If $(p)$ is a proper subset of a proper ideal $I$, then is $I$ prime?
$begingroup$
Let $R$ be the ring of algebraic integers of a quadratic imaginary number field $mathbb Q[sqrt{d}]$ for a negative square-free integer $d$. For a prime integer $p$, $(p)$ is a prime ideal or is the product $P overline P$ of some prime ideal $P$ and $overline P$, the ideal consisting of the complex conjugates of elements of $P$. Why does this mean if $(p)$ is a proper subset of a proper ideal $I$ of $R$, then $I$ is prime?
If $(p)$ is a prime ideal, then $(p)$ is a maximal ideal so $(p)=I$.
I don't know how to say $(p)=P overline P subset I subset R$ implies $I$ is a prime ideal.
Our definition of a prime ideal $P$ is that $P$ is nonzero and if the product $CD$ of two ideals $C$ and $D$ is a subset of $P$, then $C$ or $D$ is a subset of $P$.
Thanks in advance!
abstract-algebra ring-theory maximal-and-prime-ideals
asked Jan 10 at 11:47
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
$endgroup$
add a comment |
$begingroup$
Let $R$ be the ring of algebraic integers of a quadratic imaginary number field $mathbb Q[sqrt{d}]$ for a negative square-free integer $d$. For a prime integer $p$, $(p)$ is a prime ideal or is the product $P overline P$ of some prime ideal $P$ and $overline P$, the ideal consisting of the complex conjugates of elements of $P$. Why does this mean if $(p)$ is a proper subset of a proper ideal $I$ of $R$, then $I$ is prime?
If $(p)$ is a prime ideal, then $(p)$ is a maximal ideal so $(p)=I$.
I don't know how to say $(p)=P overline P subset I subset R$ implies $I$ is a prime ideal.
Our definition of a prime ideal $P$ is that $P$ is nonzero and if the product $CD$ of two ideals $C$ and $D$ is a subset of $P$, then $C$ or $D$ is a subset of $P$.
Thanks in advance!
abstract-algebra ring-theory maximal-and-prime-ideals
asked Jan 10 at 11:47
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
$endgroup$
1
$begingroup$
I'm afraid what you claim is not true. $I=(p)$ clearly contains $(p)$, but is not always prime. For an explicit example, consider $I=(2)$ in $mathbb Z[i]$. Then $(1+i)^2in I$ but $1+inotin I$, so $I$ is not prime.
$endgroup$
– Wojowu
Jan 20 at 9:29
$begingroup$
@Wojowu I changed to $(p)=P overline P subset I subset R$
$endgroup$
– Ekhin Taylor R. Wilson
Jan 20 at 10:02
1
$begingroup$
I see, now it makes more sense.
$endgroup$
– Wojowu
Jan 20 at 10:04
add a comment |
$begingroup$
Let $R$ be the ring of algebraic integers of a quadratic imaginary number field $mathbb Q[sqrt{d}]$ for a negative square-free integer $d$. For a prime integer $p$, $(p)$ is a prime ideal or is the product $P overline P$ of some prime ideal $P$ and $overline P$, the ideal consisting of the complex conjugates of elements of $P$. Why does this mean if $(p)$ is a proper subset of a proper ideal $I$ of $R$, then $I$ is prime?
If $(p)$ is a prime ideal, then $(p)$ is a maximal ideal so $(p)=I$.
I don't know how to say $(p)=P overline P subset I subset R$ implies $I$ is a prime ideal.
Our definition of a prime ideal $P$ is that $P$ is nonzero and if the product $CD$ of two ideals $C$ and $D$ is a subset of $P$, then $C$ or $D$ is a subset of $P$.
Thanks in advance!
abstract-algebra ring-theory maximal-and-prime-ideals
asked Jan 10 at 11:47
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
$endgroup$
Let $R$ be the ring of algebraic integers of a quadratic imaginary number field $mathbb Q[sqrt{d}]$ for a negative square-free integer $d$. For a prime integer $p$, $(p)$ is a prime ideal or is the product $P overline P$ of some prime ideal $P$ and $overline P$, the ideal consisting of the complex conjugates of elements of $P$. Why does this mean if $(p)$ is a proper subset of a proper ideal $I$ of $R$, then $I$ is prime?
If $(p)$ is a prime ideal, then $(p)$ is a maximal ideal so $(p)=I$.
I don't know how to say $(p)=P overline P subset I subset R$ implies $I$ is a prime ideal.
Our definition of a prime ideal $P$ is that $P$ is nonzero and if the product $CD$ of two ideals $C$ and $D$ is a subset of $P$, then $C$ or $D$ is a subset of $P$.
Thanks in advance!
abstract-algebra ring-theory maximal-and-prime-ideals
abstract-algebra ring-theory maximal-and-prime-ideals
asked Jan 10 at 11:47
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
asked Jan 10 at 11:47
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
edited Jan 20 at 10:01
Ekhin Taylor R. Wilson
asked Jan 10 at 11:47
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
asked Jan 10 at 11:47
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
asked Jan 10 at 11:47
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
558
1
$begingroup$
I'm afraid what you claim is not true. $I=(p)$ clearly contains $(p)$, but is not always prime. For an explicit example, consider $I=(2)$ in $mathbb Z[i]$. Then $(1+i)^2in I$ but $1+inotin I$, so $I$ is not prime.
$endgroup$
– Wojowu
Jan 20 at 9:29
$begingroup$
@Wojowu I changed to $(p)=P overline P subset I subset R$
$endgroup$
– Ekhin Taylor R. Wilson
Jan 20 at 10:02
1
$begingroup$
I see, now it makes more sense.
$endgroup$
– Wojowu
Jan 20 at 10:04
add a comment |
1
$begingroup$
I'm afraid what you claim is not true. $I=(p)$ clearly contains $(p)$, but is not always prime. For an explicit example, consider $I=(2)$ in $mathbb Z[i]$. Then $(1+i)^2in I$ but $1+inotin I$, so $I$ is not prime.
$endgroup$
– Wojowu
Jan 20 at 9:29
$begingroup$
@Wojowu I changed to $(p)=P overline P subset I subset R$
$endgroup$
– Ekhin Taylor R. Wilson
Jan 20 at 10:02
1
$begingroup$
I see, now it makes more sense.
$endgroup$
– Wojowu
Jan 20 at 10:04
1
1
$begingroup$
I'm afraid what you claim is not true. $I=(p)$ clearly contains $(p)$, but is not always prime. For an explicit example, consider $I=(2)$ in $mathbb Z[i]$. Then $(1+i)^2in I$ but $1+inotin I$, so $I$ is not prime.
$endgroup$
– Wojowu
Jan 20 at 9:29
$begingroup$
I'm afraid what you claim is not true. $I=(p)$ clearly contains $(p)$, but is not always prime. For an explicit example, consider $I=(2)$ in $mathbb Z[i]$. Then $(1+i)^2in I$ but $1+inotin I$, so $I$ is not prime.
$endgroup$
– Wojowu
Jan 20 at 9:29
$begingroup$
@Wojowu I changed to $(p)=P overline P subset I subset R$
$endgroup$
– Ekhin Taylor R. Wilson
Jan 20 at 10:02
$begingroup$
@Wojowu I changed to $(p)=P overline P subset I subset R$
$endgroup$
– Ekhin Taylor R. Wilson
Jan 20 at 10:02
1
1
$begingroup$
I see, now it makes more sense.
$endgroup$
– Wojowu
Jan 20 at 10:04
$begingroup$
I see, now it makes more sense.
$endgroup$
– Wojowu
Jan 20 at 10:04
add a comment |
2 Answers
2
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oldest
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$begingroup$
Here is a straightforward proof. Since we are in a quadratic field, it's not hard to see that $R/(p)$ has $p^2$ elements (since, as a group, $R$ is free abelian on two generators). If $I$ is a proper ideal properly containing $(p)$, then the quotient $R/I$ is isomorphic to a quotient of $R/(p)$ by the image of $I$ modulo $(p)$. From there it's clear $R/I$ has $p$ elements, so is a field, implying $I$ is maximal, hence prime.
answered Jan 20 at 10:10
WojowuWojowu
19.3k23274
$endgroup$
add a comment |
$begingroup$
I will focus on the case you still don't solve.
Suppose that $I$ is a proper ideal such that $Poverline Psubsetneq I$ (notice that the inclusion should be strict, otherwise $Poverline P=I$ is a counterexample). If $M$ is a maximal ideal containing $I$ then $Poverline Psubsetneq M$ implies either $Psubsetneq M$ or $overline{P}subsetneq M$. WLOG we have $Psubsetneq M$ and hence $P$ is a prime ideal which is neither $(0)$ or maximal. This contradicts the fact that the ring of integers have dimension $1$ (i.e., every nonzero prime ideal is maximal).
A more elucidative proof would be using the properties of the ideal factorization in number fields. If $Poverline{P}subsetneq I$ then we have that $I$ properly divides $Poverline{P}$ and hence either $I=P$ or $I=overline{P}$.
edited Jan 20 at 9:24
user26857
39.5k124284
answered Jan 10 at 12:35
yamete kudasaiyamete kudasai
1,160818
$endgroup$
add a comment |
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2 Answers
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$begingroup$
Here is a straightforward proof. Since we are in a quadratic field, it's not hard to see that $R/(p)$ has $p^2$ elements (since, as a group, $R$ is free abelian on two generators). If $I$ is a proper ideal properly containing $(p)$, then the quotient $R/I$ is isomorphic to a quotient of $R/(p)$ by the image of $I$ modulo $(p)$. From there it's clear $R/I$ has $p$ elements, so is a field, implying $I$ is maximal, hence prime.
answered Jan 20 at 10:10
WojowuWojowu
19.3k23274
$endgroup$
add a comment |
$begingroup$
Here is a straightforward proof. Since we are in a quadratic field, it's not hard to see that $R/(p)$ has $p^2$ elements (since, as a group, $R$ is free abelian on two generators). If $I$ is a proper ideal properly containing $(p)$, then the quotient $R/I$ is isomorphic to a quotient of $R/(p)$ by the image of $I$ modulo $(p)$. From there it's clear $R/I$ has $p$ elements, so is a field, implying $I$ is maximal, hence prime.
answered Jan 20 at 10:10
WojowuWojowu
19.3k23274
$endgroup$
add a comment |
$begingroup$
Here is a straightforward proof. Since we are in a quadratic field, it's not hard to see that $R/(p)$ has $p^2$ elements (since, as a group, $R$ is free abelian on two generators). If $I$ is a proper ideal properly containing $(p)$, then the quotient $R/I$ is isomorphic to a quotient of $R/(p)$ by the image of $I$ modulo $(p)$. From there it's clear $R/I$ has $p$ elements, so is a field, implying $I$ is maximal, hence prime.
answered Jan 20 at 10:10
WojowuWojowu
19.3k23274
$endgroup$
Here is a straightforward proof. Since we are in a quadratic field, it's not hard to see that $R/(p)$ has $p^2$ elements (since, as a group, $R$ is free abelian on two generators). If $I$ is a proper ideal properly containing $(p)$, then the quotient $R/I$ is isomorphic to a quotient of $R/(p)$ by the image of $I$ modulo $(p)$. From there it's clear $R/I$ has $p$ elements, so is a field, implying $I$ is maximal, hence prime.
answered Jan 20 at 10:10
WojowuWojowu
19.3k23274
answered Jan 20 at 10:10
WojowuWojowu
19.3k23274
answered Jan 20 at 10:10
WojowuWojowu
19.3k23274
answered Jan 20 at 10:10
WojowuWojowu
19.3k23274
19.3k23274
add a comment |
add a comment |
$begingroup$
I will focus on the case you still don't solve.
Suppose that $I$ is a proper ideal such that $Poverline Psubsetneq I$ (notice that the inclusion should be strict, otherwise $Poverline P=I$ is a counterexample). If $M$ is a maximal ideal containing $I$ then $Poverline Psubsetneq M$ implies either $Psubsetneq M$ or $overline{P}subsetneq M$. WLOG we have $Psubsetneq M$ and hence $P$ is a prime ideal which is neither $(0)$ or maximal. This contradicts the fact that the ring of integers have dimension $1$ (i.e., every nonzero prime ideal is maximal).
A more elucidative proof would be using the properties of the ideal factorization in number fields. If $Poverline{P}subsetneq I$ then we have that $I$ properly divides $Poverline{P}$ and hence either $I=P$ or $I=overline{P}$.
edited Jan 20 at 9:24
user26857
39.5k124284
answered Jan 10 at 12:35
yamete kudasaiyamete kudasai
1,160818
$endgroup$
add a comment |
$begingroup$
I will focus on the case you still don't solve.
Suppose that $I$ is a proper ideal such that $Poverline Psubsetneq I$ (notice that the inclusion should be strict, otherwise $Poverline P=I$ is a counterexample). If $M$ is a maximal ideal containing $I$ then $Poverline Psubsetneq M$ implies either $Psubsetneq M$ or $overline{P}subsetneq M$. WLOG we have $Psubsetneq M$ and hence $P$ is a prime ideal which is neither $(0)$ or maximal. This contradicts the fact that the ring of integers have dimension $1$ (i.e., every nonzero prime ideal is maximal).
A more elucidative proof would be using the properties of the ideal factorization in number fields. If $Poverline{P}subsetneq I$ then we have that $I$ properly divides $Poverline{P}$ and hence either $I=P$ or $I=overline{P}$.
edited Jan 20 at 9:24
user26857
39.5k124284
answered Jan 10 at 12:35
yamete kudasaiyamete kudasai
1,160818
$endgroup$
add a comment |
$begingroup$
I will focus on the case you still don't solve.
Suppose that $I$ is a proper ideal such that $Poverline Psubsetneq I$ (notice that the inclusion should be strict, otherwise $Poverline P=I$ is a counterexample). If $M$ is a maximal ideal containing $I$ then $Poverline Psubsetneq M$ implies either $Psubsetneq M$ or $overline{P}subsetneq M$. WLOG we have $Psubsetneq M$ and hence $P$ is a prime ideal which is neither $(0)$ or maximal. This contradicts the fact that the ring of integers have dimension $1$ (i.e., every nonzero prime ideal is maximal).
A more elucidative proof would be using the properties of the ideal factorization in number fields. If $Poverline{P}subsetneq I$ then we have that $I$ properly divides $Poverline{P}$ and hence either $I=P$ or $I=overline{P}$.
edited Jan 20 at 9:24
user26857
39.5k124284
answered Jan 10 at 12:35
yamete kudasaiyamete kudasai
1,160818
$endgroup$
I will focus on the case you still don't solve.
Suppose that $I$ is a proper ideal such that $Poverline Psubsetneq I$ (notice that the inclusion should be strict, otherwise $Poverline P=I$ is a counterexample). If $M$ is a maximal ideal containing $I$ then $Poverline Psubsetneq M$ implies either $Psubsetneq M$ or $overline{P}subsetneq M$. WLOG we have $Psubsetneq M$ and hence $P$ is a prime ideal which is neither $(0)$ or maximal. This contradicts the fact that the ring of integers have dimension $1$ (i.e., every nonzero prime ideal is maximal).
A more elucidative proof would be using the properties of the ideal factorization in number fields. If $Poverline{P}subsetneq I$ then we have that $I$ properly divides $Poverline{P}$ and hence either $I=P$ or $I=overline{P}$.
edited Jan 20 at 9:24
user26857
39.5k124284
answered Jan 10 at 12:35
yamete kudasaiyamete kudasai
1,160818
edited Jan 20 at 9:24
user26857
39.5k124284
edited Jan 20 at 9:24
user26857
39.5k124284
edited Jan 20 at 9:24
user26857
39.5k124284
39.5k124284
answered Jan 10 at 12:35
yamete kudasaiyamete kudasai
1,160818
answered Jan 10 at 12:35
yamete kudasaiyamete kudasai
1,160818
answered Jan 10 at 12:35
yamete kudasaiyamete kudasai
1,160818
1,160818
add a comment |
add a comment |
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$begingroup$
I'm afraid what you claim is not true. $I=(p)$ clearly contains $(p)$, but is not always prime. For an explicit example, consider $I=(2)$ in $mathbb Z[i]$. Then $(1+i)^2in I$ but $1+inotin I$, so $I$ is not prime.
$endgroup$
– Wojowu
Jan 20 at 9:29
$begingroup$
@Wojowu I changed to $(p)=P overline P subset I subset R$
$endgroup$
– Ekhin Taylor R. Wilson
Jan 20 at 10:02
1
$begingroup$
I see, now it makes more sense.
$endgroup$
– Wojowu
Jan 20 at 10:04