An Ideal I of $ mathbb Z[sqrt 3]$ generated by an integer prime, such that $mathbb Z[ sqrt 3]/I$ is not an integral domain. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that every nonzero prime ideal is maximal in $mathbbZ[sqrtd]$Verifying proof :an Ideal $P$ is prime Ideal if $R/P$ is an integral domain.Integral domain without prime elements which is not a field$R$ be a commutative ring with unity such that every prime ideal contains no non-zero zero divisor , then is $R$ an integral domain?Why do we require a principal ideal domain to be an integral domain?$P$ is a prime ideal of $R$ iff $R/P$ is an integral domain. : $P≠R$Is $p=257$ prime in $mathbb Z[sqrt -92]$?$(x)$ prime ideal in $R[x]$ iff $R$ integral domain by contrapositiveThe localization of the ring $mathbbZ times mathbbZ$ at every prime ideal is an integral domainIdeal with no zero divisors implies integral domain?
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An Ideal I of $ mathbb Z[sqrt 3]$ generated by an integer prime, such that $mathbb Z[ sqrt 3]/I$ is not an integral domain.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that every nonzero prime ideal is maximal in $mathbbZ[sqrtd]$Verifying proof :an Ideal $P$ is prime Ideal if $R/P$ is an integral domain.Integral domain without prime elements which is not a field$R$ be a commutative ring with unity such that every prime ideal contains no non-zero zero divisor , then is $R$ an integral domain?Why do we require a principal ideal domain to be an integral domain?$P$ is a prime ideal of $R$ iff $R/P$ is an integral domain. : $P≠R$Is $p=257$ prime in $mathbb Z[sqrt -92]$?$(x)$ prime ideal in $R[x]$ iff $R$ integral domain by contrapositiveThe localization of the ring $mathbbZ times mathbbZ$ at every prime ideal is an integral domainIdeal with no zero divisors implies integral domain?
$begingroup$
Find an ideal I of $ mathbb Z[sqrt
3]$ generated by an integer prime (i.e. a prime
number in $mathbb Z$) such that $mathbb Z[
sqrt
3]/I$ is not an integral domain.
Thoughts: We need to find some prime number that generates an ideal. This ideal should have the property that upon constructing a quotient of its ring, it should exhibit zero divisors. If we choose the prime number to be $3$, we consider $I =( 3)$. Then observe that $$sqrt3=0+ 1 cdot sqrt3 in mathbb Z[sqrt3]$$
And also that $sqrt3$ is not divisible by $3$ in $mathbb Z[
sqrt
3]$, therefore its residue class modulo 3 will be $overlinesqrt3$
$$ overlinesqrt3 cdot overlinesqrt3 =3 equiv 0 bmod 3$$
So in $mathbb Z[
sqrt
3]/(3)$ we observe that there is at least one zero divisor and therefore it cannot be an integral domain.
I'm fairly new to integral domains and ideals, is this reasoning correct/on point?
abstract-algebra proof-verification ideals integral-domain
asked Mar 26 at 17:49
Wesley StrikWesley Strik
2,201424
$endgroup$
add a comment |
$begingroup$
Find an ideal I of $ mathbb Z[sqrt
3]$ generated by an integer prime (i.e. a prime
number in $mathbb Z$) such that $mathbb Z[
sqrt
3]/I$ is not an integral domain.
Thoughts: We need to find some prime number that generates an ideal. This ideal should have the property that upon constructing a quotient of its ring, it should exhibit zero divisors. If we choose the prime number to be $3$, we consider $I =( 3)$. Then observe that $$sqrt3=0+ 1 cdot sqrt3 in mathbb Z[sqrt3]$$
And also that $sqrt3$ is not divisible by $3$ in $mathbb Z[
sqrt
3]$, therefore its residue class modulo 3 will be $overlinesqrt3$
$$ overlinesqrt3 cdot overlinesqrt3 =3 equiv 0 bmod 3$$
So in $mathbb Z[
sqrt
3]/(3)$ we observe that there is at least one zero divisor and therefore it cannot be an integral domain.
I'm fairly new to integral domains and ideals, is this reasoning correct/on point?
abstract-algebra proof-verification ideals integral-domain
asked Mar 26 at 17:49
Wesley StrikWesley Strik
2,201424
$endgroup$
1
$begingroup$
Your reasoning is correct. The important point is that $sqrt3notin I= (3)_mathbbZ[sqrt3]$, so $overlinesqrt3notequiv overline0$, but $sqrt3sqrt3=3in I$ so $mathbbZ[sqrt3]/I$ is not a domain.
$endgroup$
– Kolja
Mar 26 at 17:59
2
$begingroup$
More generally this exact reasoning will work for any prime $p$ in which $3$ is a square, because then the ideal $(p)$ will not be prime in $mathbbZ[sqrt3]$ so that the quotient is not a domain.
$endgroup$
– JonHales
Mar 26 at 18:02
add a comment |
$begingroup$
Find an ideal I of $ mathbb Z[sqrt
3]$ generated by an integer prime (i.e. a prime
number in $mathbb Z$) such that $mathbb Z[
sqrt
3]/I$ is not an integral domain.
Thoughts: We need to find some prime number that generates an ideal. This ideal should have the property that upon constructing a quotient of its ring, it should exhibit zero divisors. If we choose the prime number to be $3$, we consider $I =( 3)$. Then observe that $$sqrt3=0+ 1 cdot sqrt3 in mathbb Z[sqrt3]$$
And also that $sqrt3$ is not divisible by $3$ in $mathbb Z[
sqrt
3]$, therefore its residue class modulo 3 will be $overlinesqrt3$
$$ overlinesqrt3 cdot overlinesqrt3 =3 equiv 0 bmod 3$$
So in $mathbb Z[
sqrt
3]/(3)$ we observe that there is at least one zero divisor and therefore it cannot be an integral domain.
I'm fairly new to integral domains and ideals, is this reasoning correct/on point?
abstract-algebra proof-verification ideals integral-domain
asked Mar 26 at 17:49
Wesley StrikWesley Strik
2,201424
$endgroup$
Find an ideal I of $ mathbb Z[sqrt
3]$ generated by an integer prime (i.e. a prime
number in $mathbb Z$) such that $mathbb Z[
sqrt
3]/I$ is not an integral domain.
Thoughts: We need to find some prime number that generates an ideal. This ideal should have the property that upon constructing a quotient of its ring, it should exhibit zero divisors. If we choose the prime number to be $3$, we consider $I =( 3)$. Then observe that $$sqrt3=0+ 1 cdot sqrt3 in mathbb Z[sqrt3]$$
And also that $sqrt3$ is not divisible by $3$ in $mathbb Z[
sqrt
3]$, therefore its residue class modulo 3 will be $overlinesqrt3$
$$ overlinesqrt3 cdot overlinesqrt3 =3 equiv 0 bmod 3$$
So in $mathbb Z[
sqrt
3]/(3)$ we observe that there is at least one zero divisor and therefore it cannot be an integral domain.
I'm fairly new to integral domains and ideals, is this reasoning correct/on point?
abstract-algebra proof-verification ideals integral-domain
abstract-algebra proof-verification ideals integral-domain
asked Mar 26 at 17:49
Wesley StrikWesley Strik
2,201424
asked Mar 26 at 17:49
Wesley StrikWesley Strik
2,201424
asked Mar 26 at 17:49
Wesley StrikWesley Strik
2,201424
asked Mar 26 at 17:49
Wesley StrikWesley Strik
2,201424
asked Mar 26 at 17:49
Wesley StrikWesley Strik
2,201424
2,201424
1
$begingroup$
Your reasoning is correct. The important point is that $sqrt3notin I= (3)_mathbbZ[sqrt3]$, so $overlinesqrt3notequiv overline0$, but $sqrt3sqrt3=3in I$ so $mathbbZ[sqrt3]/I$ is not a domain.
$endgroup$
– Kolja
Mar 26 at 17:59
2
$begingroup$
More generally this exact reasoning will work for any prime $p$ in which $3$ is a square, because then the ideal $(p)$ will not be prime in $mathbbZ[sqrt3]$ so that the quotient is not a domain.
$endgroup$
– JonHales
Mar 26 at 18:02
add a comment |
1
$begingroup$
Your reasoning is correct. The important point is that $sqrt3notin I= (3)_mathbbZ[sqrt3]$, so $overlinesqrt3notequiv overline0$, but $sqrt3sqrt3=3in I$ so $mathbbZ[sqrt3]/I$ is not a domain.
$endgroup$
– Kolja
Mar 26 at 17:59
2
$begingroup$
More generally this exact reasoning will work for any prime $p$ in which $3$ is a square, because then the ideal $(p)$ will not be prime in $mathbbZ[sqrt3]$ so that the quotient is not a domain.
$endgroup$
– JonHales
Mar 26 at 18:02
1
1
$begingroup$
Your reasoning is correct. The important point is that $sqrt3notin I= (3)_mathbbZ[sqrt3]$, so $overlinesqrt3notequiv overline0$, but $sqrt3sqrt3=3in I$ so $mathbbZ[sqrt3]/I$ is not a domain.
$endgroup$
– Kolja
Mar 26 at 17:59
$begingroup$
Your reasoning is correct. The important point is that $sqrt3notin I= (3)_mathbbZ[sqrt3]$, so $overlinesqrt3notequiv overline0$, but $sqrt3sqrt3=3in I$ so $mathbbZ[sqrt3]/I$ is not a domain.
$endgroup$
– Kolja
Mar 26 at 17:59
2
2
$begingroup$
More generally this exact reasoning will work for any prime $p$ in which $3$ is a square, because then the ideal $(p)$ will not be prime in $mathbbZ[sqrt3]$ so that the quotient is not a domain.
$endgroup$
– JonHales
Mar 26 at 18:02
$begingroup$
More generally this exact reasoning will work for any prime $p$ in which $3$ is a square, because then the ideal $(p)$ will not be prime in $mathbbZ[sqrt3]$ so that the quotient is not a domain.
$endgroup$
– JonHales
Mar 26 at 18:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Couldn't fit this in a comment so I will write it here. Your reasoning is correct.
Furhtermore for all primes $pinmathbbZsetminus2,3$, let $mathfrakp=(p)_mathbbZ[sqrt3]$.
Then $mathfrakp$ is a prime ideal if and only if $x^2-3$ has no roots modulo $p$, in which case $mathbbZ[sqrt3]/mathfrakpcong mathbbF_p^2$.
On the other hand $mathfrakp$ is not prime if and only if $x^2-3$ has a root $r$ modulo $p$ (so it has two roots, $-r$ is the other one), and in this case $mathbbZ[sqrt3]/mathfrakpcong mathbbF_p times mathbbF_p$.
When the prime $p=3$ you will have $mathbbZ[sqrt3]/mathfrakpcong mathbbF_3[x]/(x^2)$.
When the prime $p=2$ you will have $mathbbZ[sqrt3]/mathfrakpcong mathbbF_2[x]/((x-1)^2)cong mathbbF_2[x]/(x^2)$.
In other words $mathbbZ[sqrt3] = mathbbZ[x]/(x^2-3)$ and $mathbbZ[sqrt3]/(p)_mathbbZ[sqrt3] cong mathbbF_p[x]/(x^2-3)$
answered Mar 26 at 18:22
KoljaKolja
625310
$endgroup$
add a comment |
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$begingroup$
Couldn't fit this in a comment so I will write it here. Your reasoning is correct.
Furhtermore for all primes $pinmathbbZsetminus2,3$, let $mathfrakp=(p)_mathbbZ[sqrt3]$.
Then $mathfrakp$ is a prime ideal if and only if $x^2-3$ has no roots modulo $p$, in which case $mathbbZ[sqrt3]/mathfrakpcong mathbbF_p^2$.
On the other hand $mathfrakp$ is not prime if and only if $x^2-3$ has a root $r$ modulo $p$ (so it has two roots, $-r$ is the other one), and in this case $mathbbZ[sqrt3]/mathfrakpcong mathbbF_p times mathbbF_p$.
When the prime $p=3$ you will have $mathbbZ[sqrt3]/mathfrakpcong mathbbF_3[x]/(x^2)$.
When the prime $p=2$ you will have $mathbbZ[sqrt3]/mathfrakpcong mathbbF_2[x]/((x-1)^2)cong mathbbF_2[x]/(x^2)$.
In other words $mathbbZ[sqrt3] = mathbbZ[x]/(x^2-3)$ and $mathbbZ[sqrt3]/(p)_mathbbZ[sqrt3] cong mathbbF_p[x]/(x^2-3)$
answered Mar 26 at 18:22
KoljaKolja
625310
$endgroup$
add a comment |
$begingroup$
Couldn't fit this in a comment so I will write it here. Your reasoning is correct.
Furhtermore for all primes $pinmathbbZsetminus2,3$, let $mathfrakp=(p)_mathbbZ[sqrt3]$.
Then $mathfrakp$ is a prime ideal if and only if $x^2-3$ has no roots modulo $p$, in which case $mathbbZ[sqrt3]/mathfrakpcong mathbbF_p^2$.
On the other hand $mathfrakp$ is not prime if and only if $x^2-3$ has a root $r$ modulo $p$ (so it has two roots, $-r$ is the other one), and in this case $mathbbZ[sqrt3]/mathfrakpcong mathbbF_p times mathbbF_p$.
When the prime $p=3$ you will have $mathbbZ[sqrt3]/mathfrakpcong mathbbF_3[x]/(x^2)$.
When the prime $p=2$ you will have $mathbbZ[sqrt3]/mathfrakpcong mathbbF_2[x]/((x-1)^2)cong mathbbF_2[x]/(x^2)$.
In other words $mathbbZ[sqrt3] = mathbbZ[x]/(x^2-3)$ and $mathbbZ[sqrt3]/(p)_mathbbZ[sqrt3] cong mathbbF_p[x]/(x^2-3)$
answered Mar 26 at 18:22
KoljaKolja
625310
$endgroup$
add a comment |
$begingroup$
Couldn't fit this in a comment so I will write it here. Your reasoning is correct.
Furhtermore for all primes $pinmathbbZsetminus2,3$, let $mathfrakp=(p)_mathbbZ[sqrt3]$.
Then $mathfrakp$ is a prime ideal if and only if $x^2-3$ has no roots modulo $p$, in which case $mathbbZ[sqrt3]/mathfrakpcong mathbbF_p^2$.
On the other hand $mathfrakp$ is not prime if and only if $x^2-3$ has a root $r$ modulo $p$ (so it has two roots, $-r$ is the other one), and in this case $mathbbZ[sqrt3]/mathfrakpcong mathbbF_p times mathbbF_p$.
When the prime $p=3$ you will have $mathbbZ[sqrt3]/mathfrakpcong mathbbF_3[x]/(x^2)$.
When the prime $p=2$ you will have $mathbbZ[sqrt3]/mathfrakpcong mathbbF_2[x]/((x-1)^2)cong mathbbF_2[x]/(x^2)$.
In other words $mathbbZ[sqrt3] = mathbbZ[x]/(x^2-3)$ and $mathbbZ[sqrt3]/(p)_mathbbZ[sqrt3] cong mathbbF_p[x]/(x^2-3)$
answered Mar 26 at 18:22
KoljaKolja
625310
$endgroup$
Couldn't fit this in a comment so I will write it here. Your reasoning is correct.
Furhtermore for all primes $pinmathbbZsetminus2,3$, let $mathfrakp=(p)_mathbbZ[sqrt3]$.
Then $mathfrakp$ is a prime ideal if and only if $x^2-3$ has no roots modulo $p$, in which case $mathbbZ[sqrt3]/mathfrakpcong mathbbF_p^2$.
On the other hand $mathfrakp$ is not prime if and only if $x^2-3$ has a root $r$ modulo $p$ (so it has two roots, $-r$ is the other one), and in this case $mathbbZ[sqrt3]/mathfrakpcong mathbbF_p times mathbbF_p$.
When the prime $p=3$ you will have $mathbbZ[sqrt3]/mathfrakpcong mathbbF_3[x]/(x^2)$.
When the prime $p=2$ you will have $mathbbZ[sqrt3]/mathfrakpcong mathbbF_2[x]/((x-1)^2)cong mathbbF_2[x]/(x^2)$.
In other words $mathbbZ[sqrt3] = mathbbZ[x]/(x^2-3)$ and $mathbbZ[sqrt3]/(p)_mathbbZ[sqrt3] cong mathbbF_p[x]/(x^2-3)$
answered Mar 26 at 18:22
KoljaKolja
625310
answered Mar 26 at 18:22
KoljaKolja
625310
answered Mar 26 at 18:22
KoljaKolja
625310
answered Mar 26 at 18:22
KoljaKolja
625310
625310
add a comment |
add a comment |
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$begingroup$
Your reasoning is correct. The important point is that $sqrt3notin I= (3)_mathbbZ[sqrt3]$, so $overlinesqrt3notequiv overline0$, but $sqrt3sqrt3=3in I$ so $mathbbZ[sqrt3]/I$ is not a domain.
$endgroup$
– Kolja
Mar 26 at 17:59
2
$begingroup$
More generally this exact reasoning will work for any prime $p$ in which $3$ is a square, because then the ideal $(p)$ will not be prime in $mathbbZ[sqrt3]$ so that the quotient is not a domain.
$endgroup$
– JonHales
Mar 26 at 18:02