Isomorphism between quotient fields of polynomial rings [closed]
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Let $R$ be an integral domain and $F$ be its field of fractions. If $X$ is a nonempty set, prove that there is a ring monomorphism from $R[X]$ to $F[X]$ that extends to an isomorphism of their quotient fields.
abstract-algebra field-theory quotient-spaces integral-domain polynomial-rings
asked Dec 2 '18 at 4:17
Sierra ThornleySierra Thornley
62
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closed as off-topic by Saad, caverac, amWhy, Cesareo, user416281 Dec 2 '18 at 14:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, caverac, amWhy, Cesareo, Yanko
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
1
$begingroup$
Let $R$ be an integral domain and $F$ be its field of fractions. If $X$ is a nonempty set, prove that there is a ring monomorphism from $R[X]$ to $F[X]$ that extends to an isomorphism of their quotient fields.
abstract-algebra field-theory quotient-spaces integral-domain polynomial-rings
asked Dec 2 '18 at 4:17
Sierra ThornleySierra Thornley
62
$endgroup$
closed as off-topic by Saad, caverac, amWhy, Cesareo, user416281 Dec 2 '18 at 14:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, caverac, amWhy, Cesareo, Yanko
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Integral domain means defining $Frac(R)$ is easy, and $R[X]$ is again an integral domain.
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– reuns
Dec 2 '18 at 6:11
$begingroup$
I recommend that you take a look at our guide for new askers. While that guide is mostly targeting freshmen and high schoolers, it does apply to all (and it is not kosher to discriminate based on the level anyway).
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 8:36
add a comment |
1
1
1
$begingroup$
Let $R$ be an integral domain and $F$ be its field of fractions. If $X$ is a nonempty set, prove that there is a ring monomorphism from $R[X]$ to $F[X]$ that extends to an isomorphism of their quotient fields.
abstract-algebra field-theory quotient-spaces integral-domain polynomial-rings
asked Dec 2 '18 at 4:17
Sierra ThornleySierra Thornley
62
$endgroup$
Let $R$ be an integral domain and $F$ be its field of fractions. If $X$ is a nonempty set, prove that there is a ring monomorphism from $R[X]$ to $F[X]$ that extends to an isomorphism of their quotient fields.
abstract-algebra field-theory quotient-spaces integral-domain polynomial-rings
abstract-algebra field-theory quotient-spaces integral-domain polynomial-rings
asked Dec 2 '18 at 4:17
Sierra ThornleySierra Thornley
62
asked Dec 2 '18 at 4:17
Sierra ThornleySierra Thornley
62
asked Dec 2 '18 at 4:17
Sierra ThornleySierra Thornley
62
asked Dec 2 '18 at 4:17
Sierra ThornleySierra Thornley
62
asked Dec 2 '18 at 4:17
Sierra ThornleySierra Thornley
62
62
closed as off-topic by Saad, caverac, amWhy, Cesareo, user416281 Dec 2 '18 at 14:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, caverac, amWhy, Cesareo, Yanko
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, caverac, amWhy, Cesareo, user416281 Dec 2 '18 at 14:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, caverac, amWhy, Cesareo, Yanko
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Integral domain means defining $Frac(R)$ is easy, and $R[X]$ is again an integral domain.
$endgroup$
– reuns
Dec 2 '18 at 6:11
$begingroup$
I recommend that you take a look at our guide for new askers. While that guide is mostly targeting freshmen and high schoolers, it does apply to all (and it is not kosher to discriminate based on the level anyway).
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 8:36
add a comment |
$begingroup$
Integral domain means defining $Frac(R)$ is easy, and $R[X]$ is again an integral domain.
$endgroup$
– reuns
Dec 2 '18 at 6:11
$begingroup$
I recommend that you take a look at our guide for new askers. While that guide is mostly targeting freshmen and high schoolers, it does apply to all (and it is not kosher to discriminate based on the level anyway).
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 8:36
$begingroup$
Integral domain means defining $Frac(R)$ is easy, and $R[X]$ is again an integral domain.
$endgroup$
– reuns
Dec 2 '18 at 6:11
$begingroup$
Integral domain means defining $Frac(R)$ is easy, and $R[X]$ is again an integral domain.
$endgroup$
– reuns
Dec 2 '18 at 6:11
$begingroup$
I recommend that you take a look at our guide for new askers. While that guide is mostly targeting freshmen and high schoolers, it does apply to all (and it is not kosher to discriminate based on the level anyway).
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 8:36
$begingroup$
I recommend that you take a look at our guide for new askers. While that guide is mostly targeting freshmen and high schoolers, it does apply to all (and it is not kosher to discriminate based on the level anyway).
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 8:36
add a comment |
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$begingroup$
Integral domain means defining $Frac(R)$ is easy, and $R[X]$ is again an integral domain.
$endgroup$
– reuns
Dec 2 '18 at 6:11
$begingroup$
I recommend that you take a look at our guide for new askers. While that guide is mostly targeting freshmen and high schoolers, it does apply to all (and it is not kosher to discriminate based on the level anyway).
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 8:36