if $K$ is a field then $K[x]$ is principal
Here is what I did:
I took 2 polynomial p(x) and q(x). Since we're on a field we can do a euclidean division of polynomials:
$q(x) = p(x)a(x) + r(x)$ with $deg(r) < deg(p)$
Now if $r= 0$ we obviously have $p in (q)$
But I don't know how to show that it is principal if $r neq 0$
Also I would like to show that the assertion is false if we replace "$K$ is a field" by "$K$ is a principal domain and an integral domain". I can't figure out how to do that
This question already has an answer here:
$F$ is a field iff $F[x]$ is a Principal Ideal Domain
1 answer
abstract-algebra ring-theory field-theory
abstract-algebra ring-theory field-theory
asked Mar 12 '17 at 15:56
Sylvester StalloneSylvester Stallone
515311
asked Mar 12 '17 at 15:56
Sylvester StalloneSylvester Stallone
515311
asked Mar 12 '17 at 15:56
Sylvester StalloneSylvester Stallone
515311
asked Mar 12 '17 at 15:56
Sylvester StalloneSylvester Stallone
515311
asked Mar 12 '17 at 15:56
Sylvester StalloneSylvester Stallone
515311
515311
marked as duplicate by rschwieb abstract-algebra
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Mar 12 '17 at 16:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by rschwieb abstract-algebra
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Mar 12 '17 at 16:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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This and variants have been asked before. Check for example these: math.stackexchange.com/q/1214667/133781 math.stackexchange.com/q/2014146/133781 and math.stackexchange.com/q/873088/133781
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– Xam
Mar 12 '17 at 16:02
add a comment |
1
$begingroup$
This and variants have been asked before. Check for example these: math.stackexchange.com/q/1214667/133781 math.stackexchange.com/q/2014146/133781 and math.stackexchange.com/q/873088/133781
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– Xam
Mar 12 '17 at 16:02
1
1
$begingroup$
This and variants have been asked before. Check for example these: math.stackexchange.com/q/1214667/133781 math.stackexchange.com/q/2014146/133781 and math.stackexchange.com/q/873088/133781
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– Xam
Mar 12 '17 at 16:02
$begingroup$
This and variants have been asked before. Check for example these: math.stackexchange.com/q/1214667/133781 math.stackexchange.com/q/2014146/133781 and math.stackexchange.com/q/873088/133781
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– Xam
Mar 12 '17 at 16:02
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1 Answer
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Hint: Let $0 neq I subsetneq K[x]$ be an ideal. Let $0neq fin I$ such that
$$ deg(f)= min{ deg(g) : gin I }.$$
Use your argument above to show $I=(f)$.
For your second question I'd suggest that you show that the ideal $(2, x)subseteq mathbb{Z}[x]$ is not principal.
answered Mar 12 '17 at 16:02
Severin SchravenSeverin Schraven
6,1481934
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Let $0 neq I subsetneq K[x]$ be an ideal. Let $0neq fin I$ such that
$$ deg(f)= min{ deg(g) : gin I }.$$
Use your argument above to show $I=(f)$.
For your second question I'd suggest that you show that the ideal $(2, x)subseteq mathbb{Z}[x]$ is not principal.
answered Mar 12 '17 at 16:02
Severin SchravenSeverin Schraven
6,1481934
$endgroup$
add a comment |
$begingroup$
Hint: Let $0 neq I subsetneq K[x]$ be an ideal. Let $0neq fin I$ such that
$$ deg(f)= min{ deg(g) : gin I }.$$
Use your argument above to show $I=(f)$.
For your second question I'd suggest that you show that the ideal $(2, x)subseteq mathbb{Z}[x]$ is not principal.
answered Mar 12 '17 at 16:02
Severin SchravenSeverin Schraven
6,1481934
$endgroup$
add a comment |
$begingroup$
Hint: Let $0 neq I subsetneq K[x]$ be an ideal. Let $0neq fin I$ such that
$$ deg(f)= min{ deg(g) : gin I }.$$
Use your argument above to show $I=(f)$.
For your second question I'd suggest that you show that the ideal $(2, x)subseteq mathbb{Z}[x]$ is not principal.
answered Mar 12 '17 at 16:02
Severin SchravenSeverin Schraven
6,1481934
$endgroup$
Hint: Let $0 neq I subsetneq K[x]$ be an ideal. Let $0neq fin I$ such that
$$ deg(f)= min{ deg(g) : gin I }.$$
Use your argument above to show $I=(f)$.
For your second question I'd suggest that you show that the ideal $(2, x)subseteq mathbb{Z}[x]$ is not principal.
answered Mar 12 '17 at 16:02
Severin SchravenSeverin Schraven
6,1481934
answered Mar 12 '17 at 16:02
Severin SchravenSeverin Schraven
6,1481934
answered Mar 12 '17 at 16:02
Severin SchravenSeverin Schraven
6,1481934
answered Mar 12 '17 at 16:02
Severin SchravenSeverin Schraven
6,1481934
6,1481934
add a comment |
add a comment |
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1
$begingroup$
This and variants have been asked before. Check for example these: math.stackexchange.com/q/1214667/133781 math.stackexchange.com/q/2014146/133781 and math.stackexchange.com/q/873088/133781
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– Xam
Mar 12 '17 at 16:02