Dummit 7.5.2. (Ring of fractions)
I am doing exercise problem of chapter 7 of Dummit and Foote's Algebra.
The problem states
7.5.2 Let $R$ be an integral domain and let $D$ be a nonempty subset of $R$ that is closed under multiplication. Prove that the ring of fractions $D^{-1}R$ is isomorphic to a subring of the quotient field of $R$. (hence is also an integral domain.).
In this problem, there is no restriction for $D$. For example, $D$ can contain $0$. But, when Dummit and Foote were constructing ring of fraction (Theorem 15), they specifically said $D$ should not contain $0$. Is it just wrong statement or I am missing something?
abstract-algebra ring-theory
asked Nov 30 '18 at 2:07
LeBLeB
1,061217
add a comment |
I am doing exercise problem of chapter 7 of Dummit and Foote's Algebra.
The problem states
7.5.2 Let $R$ be an integral domain and let $D$ be a nonempty subset of $R$ that is closed under multiplication. Prove that the ring of fractions $D^{-1}R$ is isomorphic to a subring of the quotient field of $R$. (hence is also an integral domain.).
In this problem, there is no restriction for $D$. For example, $D$ can contain $0$. But, when Dummit and Foote were constructing ring of fraction (Theorem 15), they specifically said $D$ should not contain $0$. Is it just wrong statement or I am missing something?
abstract-algebra ring-theory
asked Nov 30 '18 at 2:07
LeBLeB
1,061217
In the definition of "ring of fractions", the authors assumed $D$ does not contain $0$, does not contain any zero divisors, and closed under multiplication (assumptions in Theorem 15). So you can safely assume that $0notin D$.
– Krish
Nov 30 '18 at 3:18
add a comment |
I am doing exercise problem of chapter 7 of Dummit and Foote's Algebra.
The problem states
7.5.2 Let $R$ be an integral domain and let $D$ be a nonempty subset of $R$ that is closed under multiplication. Prove that the ring of fractions $D^{-1}R$ is isomorphic to a subring of the quotient field of $R$. (hence is also an integral domain.).
In this problem, there is no restriction for $D$. For example, $D$ can contain $0$. But, when Dummit and Foote were constructing ring of fraction (Theorem 15), they specifically said $D$ should not contain $0$. Is it just wrong statement or I am missing something?
abstract-algebra ring-theory
asked Nov 30 '18 at 2:07
LeBLeB
1,061217
I am doing exercise problem of chapter 7 of Dummit and Foote's Algebra.
The problem states
7.5.2 Let $R$ be an integral domain and let $D$ be a nonempty subset of $R$ that is closed under multiplication. Prove that the ring of fractions $D^{-1}R$ is isomorphic to a subring of the quotient field of $R$. (hence is also an integral domain.).
In this problem, there is no restriction for $D$. For example, $D$ can contain $0$. But, when Dummit and Foote were constructing ring of fraction (Theorem 15), they specifically said $D$ should not contain $0$. Is it just wrong statement or I am missing something?
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Nov 30 '18 at 2:07
LeBLeB
1,061217
asked Nov 30 '18 at 2:07
LeBLeB
1,061217
asked Nov 30 '18 at 2:07
LeBLeB
1,061217
asked Nov 30 '18 at 2:07
LeBLeB
1,061217
asked Nov 30 '18 at 2:07
LeBLeB
1,061217
1,061217
In the definition of "ring of fractions", the authors assumed $D$ does not contain $0$, does not contain any zero divisors, and closed under multiplication (assumptions in Theorem 15). So you can safely assume that $0notin D$.
– Krish
Nov 30 '18 at 3:18
add a comment |
In the definition of "ring of fractions", the authors assumed $D$ does not contain $0$, does not contain any zero divisors, and closed under multiplication (assumptions in Theorem 15). So you can safely assume that $0notin D$.
– Krish
Nov 30 '18 at 3:18
In the definition of "ring of fractions", the authors assumed $D$ does not contain $0$, does not contain any zero divisors, and closed under multiplication (assumptions in Theorem 15). So you can safely assume that $0notin D$.
– Krish
Nov 30 '18 at 3:18
In the definition of "ring of fractions", the authors assumed $D$ does not contain $0$, does not contain any zero divisors, and closed under multiplication (assumptions in Theorem 15). So you can safely assume that $0notin D$.
– Krish
Nov 30 '18 at 3:18
add a comment |
1 Answer
1
active
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votes
$ D $ obviously cannot have zero because these are the denominators of the quotient field you are constructing! They are using the notation of Theorem 15, I think.
answered Nov 30 '18 at 2:55
hhp2122hhp2122
163
In general $D$ may contain $0$. In that case $D^{-1}R$ will be the zero ring. It's a necessary and sufficient condition. But here you are right, the assumption is $0notin D$.
– Krish
Nov 30 '18 at 3:21
You are right, but in the context of this particular book, they are likely referring to previously used notation.
– hhp2122
Nov 30 '18 at 3:24
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$ D $ obviously cannot have zero because these are the denominators of the quotient field you are constructing! They are using the notation of Theorem 15, I think.
answered Nov 30 '18 at 2:55
hhp2122hhp2122
163
In general $D$ may contain $0$. In that case $D^{-1}R$ will be the zero ring. It's a necessary and sufficient condition. But here you are right, the assumption is $0notin D$.
– Krish
Nov 30 '18 at 3:21
You are right, but in the context of this particular book, they are likely referring to previously used notation.
– hhp2122
Nov 30 '18 at 3:24
add a comment |
$ D $ obviously cannot have zero because these are the denominators of the quotient field you are constructing! They are using the notation of Theorem 15, I think.
answered Nov 30 '18 at 2:55
hhp2122hhp2122
163
In general $D$ may contain $0$. In that case $D^{-1}R$ will be the zero ring. It's a necessary and sufficient condition. But here you are right, the assumption is $0notin D$.
– Krish
Nov 30 '18 at 3:21
You are right, but in the context of this particular book, they are likely referring to previously used notation.
– hhp2122
Nov 30 '18 at 3:24
add a comment |
$ D $ obviously cannot have zero because these are the denominators of the quotient field you are constructing! They are using the notation of Theorem 15, I think.
answered Nov 30 '18 at 2:55
hhp2122hhp2122
163
$ D $ obviously cannot have zero because these are the denominators of the quotient field you are constructing! They are using the notation of Theorem 15, I think.
answered Nov 30 '18 at 2:55
hhp2122hhp2122
163
edited Nov 30 '18 at 3:06
answered Nov 30 '18 at 2:55
hhp2122hhp2122
163
answered Nov 30 '18 at 2:55
hhp2122hhp2122
163
answered Nov 30 '18 at 2:55
hhp2122hhp2122
163
163
In general $D$ may contain $0$. In that case $D^{-1}R$ will be the zero ring. It's a necessary and sufficient condition. But here you are right, the assumption is $0notin D$.
– Krish
Nov 30 '18 at 3:21
You are right, but in the context of this particular book, they are likely referring to previously used notation.
– hhp2122
Nov 30 '18 at 3:24
add a comment |
In general $D$ may contain $0$. In that case $D^{-1}R$ will be the zero ring. It's a necessary and sufficient condition. But here you are right, the assumption is $0notin D$.
– Krish
Nov 30 '18 at 3:21
You are right, but in the context of this particular book, they are likely referring to previously used notation.
– hhp2122
Nov 30 '18 at 3:24
In general $D$ may contain $0$. In that case $D^{-1}R$ will be the zero ring. It's a necessary and sufficient condition. But here you are right, the assumption is $0notin D$.
– Krish
Nov 30 '18 at 3:21
In general $D$ may contain $0$. In that case $D^{-1}R$ will be the zero ring. It's a necessary and sufficient condition. But here you are right, the assumption is $0notin D$.
– Krish
Nov 30 '18 at 3:21
You are right, but in the context of this particular book, they are likely referring to previously used notation.
– hhp2122
Nov 30 '18 at 3:24
You are right, but in the context of this particular book, they are likely referring to previously used notation.
– hhp2122
Nov 30 '18 at 3:24
add a comment |
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In the definition of "ring of fractions", the authors assumed $D$ does not contain $0$, does not contain any zero divisors, and closed under multiplication (assumptions in Theorem 15). So you can safely assume that $0notin D$.
– Krish
Nov 30 '18 at 3:18