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Given an integral domain $R$ and a subset $S$ of $R$, does there exist a smallest subfield of $R$ that contains $S$?
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Given an integral domain $R$ and a subset $S$ of $R$, does there exist a smallest subring $T$ of $R$ such that $T$ is a field and $S subseteq T$ ? ($S$ need not be a subring of $R$.)
I proved that this is true if there exists at least one subring $F$ of $R$ such that $F$ is a field and $S subseteq F$. In that case, the smallest subfield $T$ is $cap_F in mathcalF F$, where $mathcalF = F$ is a subring of $R$, $F$ is a field, and $S subseteq F$$$. But I don't know if this is true even when there is no subring $F$ of $R$ such that $F$ is a field and $S subseteq F$.
I need the answer to this question because of the following reason:
Let $R$ be an integral domain and let $F$ and $S$ be a subfield and a subset of $R$, respectively. I learned that $F(S)$ means the smallest subfield of $R$ that contains $F$ and $S$. But I want to know if $F(S)$ always exists.
abstract-algebra field-theory
asked Jul 25 at 3:11
zxcv
376
add a comment |Â
up vote
1
down vote
favorite
Given an integral domain $R$ and a subset $S$ of $R$, does there exist a smallest subring $T$ of $R$ such that $T$ is a field and $S subseteq T$ ? ($S$ need not be a subring of $R$.)
I proved that this is true if there exists at least one subring $F$ of $R$ such that $F$ is a field and $S subseteq F$. In that case, the smallest subfield $T$ is $cap_F in mathcalF F$, where $mathcalF = F$ is a subring of $R$, $F$ is a field, and $S subseteq F$$$. But I don't know if this is true even when there is no subring $F$ of $R$ such that $F$ is a field and $S subseteq F$.
I need the answer to this question because of the following reason:
Let $R$ be an integral domain and let $F$ and $S$ be a subfield and a subset of $R$, respectively. I learned that $F(S)$ means the smallest subfield of $R$ that contains $F$ and $S$. But I want to know if $F(S)$ always exists.
abstract-algebra field-theory
asked Jul 25 at 3:11
zxcv
376
3
Tried the integers?
– Pedro Tamaroff♦
Jul 25 at 3:12
1
@PedroTamaroff Oh, I got it. If R is Z and S is 2, then there is no such subfield. Thank you!
– zxcv
Jul 25 at 3:19
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given an integral domain $R$ and a subset $S$ of $R$, does there exist a smallest subring $T$ of $R$ such that $T$ is a field and $S subseteq T$ ? ($S$ need not be a subring of $R$.)
I proved that this is true if there exists at least one subring $F$ of $R$ such that $F$ is a field and $S subseteq F$. In that case, the smallest subfield $T$ is $cap_F in mathcalF F$, where $mathcalF = F$ is a subring of $R$, $F$ is a field, and $S subseteq F$$$. But I don't know if this is true even when there is no subring $F$ of $R$ such that $F$ is a field and $S subseteq F$.
I need the answer to this question because of the following reason:
Let $R$ be an integral domain and let $F$ and $S$ be a subfield and a subset of $R$, respectively. I learned that $F(S)$ means the smallest subfield of $R$ that contains $F$ and $S$. But I want to know if $F(S)$ always exists.
abstract-algebra field-theory
asked Jul 25 at 3:11
zxcv
376
Given an integral domain $R$ and a subset $S$ of $R$, does there exist a smallest subring $T$ of $R$ such that $T$ is a field and $S subseteq T$ ? ($S$ need not be a subring of $R$.)
I proved that this is true if there exists at least one subring $F$ of $R$ such that $F$ is a field and $S subseteq F$. In that case, the smallest subfield $T$ is $cap_F in mathcalF F$, where $mathcalF = F$ is a subring of $R$, $F$ is a field, and $S subseteq F$$$. But I don't know if this is true even when there is no subring $F$ of $R$ such that $F$ is a field and $S subseteq F$.
I need the answer to this question because of the following reason:
Let $R$ be an integral domain and let $F$ and $S$ be a subfield and a subset of $R$, respectively. I learned that $F(S)$ means the smallest subfield of $R$ that contains $F$ and $S$. But I want to know if $F(S)$ always exists.
abstract-algebra field-theory
asked Jul 25 at 3:11
zxcv
376
edited Jul 25 at 7:50
asked Jul 25 at 3:11
zxcv
376
asked Jul 25 at 3:11
zxcv
376
asked Jul 25 at 3:11
zxcv
376
376
3
Tried the integers?
– Pedro Tamaroff♦
Jul 25 at 3:12
1
@PedroTamaroff Oh, I got it. If R is Z and S is 2, then there is no such subfield. Thank you!
– zxcv
Jul 25 at 3:19
add a comment |Â
3
Tried the integers?
– Pedro Tamaroff♦
Jul 25 at 3:12
1
@PedroTamaroff Oh, I got it. If R is Z and S is 2, then there is no such subfield. Thank you!
– zxcv
Jul 25 at 3:19
3
3
Tried the integers?
– Pedro Tamaroff♦
Jul 25 at 3:12
Tried the integers?
– Pedro Tamaroff♦
Jul 25 at 3:12
1
1
@PedroTamaroff Oh, I got it. If R is Z and S is 2, then there is no such subfield. Thank you!
– zxcv
Jul 25 at 3:19
@PedroTamaroff Oh, I got it. If R is Z and S is 2, then there is no such subfield. Thank you!
– zxcv
Jul 25 at 3:19
add a comment |Â
1 Answer
1
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0
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accepted
From the comments above.
There does not always exist such a subring $T$. For example, take $R = mathbbZ$ and $S$ to be any subset, say $S = 2 $. There is clearly no subring of $mathbbZ$ that is a field, and in particular no subring of $mathbbZ$ that is a field and contains $2$.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
From the comments above.
There does not always exist such a subring $T$. For example, take $R = mathbbZ$ and $S$ to be any subset, say $S = 2 $. There is clearly no subring of $mathbbZ$ that is a field, and in particular no subring of $mathbbZ$ that is a field and contains $2$.
add a comment |Â
up vote
0
down vote
accepted
From the comments above.
There does not always exist such a subring $T$. For example, take $R = mathbbZ$ and $S$ to be any subset, say $S = 2 $. There is clearly no subring of $mathbbZ$ that is a field, and in particular no subring of $mathbbZ$ that is a field and contains $2$.
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
From the comments above.
There does not always exist such a subring $T$. For example, take $R = mathbbZ$ and $S$ to be any subset, say $S = 2 $. There is clearly no subring of $mathbbZ$ that is a field, and in particular no subring of $mathbbZ$ that is a field and contains $2$.
From the comments above.
There does not always exist such a subring $T$. For example, take $R = mathbbZ$ and $S$ to be any subset, say $S = 2 $. There is clearly no subring of $mathbbZ$ that is a field, and in particular no subring of $mathbbZ$ that is a field and contains $2$.
answered Jul 31 at 5:19
community wiki
Brahadeesh
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3
Tried the integers?
– Pedro Tamaroff♦
Jul 25 at 3:12
1
@PedroTamaroff Oh, I got it. If R is Z and S is 2, then there is no such subfield. Thank you!
– zxcv
Jul 25 at 3:19