How to show that all prime ideals of $mathbb Z[sqrt -3]$ are maximal?Why are maximal ideals prime?Let $R$ be a finite commutative ring. Show that an ideal is maximal if and only if it is prime.Maximal ideals and Prime ideals.Prime and Maximal Ideals of $mathbbZ[x]$Prime and Maximal ideals in the polynomial ringsideal behind Maximal ideals are Prime idealsPrime ideals of $mathbbZ[sqrtd]$Prime/Maximal Ideals in $mathbbZ[sqrt d]$Describe all prime and maximal ideals of $mathbbZ_n$Proof of Maximal Ideals are prime ideals
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How to show that all prime ideals of $mathbb Z[sqrt -3]$ are maximal?
Why are maximal ideals prime?Let $R$ be a finite commutative ring. Show that an ideal is maximal if and only if it is prime.Maximal ideals and Prime ideals.Prime and Maximal Ideals of $mathbbZ[x]$Prime and Maximal ideals in the polynomial ringsideal behind Maximal ideals are Prime idealsPrime ideals of $mathbbZ[sqrtd]$Prime/Maximal Ideals in $mathbbZ[sqrt d]$Describe all prime and maximal ideals of $mathbbZ_n$Proof of Maximal Ideals are prime ideals
$begingroup$
How to show that all prime ideals of $mathbb Z[sqrt -3]$ are maximal?
My attempt:
$mathbb Z[sqrt -3]cong mathbb Z[x]/(x^2+3)$
Let p be prime ideal of $mathbb Z[sqrt -3]$
SO $mathbb Z[sqrt -3]/(p(x))$ is integral domain
I could not prove but I think it will be finite.
So it will be field so p become maximal ideal .
Please Help me show above claim
Any Help will be appreciated
abstract-algebra ring-theory
asked Mar 30 at 3:05
MathLoverMathLover
59710
$endgroup$
add a comment |
$begingroup$
How to show that all prime ideals of $mathbb Z[sqrt -3]$ are maximal?
My attempt:
$mathbb Z[sqrt -3]cong mathbb Z[x]/(x^2+3)$
Let p be prime ideal of $mathbb Z[sqrt -3]$
SO $mathbb Z[sqrt -3]/(p(x))$ is integral domain
I could not prove but I think it will be finite.
So it will be field so p become maximal ideal .
Please Help me show above claim
Any Help will be appreciated
abstract-algebra ring-theory
asked Mar 30 at 3:05
MathLoverMathLover
59710
$endgroup$
add a comment |
$begingroup$
How to show that all prime ideals of $mathbb Z[sqrt -3]$ are maximal?
My attempt:
$mathbb Z[sqrt -3]cong mathbb Z[x]/(x^2+3)$
Let p be prime ideal of $mathbb Z[sqrt -3]$
SO $mathbb Z[sqrt -3]/(p(x))$ is integral domain
I could not prove but I think it will be finite.
So it will be field so p become maximal ideal .
Please Help me show above claim
Any Help will be appreciated
abstract-algebra ring-theory
asked Mar 30 at 3:05
MathLoverMathLover
59710
$endgroup$
How to show that all prime ideals of $mathbb Z[sqrt -3]$ are maximal?
My attempt:
$mathbb Z[sqrt -3]cong mathbb Z[x]/(x^2+3)$
Let p be prime ideal of $mathbb Z[sqrt -3]$
SO $mathbb Z[sqrt -3]/(p(x))$ is integral domain
I could not prove but I think it will be finite.
So it will be field so p become maximal ideal .
Please Help me show above claim
Any Help will be appreciated
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Mar 30 at 3:05
MathLoverMathLover
59710
asked Mar 30 at 3:05
MathLoverMathLover
59710
asked Mar 30 at 3:05
MathLoverMathLover
59710
asked Mar 30 at 3:05
MathLoverMathLover
59710
asked Mar 30 at 3:05
MathLoverMathLover
59710
59710
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You need to exclude the zero ideal: it is prime but not maximal.
If $I$ is a nonzero ideal of $R=Bbb Z[sqrt-3]$, then $I$ has finite
index in $R$, so $R/I$ is a finite ring.
An ideal $I$ in a commutative ring $R$ is maximal iff $R/I$ is a field
and is prime iff $R/I$ is an integral domain.
A finite integral domain is a field, by a well-known theorem, so in
our example, if $I$ is a non-zero prime ideal, then $R/I$ is an integral
domain, so $R/I$ is a field, and $I$ must be maximal.
This argument also works when $R$ any order in an algebraic number field.
answered Mar 30 at 3:18
Lord Shark the UnknownLord Shark the Unknown
108k1162135
$endgroup$
$begingroup$
Dear Sir, Can You please tell me Why I has finite index as I had no argument for that?
$endgroup$
– MathLover
Mar 30 at 3:28
$begingroup$
@MathLover Show that every nonzero principal ideal has finite index: in fact the ideal generated by $(a+bsqrt-3)$ has index $a^2+3b^2$.
$endgroup$
– Lord Shark the Unknown
Mar 30 at 3:30
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
You need to exclude the zero ideal: it is prime but not maximal.
If $I$ is a nonzero ideal of $R=Bbb Z[sqrt-3]$, then $I$ has finite
index in $R$, so $R/I$ is a finite ring.
An ideal $I$ in a commutative ring $R$ is maximal iff $R/I$ is a field
and is prime iff $R/I$ is an integral domain.
A finite integral domain is a field, by a well-known theorem, so in
our example, if $I$ is a non-zero prime ideal, then $R/I$ is an integral
domain, so $R/I$ is a field, and $I$ must be maximal.
This argument also works when $R$ any order in an algebraic number field.
answered Mar 30 at 3:18
Lord Shark the UnknownLord Shark the Unknown
108k1162135
$endgroup$
$begingroup$
Dear Sir, Can You please tell me Why I has finite index as I had no argument for that?
$endgroup$
– MathLover
Mar 30 at 3:28
$begingroup$
@MathLover Show that every nonzero principal ideal has finite index: in fact the ideal generated by $(a+bsqrt-3)$ has index $a^2+3b^2$.
$endgroup$
– Lord Shark the Unknown
Mar 30 at 3:30
add a comment |
$begingroup$
You need to exclude the zero ideal: it is prime but not maximal.
If $I$ is a nonzero ideal of $R=Bbb Z[sqrt-3]$, then $I$ has finite
index in $R$, so $R/I$ is a finite ring.
An ideal $I$ in a commutative ring $R$ is maximal iff $R/I$ is a field
and is prime iff $R/I$ is an integral domain.
A finite integral domain is a field, by a well-known theorem, so in
our example, if $I$ is a non-zero prime ideal, then $R/I$ is an integral
domain, so $R/I$ is a field, and $I$ must be maximal.
This argument also works when $R$ any order in an algebraic number field.
answered Mar 30 at 3:18
Lord Shark the UnknownLord Shark the Unknown
108k1162135
$endgroup$
$begingroup$
Dear Sir, Can You please tell me Why I has finite index as I had no argument for that?
$endgroup$
– MathLover
Mar 30 at 3:28
$begingroup$
@MathLover Show that every nonzero principal ideal has finite index: in fact the ideal generated by $(a+bsqrt-3)$ has index $a^2+3b^2$.
$endgroup$
– Lord Shark the Unknown
Mar 30 at 3:30
add a comment |
$begingroup$
You need to exclude the zero ideal: it is prime but not maximal.
If $I$ is a nonzero ideal of $R=Bbb Z[sqrt-3]$, then $I$ has finite
index in $R$, so $R/I$ is a finite ring.
An ideal $I$ in a commutative ring $R$ is maximal iff $R/I$ is a field
and is prime iff $R/I$ is an integral domain.
A finite integral domain is a field, by a well-known theorem, so in
our example, if $I$ is a non-zero prime ideal, then $R/I$ is an integral
domain, so $R/I$ is a field, and $I$ must be maximal.
This argument also works when $R$ any order in an algebraic number field.
answered Mar 30 at 3:18
Lord Shark the UnknownLord Shark the Unknown
108k1162135
$endgroup$
You need to exclude the zero ideal: it is prime but not maximal.
If $I$ is a nonzero ideal of $R=Bbb Z[sqrt-3]$, then $I$ has finite
index in $R$, so $R/I$ is a finite ring.
An ideal $I$ in a commutative ring $R$ is maximal iff $R/I$ is a field
and is prime iff $R/I$ is an integral domain.
A finite integral domain is a field, by a well-known theorem, so in
our example, if $I$ is a non-zero prime ideal, then $R/I$ is an integral
domain, so $R/I$ is a field, and $I$ must be maximal.
This argument also works when $R$ any order in an algebraic number field.
answered Mar 30 at 3:18
Lord Shark the UnknownLord Shark the Unknown
108k1162135
answered Mar 30 at 3:18
Lord Shark the UnknownLord Shark the Unknown
108k1162135
answered Mar 30 at 3:18
Lord Shark the UnknownLord Shark the Unknown
108k1162135
answered Mar 30 at 3:18
Lord Shark the UnknownLord Shark the Unknown
108k1162135
108k1162135
$begingroup$
Dear Sir, Can You please tell me Why I has finite index as I had no argument for that?
$endgroup$
– MathLover
Mar 30 at 3:28
$begingroup$
@MathLover Show that every nonzero principal ideal has finite index: in fact the ideal generated by $(a+bsqrt-3)$ has index $a^2+3b^2$.
$endgroup$
– Lord Shark the Unknown
Mar 30 at 3:30
add a comment |
$begingroup$
Dear Sir, Can You please tell me Why I has finite index as I had no argument for that?
$endgroup$
– MathLover
Mar 30 at 3:28
$begingroup$
@MathLover Show that every nonzero principal ideal has finite index: in fact the ideal generated by $(a+bsqrt-3)$ has index $a^2+3b^2$.
$endgroup$
– Lord Shark the Unknown
Mar 30 at 3:30
$begingroup$
Dear Sir, Can You please tell me Why I has finite index as I had no argument for that?
$endgroup$
– MathLover
Mar 30 at 3:28
$begingroup$
Dear Sir, Can You please tell me Why I has finite index as I had no argument for that?
$endgroup$
– MathLover
Mar 30 at 3:28
$begingroup$
@MathLover Show that every nonzero principal ideal has finite index: in fact the ideal generated by $(a+bsqrt-3)$ has index $a^2+3b^2$.
$endgroup$
– Lord Shark the Unknown
Mar 30 at 3:30
$begingroup$
@MathLover Show that every nonzero principal ideal has finite index: in fact the ideal generated by $(a+bsqrt-3)$ has index $a^2+3b^2$.
$endgroup$
– Lord Shark the Unknown
Mar 30 at 3:30
add a comment |
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