Mixing
The cardinality of a integral domain and its quotient field.
$begingroup$
Let R1、R2 be two integral domains and F1、F2 be their quotient field. Suppose F1 is isomorphic to F2, I want to know if R1 is isomorphic to R2 or |R1|=|R2|.
If R1 and R2 are fields, then it is trivial.
If only one of them is an integral domain, then it is not necessary for R1 isomorphic to R2. Ex: R1=Q R2=Z Their quotient fields are exact the same, but clearly they are not isomorphic.
- If neither of them is a field, then we also can't say R1 is isomorphic to R2. Ex: R1={
frac{a+b*(sqrt -3)}{2}|a,b are integers and a+b is even} R2=Z[sqrt -3] - But I don't know whether |R1|=|R2|.
I think that's an interesting question to me because I find that the relationship between an integral domain and its quotient field is quite similar to a given set and the free group on it, and a basis and the free abelian group constructed by it. And all these have a certain "mapping property"(ex. free object on groups).In the free (abelian) group, we have this theorem:
Let X、Y be two sets, G1、G2 are two free groups on it, then |X|=|Y| if and only if G1 is isomorphic to G2.
So I think if the preceding 4 is true, we can get a similar conclusion about an integral domain and its quotient field.
Many thanks!
abstract-algebra
asked Jan 17 at 5:46
LikemathLikemath
11
$endgroup$
add a comment |
$begingroup$
Let R1、R2 be two integral domains and F1、F2 be their quotient field. Suppose F1 is isomorphic to F2, I want to know if R1 is isomorphic to R2 or |R1|=|R2|.
If R1 and R2 are fields, then it is trivial.
If only one of them is an integral domain, then it is not necessary for R1 isomorphic to R2. Ex: R1=Q R2=Z Their quotient fields are exact the same, but clearly they are not isomorphic.
- If neither of them is a field, then we also can't say R1 is isomorphic to R2. Ex: R1={
frac{a+b*(sqrt -3)}{2}|a,b are integers and a+b is even} R2=Z[sqrt -3] - But I don't know whether |R1|=|R2|.
I think that's an interesting question to me because I find that the relationship between an integral domain and its quotient field is quite similar to a given set and the free group on it, and a basis and the free abelian group constructed by it. And all these have a certain "mapping property"(ex. free object on groups).In the free (abelian) group, we have this theorem:
Let X、Y be two sets, G1、G2 are two free groups on it, then |X|=|Y| if and only if G1 is isomorphic to G2.
So I think if the preceding 4 is true, we can get a similar conclusion about an integral domain and its quotient field.
Many thanks!
abstract-algebra
asked Jan 17 at 5:46
LikemathLikemath
11
$endgroup$
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 In any case, if $R$ is an infinite integral domain, then its field of fractions $F$ has the same cardinality as $R$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 6:22
$begingroup$
But how to prove it？
$endgroup$
– Likemath
Jan 17 at 7:18
add a comment |
$begingroup$
Let R1、R2 be two integral domains and F1、F2 be their quotient field. Suppose F1 is isomorphic to F2, I want to know if R1 is isomorphic to R2 or |R1|=|R2|.
If R1 and R2 are fields, then it is trivial.
If only one of them is an integral domain, then it is not necessary for R1 isomorphic to R2. Ex: R1=Q R2=Z Their quotient fields are exact the same, but clearly they are not isomorphic.
- If neither of them is a field, then we also can't say R1 is isomorphic to R2. Ex: R1={
frac{a+b*(sqrt -3)}{2}|a,b are integers and a+b is even} R2=Z[sqrt -3] - But I don't know whether |R1|=|R2|.
I think that's an interesting question to me because I find that the relationship between an integral domain and its quotient field is quite similar to a given set and the free group on it, and a basis and the free abelian group constructed by it. And all these have a certain "mapping property"(ex. free object on groups).In the free (abelian) group, we have this theorem:
Let X、Y be two sets, G1、G2 are two free groups on it, then |X|=|Y| if and only if G1 is isomorphic to G2.
So I think if the preceding 4 is true, we can get a similar conclusion about an integral domain and its quotient field.
Many thanks!
abstract-algebra
asked Jan 17 at 5:46
LikemathLikemath
11
$endgroup$
Let R1、R2 be two integral domains and F1、F2 be their quotient field. Suppose F1 is isomorphic to F2, I want to know if R1 is isomorphic to R2 or |R1|=|R2|.
If R1 and R2 are fields, then it is trivial.
If only one of them is an integral domain, then it is not necessary for R1 isomorphic to R2. Ex: R1=Q R2=Z Their quotient fields are exact the same, but clearly they are not isomorphic.
- If neither of them is a field, then we also can't say R1 is isomorphic to R2. Ex: R1={
frac{a+b*(sqrt -3)}{2}|a,b are integers and a+b is even} R2=Z[sqrt -3] - But I don't know whether |R1|=|R2|.
I think that's an interesting question to me because I find that the relationship between an integral domain and its quotient field is quite similar to a given set and the free group on it, and a basis and the free abelian group constructed by it. And all these have a certain "mapping property"(ex. free object on groups).In the free (abelian) group, we have this theorem:
Let X、Y be two sets, G1、G2 are two free groups on it, then |X|=|Y| if and only if G1 is isomorphic to G2.
So I think if the preceding 4 is true, we can get a similar conclusion about an integral domain and its quotient field.
Many thanks!
abstract-algebra
abstract-algebra
asked Jan 17 at 5:46
LikemathLikemath
11
asked Jan 17 at 5:46
LikemathLikemath
11
edited Jan 17 at 11:25
Likemath
asked Jan 17 at 5:46
LikemathLikemath
11
asked Jan 17 at 5:46
LikemathLikemath
11
asked Jan 17 at 5:46
LikemathLikemath
11
11
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 In any case, if $R$ is an infinite integral domain, then its field of fractions $F$ has the same cardinality as $R$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 6:22
$begingroup$
But how to prove it？
$endgroup$
– Likemath
Jan 17 at 7:18
add a comment |
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 In any case, if $R$ is an infinite integral domain, then its field of fractions $F$ has the same cardinality as $R$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 6:22
$begingroup$
But how to prove it？
$endgroup$
– Likemath
Jan 17 at 7:18
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 In any case, if $R$ is an infinite integral domain, then its field of fractions $F$ has the same cardinality as $R$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 6:22
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 In any case, if $R$ is an infinite integral domain, then its field of fractions $F$ has the same cardinality as $R$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 6:22
$begingroup$
But how to prove it？
$endgroup$
– Likemath
Jan 17 at 7:18
$begingroup$
But how to prove it？
$endgroup$
– Likemath
Jan 17 at 7:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $R$ is an integral domain and $F$ is its field of fractions, then $|R|=|F|$. Indeed, if $R$ is finite then it must be a field, so $Rcong F$ and $|R|=|F|$ trivially. If $R$ is infinite, then since every element of $F$ is a fraction of two elements of $R$, we have $|F|leq |R|^2=|R|$. But of course $R$ naturally embeds in $F$ so $|R|leq|F|$ as well, and so $|F|=|R|$.
So, if two integral domains have isomorphic fields of fractions (or even fields of fractions of the same cardinality), they must have the same cardinality.
answered Jan 17 at 8:08
Eric WofseyEric Wofsey
187k14215344
$endgroup$
$begingroup$
OK,thank you very much! I have tried in this way, but I neglect the fact that | R | ²= | R |……
$endgroup$
– Likemath
Jan 17 at 11:08
add a comment |
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1 Answer
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$begingroup$
If $R$ is an integral domain and $F$ is its field of fractions, then $|R|=|F|$. Indeed, if $R$ is finite then it must be a field, so $Rcong F$ and $|R|=|F|$ trivially. If $R$ is infinite, then since every element of $F$ is a fraction of two elements of $R$, we have $|F|leq |R|^2=|R|$. But of course $R$ naturally embeds in $F$ so $|R|leq|F|$ as well, and so $|F|=|R|$.
So, if two integral domains have isomorphic fields of fractions (or even fields of fractions of the same cardinality), they must have the same cardinality.
answered Jan 17 at 8:08
Eric WofseyEric Wofsey
187k14215344
$endgroup$
$begingroup$
OK,thank you very much! I have tried in this way, but I neglect the fact that | R | ²= | R |……
$endgroup$
– Likemath
Jan 17 at 11:08
add a comment |
$begingroup$
If $R$ is an integral domain and $F$ is its field of fractions, then $|R|=|F|$. Indeed, if $R$ is finite then it must be a field, so $Rcong F$ and $|R|=|F|$ trivially. If $R$ is infinite, then since every element of $F$ is a fraction of two elements of $R$, we have $|F|leq |R|^2=|R|$. But of course $R$ naturally embeds in $F$ so $|R|leq|F|$ as well, and so $|F|=|R|$.
So, if two integral domains have isomorphic fields of fractions (or even fields of fractions of the same cardinality), they must have the same cardinality.
answered Jan 17 at 8:08
Eric WofseyEric Wofsey
187k14215344
$endgroup$
$begingroup$
OK,thank you very much! I have tried in this way, but I neglect the fact that | R | ²= | R |……
$endgroup$
– Likemath
Jan 17 at 11:08
add a comment |
$begingroup$
If $R$ is an integral domain and $F$ is its field of fractions, then $|R|=|F|$. Indeed, if $R$ is finite then it must be a field, so $Rcong F$ and $|R|=|F|$ trivially. If $R$ is infinite, then since every element of $F$ is a fraction of two elements of $R$, we have $|F|leq |R|^2=|R|$. But of course $R$ naturally embeds in $F$ so $|R|leq|F|$ as well, and so $|F|=|R|$.
So, if two integral domains have isomorphic fields of fractions (or even fields of fractions of the same cardinality), they must have the same cardinality.
answered Jan 17 at 8:08
Eric WofseyEric Wofsey
187k14215344
$endgroup$
If $R$ is an integral domain and $F$ is its field of fractions, then $|R|=|F|$. Indeed, if $R$ is finite then it must be a field, so $Rcong F$ and $|R|=|F|$ trivially. If $R$ is infinite, then since every element of $F$ is a fraction of two elements of $R$, we have $|F|leq |R|^2=|R|$. But of course $R$ naturally embeds in $F$ so $|R|leq|F|$ as well, and so $|F|=|R|$.
So, if two integral domains have isomorphic fields of fractions (or even fields of fractions of the same cardinality), they must have the same cardinality.
answered Jan 17 at 8:08
Eric WofseyEric Wofsey
187k14215344
answered Jan 17 at 8:08
Eric WofseyEric Wofsey
187k14215344
answered Jan 17 at 8:08
Eric WofseyEric Wofsey
187k14215344
answered Jan 17 at 8:08
Eric WofseyEric Wofsey
187k14215344
187k14215344
$begingroup$
OK,thank you very much! I have tried in this way, but I neglect the fact that | R | ²= | R |……
$endgroup$
– Likemath
Jan 17 at 11:08
add a comment |
$begingroup$
OK,thank you very much! I have tried in this way, but I neglect the fact that | R | ²= | R |……
$endgroup$
– Likemath
Jan 17 at 11:08
$begingroup$
OK,thank you very much! I have tried in this way, but I neglect the fact that | R | ²= | R |……
$endgroup$
– Likemath
Jan 17 at 11:08
$begingroup$
OK,thank you very much! I have tried in this way, but I neglect the fact that | R | ²= | R |……
$endgroup$
– Likemath
Jan 17 at 11:08
add a comment |
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$begingroup$
Please see math.meta.stackexchange.com/questions/5020 In any case, if $R$ is an infinite integral domain, then its field of fractions $F$ has the same cardinality as $R$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 6:22
$begingroup$
But how to prove it？
$endgroup$
– Likemath
Jan 17 at 7:18