Construction of $F(a)$ as a field of fractions.
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$F$ is a field. $a$ is in some extention $E$ of $F$ and $a$ is transcendental over $F$. I think that $F(a)$, the smallest subfield of E that contains both $F$ and $a$ can be constructed in such a way.
Let $A=f(x) in F[x]$. If we view E as a ring, it can be verified that A is a subring of E. It can also be verified that A is an integral domain(since E is a field). As an integral domain, A can be extended to it's field of fractions, say B. Since B obviously contains $F$ and $a$, and that $F(a)$ is smallest, we claim that $B=F(a)$. This a a fact noticed by me, I wonder if it's correct.
edit: $a$ doesn't need to be transcendental over $F$.
abstract-algebra
asked Sep 2 at 6:00
qiang heng
17715
add a comment |Â
up vote
3
down vote
favorite
$F$ is a field. $a$ is in some extention $E$ of $F$ and $a$ is transcendental over $F$. I think that $F(a)$, the smallest subfield of E that contains both $F$ and $a$ can be constructed in such a way.
Let $A=f(x) in F[x]$. If we view E as a ring, it can be verified that A is a subring of E. It can also be verified that A is an integral domain(since E is a field). As an integral domain, A can be extended to it's field of fractions, say B. Since B obviously contains $F$ and $a$, and that $F(a)$ is smallest, we claim that $B=F(a)$. This a a fact noticed by me, I wonder if it's correct.
edit: $a$ doesn't need to be transcendental over $F$.
abstract-algebra
asked Sep 2 at 6:00
qiang heng
17715
1
Yes, it is correct.
– SMM
Sep 2 at 6:05
1
Yes, it's correct. But I think that the usual English term is field of fractions as opposed to quotient field. We often get quotient fields when forming the quotient ring $R/M$ of a commutative ring modulo a maximal ideal $M$. In the context of extension fields typically $R$ is the ring of univariate polynomials $K[x]$ over some field $K$, and $M$ is the ideal generated by an irreducible polynomial.
– Jyrki Lahtonen
Sep 2 at 6:36
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$F$ is a field. $a$ is in some extention $E$ of $F$ and $a$ is transcendental over $F$. I think that $F(a)$, the smallest subfield of E that contains both $F$ and $a$ can be constructed in such a way.
Let $A=f(x) in F[x]$. If we view E as a ring, it can be verified that A is a subring of E. It can also be verified that A is an integral domain(since E is a field). As an integral domain, A can be extended to it's field of fractions, say B. Since B obviously contains $F$ and $a$, and that $F(a)$ is smallest, we claim that $B=F(a)$. This a a fact noticed by me, I wonder if it's correct.
edit: $a$ doesn't need to be transcendental over $F$.
abstract-algebra
asked Sep 2 at 6:00
qiang heng
17715
$F$ is a field. $a$ is in some extention $E$ of $F$ and $a$ is transcendental over $F$. I think that $F(a)$, the smallest subfield of E that contains both $F$ and $a$ can be constructed in such a way.
Let $A=f(x) in F[x]$. If we view E as a ring, it can be verified that A is a subring of E. It can also be verified that A is an integral domain(since E is a field). As an integral domain, A can be extended to it's field of fractions, say B. Since B obviously contains $F$ and $a$, and that $F(a)$ is smallest, we claim that $B=F(a)$. This a a fact noticed by me, I wonder if it's correct.
edit: $a$ doesn't need to be transcendental over $F$.
abstract-algebra
abstract-algebra
asked Sep 2 at 6:00
qiang heng
17715
asked Sep 2 at 6:00
qiang heng
17715
edited Sep 2 at 7:21
asked Sep 2 at 6:00
qiang heng
17715
asked Sep 2 at 6:00
qiang heng
17715
asked Sep 2 at 6:00
qiang heng
17715
17715
1
Yes, it is correct.
– SMM
Sep 2 at 6:05
1
Yes, it's correct. But I think that the usual English term is field of fractions as opposed to quotient field. We often get quotient fields when forming the quotient ring $R/M$ of a commutative ring modulo a maximal ideal $M$. In the context of extension fields typically $R$ is the ring of univariate polynomials $K[x]$ over some field $K$, and $M$ is the ideal generated by an irreducible polynomial.
– Jyrki Lahtonen
Sep 2 at 6:36
add a comment |Â
1
Yes, it is correct.
– SMM
Sep 2 at 6:05
1
Yes, it's correct. But I think that the usual English term is field of fractions as opposed to quotient field. We often get quotient fields when forming the quotient ring $R/M$ of a commutative ring modulo a maximal ideal $M$. In the context of extension fields typically $R$ is the ring of univariate polynomials $K[x]$ over some field $K$, and $M$ is the ideal generated by an irreducible polynomial.
– Jyrki Lahtonen
Sep 2 at 6:36
1
1
Yes, it is correct.
– SMM
Sep 2 at 6:05
Yes, it is correct.
– SMM
Sep 2 at 6:05
1
1
Yes, it's correct. But I think that the usual English term is field of fractions as opposed to quotient field. We often get quotient fields when forming the quotient ring $R/M$ of a commutative ring modulo a maximal ideal $M$. In the context of extension fields typically $R$ is the ring of univariate polynomials $K[x]$ over some field $K$, and $M$ is the ideal generated by an irreducible polynomial.
– Jyrki Lahtonen
Sep 2 at 6:36
Yes, it's correct. But I think that the usual English term is field of fractions as opposed to quotient field. We often get quotient fields when forming the quotient ring $R/M$ of a commutative ring modulo a maximal ideal $M$. In the context of extension fields typically $R$ is the ring of univariate polynomials $K[x]$ over some field $K$, and $M$ is the ideal generated by an irreducible polynomial.
– Jyrki Lahtonen
Sep 2 at 6:36
add a comment |Â
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1
Yes, it is correct.
– SMM
Sep 2 at 6:05
1
Yes, it's correct. But I think that the usual English term is field of fractions as opposed to quotient field. We often get quotient fields when forming the quotient ring $R/M$ of a commutative ring modulo a maximal ideal $M$. In the context of extension fields typically $R$ is the ring of univariate polynomials $K[x]$ over some field $K$, and $M$ is the ideal generated by an irreducible polynomial.
– Jyrki Lahtonen
Sep 2 at 6:36