All ideals in $mathbb{Z}[frac{1}{2}]$ are finitely-generated
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The question comes from a problem, which asks to find all prime and maximal ideals in $mathbb{Z}[frac{1}{2}] := {frac{a}{2^k} : kinmathbb{Z} , ain 2mathbb{Z}+1 } $. Well I solved it by assuiming that $mathbb{Z}[frac{1}{2}]$ is a $PID$, but I didn't succeed to prove that all the ideals are finitely generated (for those which are finitely generated I already demonstrated that they are generated by one element).
ring-theory
asked Dec 9 '18 at 10:14
dandan
512513
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add a comment |
$begingroup$
The question comes from a problem, which asks to find all prime and maximal ideals in $mathbb{Z}[frac{1}{2}] := {frac{a}{2^k} : kinmathbb{Z} , ain 2mathbb{Z}+1 } $. Well I solved it by assuiming that $mathbb{Z}[frac{1}{2}]$ is a $PID$, but I didn't succeed to prove that all the ideals are finitely generated (for those which are finitely generated I already demonstrated that they are generated by one element).
ring-theory
asked Dec 9 '18 at 10:14
dandan
512513
$endgroup$
1
$begingroup$
You want to show that $mathbb{Z}[1/2]$ is noetherian. First recall that the integers are noetherian and thus by the Gauss lemma so is the polynomial ring over the integers. Recall that quotient rings of noetherian rings are noetherian. I leave you to find an ideal $I$ such that $mathbb{Z}[1/2]cong mathbb{Z}[x]/I$.
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– Severin Schraven
Dec 9 '18 at 10:38
add a comment |
$begingroup$
The question comes from a problem, which asks to find all prime and maximal ideals in $mathbb{Z}[frac{1}{2}] := {frac{a}{2^k} : kinmathbb{Z} , ain 2mathbb{Z}+1 } $. Well I solved it by assuiming that $mathbb{Z}[frac{1}{2}]$ is a $PID$, but I didn't succeed to prove that all the ideals are finitely generated (for those which are finitely generated I already demonstrated that they are generated by one element).
ring-theory
asked Dec 9 '18 at 10:14
dandan
512513
$endgroup$
The question comes from a problem, which asks to find all prime and maximal ideals in $mathbb{Z}[frac{1}{2}] := {frac{a}{2^k} : kinmathbb{Z} , ain 2mathbb{Z}+1 } $. Well I solved it by assuiming that $mathbb{Z}[frac{1}{2}]$ is a $PID$, but I didn't succeed to prove that all the ideals are finitely generated (for those which are finitely generated I already demonstrated that they are generated by one element).
ring-theory
ring-theory
asked Dec 9 '18 at 10:14
dandan
512513
asked Dec 9 '18 at 10:14
dandan
512513
asked Dec 9 '18 at 10:14
dandan
512513
asked Dec 9 '18 at 10:14
dandan
512513
asked Dec 9 '18 at 10:14
dandan
512513
512513
1
$begingroup$
You want to show that $mathbb{Z}[1/2]$ is noetherian. First recall that the integers are noetherian and thus by the Gauss lemma so is the polynomial ring over the integers. Recall that quotient rings of noetherian rings are noetherian. I leave you to find an ideal $I$ such that $mathbb{Z}[1/2]cong mathbb{Z}[x]/I$.
$endgroup$
– Severin Schraven
Dec 9 '18 at 10:38
add a comment |
1
$begingroup$
You want to show that $mathbb{Z}[1/2]$ is noetherian. First recall that the integers are noetherian and thus by the Gauss lemma so is the polynomial ring over the integers. Recall that quotient rings of noetherian rings are noetherian. I leave you to find an ideal $I$ such that $mathbb{Z}[1/2]cong mathbb{Z}[x]/I$.
$endgroup$
– Severin Schraven
Dec 9 '18 at 10:38
1
1
$begingroup$
You want to show that $mathbb{Z}[1/2]$ is noetherian. First recall that the integers are noetherian and thus by the Gauss lemma so is the polynomial ring over the integers. Recall that quotient rings of noetherian rings are noetherian. I leave you to find an ideal $I$ such that $mathbb{Z}[1/2]cong mathbb{Z}[x]/I$.
$endgroup$
– Severin Schraven
Dec 9 '18 at 10:38
$begingroup$
You want to show that $mathbb{Z}[1/2]$ is noetherian. First recall that the integers are noetherian and thus by the Gauss lemma so is the polynomial ring over the integers. Recall that quotient rings of noetherian rings are noetherian. I leave you to find an ideal $I$ such that $mathbb{Z}[1/2]cong mathbb{Z}[x]/I$.
$endgroup$
– Severin Schraven
Dec 9 '18 at 10:38
add a comment |
1 Answer
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Take some ideal $I leq mathbb{Z}[frac{1}{2}]$. Now consider the following ideal of the integers $J_I leq mathbb{Z}$:
$$ J_I = <{a in mathbb{Z} : exists k in mathbb{Z}, ; ; frac{a}{2^k} in I }>.$$
(Here the reason we're taking the ideal generated by this set is because technically according to the definition you gave, we can divide by $2^k$ for some negative $k$.)
Note that $J_I$ is an ideal of $mathbb{Z}$, hence it is generated by an element $a^* in mathbb{Z}$. We show that $a^*$ generates $I$ (just multiply by $frac{1}{2^k}$ for the appropriate $k$). Obviously $a^* in I$, so $(a^*) subseteq I$.
On the other hand, take any element $i = frac{b}{2^k} in I$. It's Numerator can be written as $ b = a^*z$ for some $z in mathbb{Z}$. Hence $a^*frac{z}{2^k} = i$ and thus $I subseteq (a^*)$.
We conclude that every ideal is generated by one element.
answered Dec 9 '18 at 10:49
Yotam DYotam D
18212
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$begingroup$
Take some ideal $I leq mathbb{Z}[frac{1}{2}]$. Now consider the following ideal of the integers $J_I leq mathbb{Z}$:
$$ J_I = <{a in mathbb{Z} : exists k in mathbb{Z}, ; ; frac{a}{2^k} in I }>.$$
(Here the reason we're taking the ideal generated by this set is because technically according to the definition you gave, we can divide by $2^k$ for some negative $k$.)
Note that $J_I$ is an ideal of $mathbb{Z}$, hence it is generated by an element $a^* in mathbb{Z}$. We show that $a^*$ generates $I$ (just multiply by $frac{1}{2^k}$ for the appropriate $k$). Obviously $a^* in I$, so $(a^*) subseteq I$.
On the other hand, take any element $i = frac{b}{2^k} in I$. It's Numerator can be written as $ b = a^*z$ for some $z in mathbb{Z}$. Hence $a^*frac{z}{2^k} = i$ and thus $I subseteq (a^*)$.
We conclude that every ideal is generated by one element.
answered Dec 9 '18 at 10:49
Yotam DYotam D
18212
$endgroup$
add a comment |
$begingroup$
Take some ideal $I leq mathbb{Z}[frac{1}{2}]$. Now consider the following ideal of the integers $J_I leq mathbb{Z}$:
$$ J_I = <{a in mathbb{Z} : exists k in mathbb{Z}, ; ; frac{a}{2^k} in I }>.$$
(Here the reason we're taking the ideal generated by this set is because technically according to the definition you gave, we can divide by $2^k$ for some negative $k$.)
Note that $J_I$ is an ideal of $mathbb{Z}$, hence it is generated by an element $a^* in mathbb{Z}$. We show that $a^*$ generates $I$ (just multiply by $frac{1}{2^k}$ for the appropriate $k$). Obviously $a^* in I$, so $(a^*) subseteq I$.
On the other hand, take any element $i = frac{b}{2^k} in I$. It's Numerator can be written as $ b = a^*z$ for some $z in mathbb{Z}$. Hence $a^*frac{z}{2^k} = i$ and thus $I subseteq (a^*)$.
We conclude that every ideal is generated by one element.
answered Dec 9 '18 at 10:49
Yotam DYotam D
18212
$endgroup$
add a comment |
$begingroup$
Take some ideal $I leq mathbb{Z}[frac{1}{2}]$. Now consider the following ideal of the integers $J_I leq mathbb{Z}$:
$$ J_I = <{a in mathbb{Z} : exists k in mathbb{Z}, ; ; frac{a}{2^k} in I }>.$$
(Here the reason we're taking the ideal generated by this set is because technically according to the definition you gave, we can divide by $2^k$ for some negative $k$.)
Note that $J_I$ is an ideal of $mathbb{Z}$, hence it is generated by an element $a^* in mathbb{Z}$. We show that $a^*$ generates $I$ (just multiply by $frac{1}{2^k}$ for the appropriate $k$). Obviously $a^* in I$, so $(a^*) subseteq I$.
On the other hand, take any element $i = frac{b}{2^k} in I$. It's Numerator can be written as $ b = a^*z$ for some $z in mathbb{Z}$. Hence $a^*frac{z}{2^k} = i$ and thus $I subseteq (a^*)$.
We conclude that every ideal is generated by one element.
answered Dec 9 '18 at 10:49
Yotam DYotam D
18212
$endgroup$
Take some ideal $I leq mathbb{Z}[frac{1}{2}]$. Now consider the following ideal of the integers $J_I leq mathbb{Z}$:
$$ J_I = <{a in mathbb{Z} : exists k in mathbb{Z}, ; ; frac{a}{2^k} in I }>.$$
(Here the reason we're taking the ideal generated by this set is because technically according to the definition you gave, we can divide by $2^k$ for some negative $k$.)
Note that $J_I$ is an ideal of $mathbb{Z}$, hence it is generated by an element $a^* in mathbb{Z}$. We show that $a^*$ generates $I$ (just multiply by $frac{1}{2^k}$ for the appropriate $k$). Obviously $a^* in I$, so $(a^*) subseteq I$.
On the other hand, take any element $i = frac{b}{2^k} in I$. It's Numerator can be written as $ b = a^*z$ for some $z in mathbb{Z}$. Hence $a^*frac{z}{2^k} = i$ and thus $I subseteq (a^*)$.
We conclude that every ideal is generated by one element.
answered Dec 9 '18 at 10:49
Yotam DYotam D
18212
answered Dec 9 '18 at 10:49
Yotam DYotam D
18212
answered Dec 9 '18 at 10:49
Yotam DYotam D
18212
answered Dec 9 '18 at 10:49
Yotam DYotam D
18212
18212
add a comment |
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$begingroup$
You want to show that $mathbb{Z}[1/2]$ is noetherian. First recall that the integers are noetherian and thus by the Gauss lemma so is the polynomial ring over the integers. Recall that quotient rings of noetherian rings are noetherian. I leave you to find an ideal $I$ such that $mathbb{Z}[1/2]cong mathbb{Z}[x]/I$.
$endgroup$
– Severin Schraven
Dec 9 '18 at 10:38