Expression of elements in splitting field
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I was reading $ 16.3 $ SPLITTING FIELD in Algebra by Artin, where I met the following:
For every element $ beta $ of $ K $, there is a polynomial $ p(u_1,cdots, u_n) $ with coefficients in $ F $, such that $ p(alpha_1, cdots, alpha_n)=beta $, where field $ K=F(alpha_1, cdots, alpha_n) $ is the splitting field for $ f $ over field $ F $, and $ f(x)=(x-alpha_1)cdots (x-alpha_n) $ with $ alpha_iin K $.
Why can we express $ beta $ by $ p(alpha_1, cdots, alpha_n) $ ?
abstract-algebra field-theory splitting-field
asked Aug 22 at 5:47
A beginer
406
add a comment |Â
up vote
1
down vote
favorite
I was reading $ 16.3 $ SPLITTING FIELD in Algebra by Artin, where I met the following:
For every element $ beta $ of $ K $, there is a polynomial $ p(u_1,cdots, u_n) $ with coefficients in $ F $, such that $ p(alpha_1, cdots, alpha_n)=beta $, where field $ K=F(alpha_1, cdots, alpha_n) $ is the splitting field for $ f $ over field $ F $, and $ f(x)=(x-alpha_1)cdots (x-alpha_n) $ with $ alpha_iin K $.
Why can we express $ beta $ by $ p(alpha_1, cdots, alpha_n) $ ?
abstract-algebra field-theory splitting-field
asked Aug 22 at 5:47
A beginer
406
What do you know about $K$? I suppose that it is an algebraic extension gotten by adjoining the $alpha_i$. Then every element of $F(alpha_i)$ is of the form $p(alpha_i)$.
– 4-ier
Aug 22 at 5:54
Is it clear that $beta$ can be expressed by a rational function $r(alpha_1,ldots, alpha_n)$ and the question is why we can take polynomial expression?
– Lior B-S
Aug 22 at 6:11
1
To put it succinctly...because all rings $R$ with $Fsubseteq Rsubseteq K$ are fields.
– C Monsour
Aug 22 at 6:43
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was reading $ 16.3 $ SPLITTING FIELD in Algebra by Artin, where I met the following:
For every element $ beta $ of $ K $, there is a polynomial $ p(u_1,cdots, u_n) $ with coefficients in $ F $, such that $ p(alpha_1, cdots, alpha_n)=beta $, where field $ K=F(alpha_1, cdots, alpha_n) $ is the splitting field for $ f $ over field $ F $, and $ f(x)=(x-alpha_1)cdots (x-alpha_n) $ with $ alpha_iin K $.
Why can we express $ beta $ by $ p(alpha_1, cdots, alpha_n) $ ?
abstract-algebra field-theory splitting-field
asked Aug 22 at 5:47
A beginer
406
I was reading $ 16.3 $ SPLITTING FIELD in Algebra by Artin, where I met the following:
For every element $ beta $ of $ K $, there is a polynomial $ p(u_1,cdots, u_n) $ with coefficients in $ F $, such that $ p(alpha_1, cdots, alpha_n)=beta $, where field $ K=F(alpha_1, cdots, alpha_n) $ is the splitting field for $ f $ over field $ F $, and $ f(x)=(x-alpha_1)cdots (x-alpha_n) $ with $ alpha_iin K $.
Why can we express $ beta $ by $ p(alpha_1, cdots, alpha_n) $ ?
abstract-algebra field-theory splitting-field
asked Aug 22 at 5:47
A beginer
406
asked Aug 22 at 5:47
A beginer
406
asked Aug 22 at 5:47
A beginer
406
asked Aug 22 at 5:47
A beginer
406
406
What do you know about $K$? I suppose that it is an algebraic extension gotten by adjoining the $alpha_i$. Then every element of $F(alpha_i)$ is of the form $p(alpha_i)$.
– 4-ier
Aug 22 at 5:54
Is it clear that $beta$ can be expressed by a rational function $r(alpha_1,ldots, alpha_n)$ and the question is why we can take polynomial expression?
– Lior B-S
Aug 22 at 6:11
1
To put it succinctly...because all rings $R$ with $Fsubseteq Rsubseteq K$ are fields.
– C Monsour
Aug 22 at 6:43
add a comment |Â
What do you know about $K$? I suppose that it is an algebraic extension gotten by adjoining the $alpha_i$. Then every element of $F(alpha_i)$ is of the form $p(alpha_i)$.
– 4-ier
Aug 22 at 5:54
Is it clear that $beta$ can be expressed by a rational function $r(alpha_1,ldots, alpha_n)$ and the question is why we can take polynomial expression?
– Lior B-S
Aug 22 at 6:11
1
To put it succinctly...because all rings $R$ with $Fsubseteq Rsubseteq K$ are fields.
– C Monsour
Aug 22 at 6:43
What do you know about $K$? I suppose that it is an algebraic extension gotten by adjoining the $alpha_i$. Then every element of $F(alpha_i)$ is of the form $p(alpha_i)$.
– 4-ier
Aug 22 at 5:54
What do you know about $K$? I suppose that it is an algebraic extension gotten by adjoining the $alpha_i$. Then every element of $F(alpha_i)$ is of the form $p(alpha_i)$.
– 4-ier
Aug 22 at 5:54
Is it clear that $beta$ can be expressed by a rational function $r(alpha_1,ldots, alpha_n)$ and the question is why we can take polynomial expression?
– Lior B-S
Aug 22 at 6:11
Is it clear that $beta$ can be expressed by a rational function $r(alpha_1,ldots, alpha_n)$ and the question is why we can take polynomial expression?
– Lior B-S
Aug 22 at 6:11
1
1
To put it succinctly...because all rings $R$ with $Fsubseteq Rsubseteq K$ are fields.
– C Monsour
Aug 22 at 6:43
To put it succinctly...because all rings $R$ with $Fsubseteq Rsubseteq K$ are fields.
– C Monsour
Aug 22 at 6:43
add a comment |Â
1 Answer
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1
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$K$ is the minimal field containing all the $alpha_i$. The subset of all elements that can be expressed as $p(alpha_1,dots,alpha_n)$ is a subfield of $K$. (It's easy to check that it is closed under addition and multiplication and additive inverse, since so are polynomials with coefficients in $F$. It's closed under taking multiplicative inverses because in any algebraic extension any element's inverse is an $F$-linear combination of its positive powers...see the next paragraph if this is not a familiar fact.) But since $K$ is minimal with this property, a subfield with this property must be all of $K$.
To explain the argument about not needing to be able to explicitly take inverses, note that if, for $gammain K$, $x^m+b_n-1x^m-1+cdotcdotcdot+b_0$ is the minimal polynomial for $gamma$ over $F$, then $gamma^-1=-(1/b_0)gamma^m-1-(b_m-1/b_0)gamma^m-2-cdotcdotcdot-(b_1/b_0)$. So closure of a subring of $K$ under addition and multiplication also gives closure under multiplicative inverse. (The inverses of $b_0$ are not a problem since $b_0in F$ and we already have inverses of everything in $F$.)
Or to put it another way, any subring of $K$ containing $F$ is also a subfield. Thus, since the set of all $p(alpha_1,dots,alpha_n)$ with $pin F[x]$ is obviously a subring it is also a subfield.
answered Aug 22 at 6:00
C Monsour
4,651221
Thank you! So actually the second paragraph is no longer necessary, we only need to show $ Fsubset F[ alpha_1, cdots, alpha_n ]subset K $ and because $ F[ alpha_1, cdots, alpha_n ] $ is a ring and $ F $ and $ K $ are fields and we are done.
– A beginer
Aug 22 at 6:59
If you had said "....because $F$ is a field and $K$ is an algebraic extension", you'd be right. (The theorem that all intermediate rings are fields is only for algebraic extensions, not transcendental ones.) The second paragraph is a proof of this result for those unfamiliar with it, so whether you need it depends on how much you already know.
– C Monsour
Aug 22 at 7:03
Yes, but I think $ K $ should be a finite extension of $ F $ to guarantee it is a finite dimensional vector space over $ F $. See: math.stackexchange.com/q/2879987/578483
– A beginer
Aug 22 at 7:31
It's a mistake for you to assume that just because a property holds for finite extensions that it does not also hold in some other cases. In this instance, the fact that every intermediate ring of an infinite-dimensional algebraic field extension is a field is an easy corollary of that fact for finite extensions. Not that it matters for solving your posted problem, where you only need the fact for finite extensions.
– C Monsour
Aug 22 at 7:37
I am sorry, I don't think the second paragraph is making any point. To show that $ F[alpha_1, cdots, alpha_n] $ is a field, we should see that $ F[alpha_1] $ is a field and $ F[alpha_1, cdots, alpha_n]=F[alpha_1][alpha_2, cdots alpha_n] $, by induction we are done.
– A beginer
Aug 22 at 7:42
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$K$ is the minimal field containing all the $alpha_i$. The subset of all elements that can be expressed as $p(alpha_1,dots,alpha_n)$ is a subfield of $K$. (It's easy to check that it is closed under addition and multiplication and additive inverse, since so are polynomials with coefficients in $F$. It's closed under taking multiplicative inverses because in any algebraic extension any element's inverse is an $F$-linear combination of its positive powers...see the next paragraph if this is not a familiar fact.) But since $K$ is minimal with this property, a subfield with this property must be all of $K$.
To explain the argument about not needing to be able to explicitly take inverses, note that if, for $gammain K$, $x^m+b_n-1x^m-1+cdotcdotcdot+b_0$ is the minimal polynomial for $gamma$ over $F$, then $gamma^-1=-(1/b_0)gamma^m-1-(b_m-1/b_0)gamma^m-2-cdotcdotcdot-(b_1/b_0)$. So closure of a subring of $K$ under addition and multiplication also gives closure under multiplicative inverse. (The inverses of $b_0$ are not a problem since $b_0in F$ and we already have inverses of everything in $F$.)
Or to put it another way, any subring of $K$ containing $F$ is also a subfield. Thus, since the set of all $p(alpha_1,dots,alpha_n)$ with $pin F[x]$ is obviously a subring it is also a subfield.
answered Aug 22 at 6:00
C Monsour
4,651221
Thank you! So actually the second paragraph is no longer necessary, we only need to show $ Fsubset F[ alpha_1, cdots, alpha_n ]subset K $ and because $ F[ alpha_1, cdots, alpha_n ] $ is a ring and $ F $ and $ K $ are fields and we are done.
– A beginer
Aug 22 at 6:59
If you had said "....because $F$ is a field and $K$ is an algebraic extension", you'd be right. (The theorem that all intermediate rings are fields is only for algebraic extensions, not transcendental ones.) The second paragraph is a proof of this result for those unfamiliar with it, so whether you need it depends on how much you already know.
– C Monsour
Aug 22 at 7:03
Yes, but I think $ K $ should be a finite extension of $ F $ to guarantee it is a finite dimensional vector space over $ F $. See: math.stackexchange.com/q/2879987/578483
– A beginer
Aug 22 at 7:31
It's a mistake for you to assume that just because a property holds for finite extensions that it does not also hold in some other cases. In this instance, the fact that every intermediate ring of an infinite-dimensional algebraic field extension is a field is an easy corollary of that fact for finite extensions. Not that it matters for solving your posted problem, where you only need the fact for finite extensions.
– C Monsour
Aug 22 at 7:37
I am sorry, I don't think the second paragraph is making any point. To show that $ F[alpha_1, cdots, alpha_n] $ is a field, we should see that $ F[alpha_1] $ is a field and $ F[alpha_1, cdots, alpha_n]=F[alpha_1][alpha_2, cdots alpha_n] $, by induction we are done.
– A beginer
Aug 22 at 7:42
 |Â
show 3 more comments
up vote
1
down vote
accepted
$K$ is the minimal field containing all the $alpha_i$. The subset of all elements that can be expressed as $p(alpha_1,dots,alpha_n)$ is a subfield of $K$. (It's easy to check that it is closed under addition and multiplication and additive inverse, since so are polynomials with coefficients in $F$. It's closed under taking multiplicative inverses because in any algebraic extension any element's inverse is an $F$-linear combination of its positive powers...see the next paragraph if this is not a familiar fact.) But since $K$ is minimal with this property, a subfield with this property must be all of $K$.
To explain the argument about not needing to be able to explicitly take inverses, note that if, for $gammain K$, $x^m+b_n-1x^m-1+cdotcdotcdot+b_0$ is the minimal polynomial for $gamma$ over $F$, then $gamma^-1=-(1/b_0)gamma^m-1-(b_m-1/b_0)gamma^m-2-cdotcdotcdot-(b_1/b_0)$. So closure of a subring of $K$ under addition and multiplication also gives closure under multiplicative inverse. (The inverses of $b_0$ are not a problem since $b_0in F$ and we already have inverses of everything in $F$.)
Or to put it another way, any subring of $K$ containing $F$ is also a subfield. Thus, since the set of all $p(alpha_1,dots,alpha_n)$ with $pin F[x]$ is obviously a subring it is also a subfield.
answered Aug 22 at 6:00
C Monsour
4,651221
Thank you! So actually the second paragraph is no longer necessary, we only need to show $ Fsubset F[ alpha_1, cdots, alpha_n ]subset K $ and because $ F[ alpha_1, cdots, alpha_n ] $ is a ring and $ F $ and $ K $ are fields and we are done.
– A beginer
Aug 22 at 6:59
If you had said "....because $F$ is a field and $K$ is an algebraic extension", you'd be right. (The theorem that all intermediate rings are fields is only for algebraic extensions, not transcendental ones.) The second paragraph is a proof of this result for those unfamiliar with it, so whether you need it depends on how much you already know.
– C Monsour
Aug 22 at 7:03
Yes, but I think $ K $ should be a finite extension of $ F $ to guarantee it is a finite dimensional vector space over $ F $. See: math.stackexchange.com/q/2879987/578483
– A beginer
Aug 22 at 7:31
It's a mistake for you to assume that just because a property holds for finite extensions that it does not also hold in some other cases. In this instance, the fact that every intermediate ring of an infinite-dimensional algebraic field extension is a field is an easy corollary of that fact for finite extensions. Not that it matters for solving your posted problem, where you only need the fact for finite extensions.
– C Monsour
Aug 22 at 7:37
I am sorry, I don't think the second paragraph is making any point. To show that $ F[alpha_1, cdots, alpha_n] $ is a field, we should see that $ F[alpha_1] $ is a field and $ F[alpha_1, cdots, alpha_n]=F[alpha_1][alpha_2, cdots alpha_n] $, by induction we are done.
– A beginer
Aug 22 at 7:42
 |Â
show 3 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$K$ is the minimal field containing all the $alpha_i$. The subset of all elements that can be expressed as $p(alpha_1,dots,alpha_n)$ is a subfield of $K$. (It's easy to check that it is closed under addition and multiplication and additive inverse, since so are polynomials with coefficients in $F$. It's closed under taking multiplicative inverses because in any algebraic extension any element's inverse is an $F$-linear combination of its positive powers...see the next paragraph if this is not a familiar fact.) But since $K$ is minimal with this property, a subfield with this property must be all of $K$.
To explain the argument about not needing to be able to explicitly take inverses, note that if, for $gammain K$, $x^m+b_n-1x^m-1+cdotcdotcdot+b_0$ is the minimal polynomial for $gamma$ over $F$, then $gamma^-1=-(1/b_0)gamma^m-1-(b_m-1/b_0)gamma^m-2-cdotcdotcdot-(b_1/b_0)$. So closure of a subring of $K$ under addition and multiplication also gives closure under multiplicative inverse. (The inverses of $b_0$ are not a problem since $b_0in F$ and we already have inverses of everything in $F$.)
Or to put it another way, any subring of $K$ containing $F$ is also a subfield. Thus, since the set of all $p(alpha_1,dots,alpha_n)$ with $pin F[x]$ is obviously a subring it is also a subfield.
answered Aug 22 at 6:00
C Monsour
4,651221
$K$ is the minimal field containing all the $alpha_i$. The subset of all elements that can be expressed as $p(alpha_1,dots,alpha_n)$ is a subfield of $K$. (It's easy to check that it is closed under addition and multiplication and additive inverse, since so are polynomials with coefficients in $F$. It's closed under taking multiplicative inverses because in any algebraic extension any element's inverse is an $F$-linear combination of its positive powers...see the next paragraph if this is not a familiar fact.) But since $K$ is minimal with this property, a subfield with this property must be all of $K$.
To explain the argument about not needing to be able to explicitly take inverses, note that if, for $gammain K$, $x^m+b_n-1x^m-1+cdotcdotcdot+b_0$ is the minimal polynomial for $gamma$ over $F$, then $gamma^-1=-(1/b_0)gamma^m-1-(b_m-1/b_0)gamma^m-2-cdotcdotcdot-(b_1/b_0)$. So closure of a subring of $K$ under addition and multiplication also gives closure under multiplicative inverse. (The inverses of $b_0$ are not a problem since $b_0in F$ and we already have inverses of everything in $F$.)
Or to put it another way, any subring of $K$ containing $F$ is also a subfield. Thus, since the set of all $p(alpha_1,dots,alpha_n)$ with $pin F[x]$ is obviously a subring it is also a subfield.
answered Aug 22 at 6:00
C Monsour
4,651221
edited Aug 22 at 6:19
answered Aug 22 at 6:00
C Monsour
4,651221
answered Aug 22 at 6:00
C Monsour
4,651221
answered Aug 22 at 6:00
C Monsour
4,651221
4,651221
Thank you! So actually the second paragraph is no longer necessary, we only need to show $ Fsubset F[ alpha_1, cdots, alpha_n ]subset K $ and because $ F[ alpha_1, cdots, alpha_n ] $ is a ring and $ F $ and $ K $ are fields and we are done.
– A beginer
Aug 22 at 6:59
If you had said "....because $F$ is a field and $K$ is an algebraic extension", you'd be right. (The theorem that all intermediate rings are fields is only for algebraic extensions, not transcendental ones.) The second paragraph is a proof of this result for those unfamiliar with it, so whether you need it depends on how much you already know.
– C Monsour
Aug 22 at 7:03
Yes, but I think $ K $ should be a finite extension of $ F $ to guarantee it is a finite dimensional vector space over $ F $. See: math.stackexchange.com/q/2879987/578483
– A beginer
Aug 22 at 7:31
It's a mistake for you to assume that just because a property holds for finite extensions that it does not also hold in some other cases. In this instance, the fact that every intermediate ring of an infinite-dimensional algebraic field extension is a field is an easy corollary of that fact for finite extensions. Not that it matters for solving your posted problem, where you only need the fact for finite extensions.
– C Monsour
Aug 22 at 7:37
I am sorry, I don't think the second paragraph is making any point. To show that $ F[alpha_1, cdots, alpha_n] $ is a field, we should see that $ F[alpha_1] $ is a field and $ F[alpha_1, cdots, alpha_n]=F[alpha_1][alpha_2, cdots alpha_n] $, by induction we are done.
– A beginer
Aug 22 at 7:42
 |Â
show 3 more comments
Thank you! So actually the second paragraph is no longer necessary, we only need to show $ Fsubset F[ alpha_1, cdots, alpha_n ]subset K $ and because $ F[ alpha_1, cdots, alpha_n ] $ is a ring and $ F $ and $ K $ are fields and we are done.
– A beginer
Aug 22 at 6:59
If you had said "....because $F$ is a field and $K$ is an algebraic extension", you'd be right. (The theorem that all intermediate rings are fields is only for algebraic extensions, not transcendental ones.) The second paragraph is a proof of this result for those unfamiliar with it, so whether you need it depends on how much you already know.
– C Monsour
Aug 22 at 7:03
Yes, but I think $ K $ should be a finite extension of $ F $ to guarantee it is a finite dimensional vector space over $ F $. See: math.stackexchange.com/q/2879987/578483
– A beginer
Aug 22 at 7:31
It's a mistake for you to assume that just because a property holds for finite extensions that it does not also hold in some other cases. In this instance, the fact that every intermediate ring of an infinite-dimensional algebraic field extension is a field is an easy corollary of that fact for finite extensions. Not that it matters for solving your posted problem, where you only need the fact for finite extensions.
– C Monsour
Aug 22 at 7:37
I am sorry, I don't think the second paragraph is making any point. To show that $ F[alpha_1, cdots, alpha_n] $ is a field, we should see that $ F[alpha_1] $ is a field and $ F[alpha_1, cdots, alpha_n]=F[alpha_1][alpha_2, cdots alpha_n] $, by induction we are done.
– A beginer
Aug 22 at 7:42
Thank you! So actually the second paragraph is no longer necessary, we only need to show $ Fsubset F[ alpha_1, cdots, alpha_n ]subset K $ and because $ F[ alpha_1, cdots, alpha_n ] $ is a ring and $ F $ and $ K $ are fields and we are done.
– A beginer
Aug 22 at 6:59
Thank you! So actually the second paragraph is no longer necessary, we only need to show $ Fsubset F[ alpha_1, cdots, alpha_n ]subset K $ and because $ F[ alpha_1, cdots, alpha_n ] $ is a ring and $ F $ and $ K $ are fields and we are done.
– A beginer
Aug 22 at 6:59
If you had said "....because $F$ is a field and $K$ is an algebraic extension", you'd be right. (The theorem that all intermediate rings are fields is only for algebraic extensions, not transcendental ones.) The second paragraph is a proof of this result for those unfamiliar with it, so whether you need it depends on how much you already know.
– C Monsour
Aug 22 at 7:03
If you had said "....because $F$ is a field and $K$ is an algebraic extension", you'd be right. (The theorem that all intermediate rings are fields is only for algebraic extensions, not transcendental ones.) The second paragraph is a proof of this result for those unfamiliar with it, so whether you need it depends on how much you already know.
– C Monsour
Aug 22 at 7:03
Yes, but I think $ K $ should be a finite extension of $ F $ to guarantee it is a finite dimensional vector space over $ F $. See: math.stackexchange.com/q/2879987/578483
– A beginer
Aug 22 at 7:31
Yes, but I think $ K $ should be a finite extension of $ F $ to guarantee it is a finite dimensional vector space over $ F $. See: math.stackexchange.com/q/2879987/578483
– A beginer
Aug 22 at 7:31
It's a mistake for you to assume that just because a property holds for finite extensions that it does not also hold in some other cases. In this instance, the fact that every intermediate ring of an infinite-dimensional algebraic field extension is a field is an easy corollary of that fact for finite extensions. Not that it matters for solving your posted problem, where you only need the fact for finite extensions.
– C Monsour
Aug 22 at 7:37
It's a mistake for you to assume that just because a property holds for finite extensions that it does not also hold in some other cases. In this instance, the fact that every intermediate ring of an infinite-dimensional algebraic field extension is a field is an easy corollary of that fact for finite extensions. Not that it matters for solving your posted problem, where you only need the fact for finite extensions.
– C Monsour
Aug 22 at 7:37
I am sorry, I don't think the second paragraph is making any point. To show that $ F[alpha_1, cdots, alpha_n] $ is a field, we should see that $ F[alpha_1] $ is a field and $ F[alpha_1, cdots, alpha_n]=F[alpha_1][alpha_2, cdots alpha_n] $, by induction we are done.
– A beginer
Aug 22 at 7:42
I am sorry, I don't think the second paragraph is making any point. To show that $ F[alpha_1, cdots, alpha_n] $ is a field, we should see that $ F[alpha_1] $ is a field and $ F[alpha_1, cdots, alpha_n]=F[alpha_1][alpha_2, cdots alpha_n] $, by induction we are done.
– A beginer
Aug 22 at 7:42
 |Â
show 3 more comments
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What do you know about $K$? I suppose that it is an algebraic extension gotten by adjoining the $alpha_i$. Then every element of $F(alpha_i)$ is of the form $p(alpha_i)$.
– 4-ier
Aug 22 at 5:54
Is it clear that $beta$ can be expressed by a rational function $r(alpha_1,ldots, alpha_n)$ and the question is why we can take polynomial expression?
– Lior B-S
Aug 22 at 6:11
1
To put it succinctly...because all rings $R$ with $Fsubseteq Rsubseteq K$ are fields.
– C Monsour
Aug 22 at 6:43