Let $L/K$ be a field extension and let $a, b in L$ be algebraic elements over $K$ having the same minimal polynomial. Show that $K(a) simeq K(b)$.Finding the degree of a field extension over the rationalsMinimal polynomial over an extension field divides the minimal polynomial over the base fieldFinding the minimal polynomial in this field extension of $mathbb Q$?Minimal polynomial over a fieldComputing the minimal polynomial over a field $K$ using a finite field extension $Ksubseteq E$ by means of a $K$-linear map.A question on on field extensions and minimal polynomialMinimal polynomial of root of unity over quadratic fieldShow that the extension $F vert mathbbQ$ is GaloisMinimal Polynomial over an extension of $mathbb Q$ Comp QuestionConfusion over field isomorphism F(a) onto F(b) where a and b are roots of an irreducible polynomial in F[t]?
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Let $L/K$ be a field extension and let $a, b in L$ be algebraic elements over $K$ having the same minimal polynomial. Show that $K(a) simeq K(b)$.
Finding the degree of a field extension over the rationalsMinimal polynomial over an extension field divides the minimal polynomial over the base fieldFinding the minimal polynomial in this field extension of $mathbb Q$?Minimal polynomial over a fieldComputing the minimal polynomial over a field $K$ using a finite field extension $Ksubseteq E$ by means of a $K$-linear map.A question on on field extensions and minimal polynomialMinimal polynomial of root of unity over quadratic fieldShow that the extension $F vert mathbbQ$ is GaloisMinimal Polynomial over an extension of $mathbb Q$ Comp QuestionConfusion over field isomorphism F(a) onto F(b) where a and b are roots of an irreducible polynomial in F[t]?
$begingroup$
The above question was given to me as an assignment but I'm a bit stuck and I think the way I've been going about it is a dead end, or if not a dead end, I can't figure out a way to finish it off
So in a previous question we had to prove that $mathbbQ/(x^2-5) simeqmathbbQ[sqrt5]$, and I did this with the isomorphism $varphi:mathbbQ/(x^2-5)tomathbbQ[sqrt5],;ax+bmapsto asqrt5+b$
I then thought that I could do the same with this, question, isomorphically mapping $varphi:K(a) to K[x]/(f)$ where $f$ is the minimal polynomial of $a$, and then mapping $psi:K[x]/(f) to K(b)$ using something of a similar form to my above example. What I came up with was
$$varphi: K(a) to K[x]/(f),; sum_i=0^n-1c_ia^i mapsto sum_i=0^n-1c_ix^i$$
where $n = deg(f)$, but I don't know how to prove this is an isomorphism or if I was wrong and it actually isn't one. The additive property is easy and works out fine, but I don't know how to prove the multiplicative property because of when the power of $a$ goes above $n-1$. I looked around for answers to this before posting and I know that $K(a) simeq K[x]/(f) simeq K(b)$ is true, but I couldn't find a proof anywhere. Is what I'm doing a dead end or am I barking up the completely wrong tree, and if so, how should I go about proving this?
field-theory extension-field
asked Mar 17 at 23:53
Jack DohertyJack Doherty
85
$endgroup$
add a comment |
$begingroup$
The above question was given to me as an assignment but I'm a bit stuck and I think the way I've been going about it is a dead end, or if not a dead end, I can't figure out a way to finish it off
So in a previous question we had to prove that $mathbbQ/(x^2-5) simeqmathbbQ[sqrt5]$, and I did this with the isomorphism $varphi:mathbbQ/(x^2-5)tomathbbQ[sqrt5],;ax+bmapsto asqrt5+b$
I then thought that I could do the same with this, question, isomorphically mapping $varphi:K(a) to K[x]/(f)$ where $f$ is the minimal polynomial of $a$, and then mapping $psi:K[x]/(f) to K(b)$ using something of a similar form to my above example. What I came up with was
$$varphi: K(a) to K[x]/(f),; sum_i=0^n-1c_ia^i mapsto sum_i=0^n-1c_ix^i$$
where $n = deg(f)$, but I don't know how to prove this is an isomorphism or if I was wrong and it actually isn't one. The additive property is easy and works out fine, but I don't know how to prove the multiplicative property because of when the power of $a$ goes above $n-1$. I looked around for answers to this before posting and I know that $K(a) simeq K[x]/(f) simeq K(b)$ is true, but I couldn't find a proof anywhere. Is what I'm doing a dead end or am I barking up the completely wrong tree, and if so, how should I go about proving this?
field-theory extension-field
asked Mar 17 at 23:53
Jack DohertyJack Doherty
85
$endgroup$
$begingroup$
Consider the natural homomorphism from $K [x] $ to $K (a) $ given by $p (x)mapsto p (a) $.
$endgroup$
– Thomas Shelby
Mar 18 at 1:07
$begingroup$
But isn't that not an isomorphism? Does it work both ways?
$endgroup$
– Jack Doherty
Mar 18 at 1:18
$begingroup$
The induced map from $K [x]/(f) $ to $K (a) $ will be an isomorphism. Note that $K [x]/(f) $ is a field(why?).
$endgroup$
– Thomas Shelby
Mar 18 at 1:20
$begingroup$
It's a field because $f$ is irreducible
$endgroup$
– Jack Doherty
Mar 18 at 1:33
$begingroup$
You know that $phi : K[x] to K[a]$ being a surjective homomorphism implies $K[x]/ker(phi) cong K[a]$
$endgroup$
– reuns
Mar 18 at 1:44
add a comment |
$begingroup$
The above question was given to me as an assignment but I'm a bit stuck and I think the way I've been going about it is a dead end, or if not a dead end, I can't figure out a way to finish it off
So in a previous question we had to prove that $mathbbQ/(x^2-5) simeqmathbbQ[sqrt5]$, and I did this with the isomorphism $varphi:mathbbQ/(x^2-5)tomathbbQ[sqrt5],;ax+bmapsto asqrt5+b$
I then thought that I could do the same with this, question, isomorphically mapping $varphi:K(a) to K[x]/(f)$ where $f$ is the minimal polynomial of $a$, and then mapping $psi:K[x]/(f) to K(b)$ using something of a similar form to my above example. What I came up with was
$$varphi: K(a) to K[x]/(f),; sum_i=0^n-1c_ia^i mapsto sum_i=0^n-1c_ix^i$$
where $n = deg(f)$, but I don't know how to prove this is an isomorphism or if I was wrong and it actually isn't one. The additive property is easy and works out fine, but I don't know how to prove the multiplicative property because of when the power of $a$ goes above $n-1$. I looked around for answers to this before posting and I know that $K(a) simeq K[x]/(f) simeq K(b)$ is true, but I couldn't find a proof anywhere. Is what I'm doing a dead end or am I barking up the completely wrong tree, and if so, how should I go about proving this?
field-theory extension-field
asked Mar 17 at 23:53
Jack DohertyJack Doherty
85
$endgroup$
The above question was given to me as an assignment but I'm a bit stuck and I think the way I've been going about it is a dead end, or if not a dead end, I can't figure out a way to finish it off
So in a previous question we had to prove that $mathbbQ/(x^2-5) simeqmathbbQ[sqrt5]$, and I did this with the isomorphism $varphi:mathbbQ/(x^2-5)tomathbbQ[sqrt5],;ax+bmapsto asqrt5+b$
I then thought that I could do the same with this, question, isomorphically mapping $varphi:K(a) to K[x]/(f)$ where $f$ is the minimal polynomial of $a$, and then mapping $psi:K[x]/(f) to K(b)$ using something of a similar form to my above example. What I came up with was
$$varphi: K(a) to K[x]/(f),; sum_i=0^n-1c_ia^i mapsto sum_i=0^n-1c_ix^i$$
where $n = deg(f)$, but I don't know how to prove this is an isomorphism or if I was wrong and it actually isn't one. The additive property is easy and works out fine, but I don't know how to prove the multiplicative property because of when the power of $a$ goes above $n-1$. I looked around for answers to this before posting and I know that $K(a) simeq K[x]/(f) simeq K(b)$ is true, but I couldn't find a proof anywhere. Is what I'm doing a dead end or am I barking up the completely wrong tree, and if so, how should I go about proving this?
field-theory extension-field
field-theory extension-field
asked Mar 17 at 23:53
Jack DohertyJack Doherty
85
asked Mar 17 at 23:53
Jack DohertyJack Doherty
85
edited Mar 18 at 10:53
Jack Doherty
asked Mar 17 at 23:53
Jack DohertyJack Doherty
85
asked Mar 17 at 23:53
Jack DohertyJack Doherty
85
asked Mar 17 at 23:53
Jack DohertyJack Doherty
85
85
$begingroup$
Consider the natural homomorphism from $K [x] $ to $K (a) $ given by $p (x)mapsto p (a) $.
$endgroup$
– Thomas Shelby
Mar 18 at 1:07
$begingroup$
But isn't that not an isomorphism? Does it work both ways?
$endgroup$
– Jack Doherty
Mar 18 at 1:18
$begingroup$
The induced map from $K [x]/(f) $ to $K (a) $ will be an isomorphism. Note that $K [x]/(f) $ is a field(why?).
$endgroup$
– Thomas Shelby
Mar 18 at 1:20
$begingroup$
It's a field because $f$ is irreducible
$endgroup$
– Jack Doherty
Mar 18 at 1:33
$begingroup$
You know that $phi : K[x] to K[a]$ being a surjective homomorphism implies $K[x]/ker(phi) cong K[a]$
$endgroup$
– reuns
Mar 18 at 1:44
add a comment |
$begingroup$
Consider the natural homomorphism from $K [x] $ to $K (a) $ given by $p (x)mapsto p (a) $.
$endgroup$
– Thomas Shelby
Mar 18 at 1:07
$begingroup$
But isn't that not an isomorphism? Does it work both ways?
$endgroup$
– Jack Doherty
Mar 18 at 1:18
$begingroup$
The induced map from $K [x]/(f) $ to $K (a) $ will be an isomorphism. Note that $K [x]/(f) $ is a field(why?).
$endgroup$
– Thomas Shelby
Mar 18 at 1:20
$begingroup$
It's a field because $f$ is irreducible
$endgroup$
– Jack Doherty
Mar 18 at 1:33
$begingroup$
You know that $phi : K[x] to K[a]$ being a surjective homomorphism implies $K[x]/ker(phi) cong K[a]$
$endgroup$
– reuns
Mar 18 at 1:44
$begingroup$
Consider the natural homomorphism from $K [x] $ to $K (a) $ given by $p (x)mapsto p (a) $.
$endgroup$
– Thomas Shelby
Mar 18 at 1:07
$begingroup$
Consider the natural homomorphism from $K [x] $ to $K (a) $ given by $p (x)mapsto p (a) $.
$endgroup$
– Thomas Shelby
Mar 18 at 1:07
$begingroup$
But isn't that not an isomorphism? Does it work both ways?
$endgroup$
– Jack Doherty
Mar 18 at 1:18
$begingroup$
But isn't that not an isomorphism? Does it work both ways?
$endgroup$
– Jack Doherty
Mar 18 at 1:18
$begingroup$
The induced map from $K [x]/(f) $ to $K (a) $ will be an isomorphism. Note that $K [x]/(f) $ is a field(why?).
$endgroup$
– Thomas Shelby
Mar 18 at 1:20
$begingroup$
The induced map from $K [x]/(f) $ to $K (a) $ will be an isomorphism. Note that $K [x]/(f) $ is a field(why?).
$endgroup$
– Thomas Shelby
Mar 18 at 1:20
$begingroup$
It's a field because $f$ is irreducible
$endgroup$
– Jack Doherty
Mar 18 at 1:33
$begingroup$
It's a field because $f$ is irreducible
$endgroup$
– Jack Doherty
Mar 18 at 1:33
$begingroup$
You know that $phi : K[x] to K[a]$ being a surjective homomorphism implies $K[x]/ker(phi) cong K[a]$
$endgroup$
– reuns
Mar 18 at 1:44
$begingroup$
You know that $phi : K[x] to K[a]$ being a surjective homomorphism implies $K[x]/ker(phi) cong K[a]$
$endgroup$
– reuns
Mar 18 at 1:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I agree with the comments. This is just an overview of the formal proof.
Define, $phi:K[x]to K(a)$ by $phi(q(x))=q(a)$ for all $q(x)in K[x]$ where $K(a)$ is the subfield of $L$ that contains both $K$ and $a$. Now it is a homomorphism (check that!) and for any $rin K(a)$ we set the constant polynomial $q(x)=r,forall x$ we see $phi(q(x))=q(a)=r$ so $phi$ is onto-homomorphism with kernel $Ker(phi)=q(x)in K[x]:q(a)=0=q(x):atext is a root of q(x)=(p(x))$, ideal generated by $p(x)$, $p(x)$ being the minimal polynomial having zero at $a$.
Now by, First isomorphism theorem, $$K[x]/(p(x))cong K(a)$$
Now $a,b$ both have same minimal polynomial so, $$K[x]/(p(x)cong K(b))$$ and thus the results follows.
answered Mar 18 at 11:26
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
$endgroup$
1
$begingroup$
Yes, thank you. I had mostly figured it out from the comments. It looks like I was just way overcomplicating it for myself, it's something I tend to have problems with quite a lot. This makes a lot more sense than whatever I was trying to do.
$endgroup$
– Jack Doherty
Mar 18 at 13:31
add a comment |
Your Answer
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1 Answer
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$begingroup$
I agree with the comments. This is just an overview of the formal proof.
Define, $phi:K[x]to K(a)$ by $phi(q(x))=q(a)$ for all $q(x)in K[x]$ where $K(a)$ is the subfield of $L$ that contains both $K$ and $a$. Now it is a homomorphism (check that!) and for any $rin K(a)$ we set the constant polynomial $q(x)=r,forall x$ we see $phi(q(x))=q(a)=r$ so $phi$ is onto-homomorphism with kernel $Ker(phi)=q(x)in K[x]:q(a)=0=q(x):atext is a root of q(x)=(p(x))$, ideal generated by $p(x)$, $p(x)$ being the minimal polynomial having zero at $a$.
Now by, First isomorphism theorem, $$K[x]/(p(x))cong K(a)$$
Now $a,b$ both have same minimal polynomial so, $$K[x]/(p(x)cong K(b))$$ and thus the results follows.
answered Mar 18 at 11:26
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
$endgroup$
1
$begingroup$
Yes, thank you. I had mostly figured it out from the comments. It looks like I was just way overcomplicating it for myself, it's something I tend to have problems with quite a lot. This makes a lot more sense than whatever I was trying to do.
$endgroup$
– Jack Doherty
Mar 18 at 13:31
add a comment |
$begingroup$
I agree with the comments. This is just an overview of the formal proof.
Define, $phi:K[x]to K(a)$ by $phi(q(x))=q(a)$ for all $q(x)in K[x]$ where $K(a)$ is the subfield of $L$ that contains both $K$ and $a$. Now it is a homomorphism (check that!) and for any $rin K(a)$ we set the constant polynomial $q(x)=r,forall x$ we see $phi(q(x))=q(a)=r$ so $phi$ is onto-homomorphism with kernel $Ker(phi)=q(x)in K[x]:q(a)=0=q(x):atext is a root of q(x)=(p(x))$, ideal generated by $p(x)$, $p(x)$ being the minimal polynomial having zero at $a$.
Now by, First isomorphism theorem, $$K[x]/(p(x))cong K(a)$$
Now $a,b$ both have same minimal polynomial so, $$K[x]/(p(x)cong K(b))$$ and thus the results follows.
answered Mar 18 at 11:26
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
$endgroup$
1
$begingroup$
Yes, thank you. I had mostly figured it out from the comments. It looks like I was just way overcomplicating it for myself, it's something I tend to have problems with quite a lot. This makes a lot more sense than whatever I was trying to do.
$endgroup$
– Jack Doherty
Mar 18 at 13:31
add a comment |
$begingroup$
I agree with the comments. This is just an overview of the formal proof.
Define, $phi:K[x]to K(a)$ by $phi(q(x))=q(a)$ for all $q(x)in K[x]$ where $K(a)$ is the subfield of $L$ that contains both $K$ and $a$. Now it is a homomorphism (check that!) and for any $rin K(a)$ we set the constant polynomial $q(x)=r,forall x$ we see $phi(q(x))=q(a)=r$ so $phi$ is onto-homomorphism with kernel $Ker(phi)=q(x)in K[x]:q(a)=0=q(x):atext is a root of q(x)=(p(x))$, ideal generated by $p(x)$, $p(x)$ being the minimal polynomial having zero at $a$.
Now by, First isomorphism theorem, $$K[x]/(p(x))cong K(a)$$
Now $a,b$ both have same minimal polynomial so, $$K[x]/(p(x)cong K(b))$$ and thus the results follows.
answered Mar 18 at 11:26
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
$endgroup$
I agree with the comments. This is just an overview of the formal proof.
Define, $phi:K[x]to K(a)$ by $phi(q(x))=q(a)$ for all $q(x)in K[x]$ where $K(a)$ is the subfield of $L$ that contains both $K$ and $a$. Now it is a homomorphism (check that!) and for any $rin K(a)$ we set the constant polynomial $q(x)=r,forall x$ we see $phi(q(x))=q(a)=r$ so $phi$ is onto-homomorphism with kernel $Ker(phi)=q(x)in K[x]:q(a)=0=q(x):atext is a root of q(x)=(p(x))$, ideal generated by $p(x)$, $p(x)$ being the minimal polynomial having zero at $a$.
Now by, First isomorphism theorem, $$K[x]/(p(x))cong K(a)$$
Now $a,b$ both have same minimal polynomial so, $$K[x]/(p(x)cong K(b))$$ and thus the results follows.
answered Mar 18 at 11:26
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
answered Mar 18 at 11:26
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
answered Mar 18 at 11:26
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
answered Mar 18 at 11:26
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
1,580519
1
$begingroup$
Yes, thank you. I had mostly figured it out from the comments. It looks like I was just way overcomplicating it for myself, it's something I tend to have problems with quite a lot. This makes a lot more sense than whatever I was trying to do.
$endgroup$
– Jack Doherty
Mar 18 at 13:31
add a comment |
1
$begingroup$
Yes, thank you. I had mostly figured it out from the comments. It looks like I was just way overcomplicating it for myself, it's something I tend to have problems with quite a lot. This makes a lot more sense than whatever I was trying to do.
$endgroup$
– Jack Doherty
Mar 18 at 13:31
1
1
$begingroup$
Yes, thank you. I had mostly figured it out from the comments. It looks like I was just way overcomplicating it for myself, it's something I tend to have problems with quite a lot. This makes a lot more sense than whatever I was trying to do.
$endgroup$
– Jack Doherty
Mar 18 at 13:31
$begingroup$
Yes, thank you. I had mostly figured it out from the comments. It looks like I was just way overcomplicating it for myself, it's something I tend to have problems with quite a lot. This makes a lot more sense than whatever I was trying to do.
$endgroup$
– Jack Doherty
Mar 18 at 13:31
add a comment |
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$begingroup$
Consider the natural homomorphism from $K [x] $ to $K (a) $ given by $p (x)mapsto p (a) $.
$endgroup$
– Thomas Shelby
Mar 18 at 1:07
$begingroup$
But isn't that not an isomorphism? Does it work both ways?
$endgroup$
– Jack Doherty
Mar 18 at 1:18
$begingroup$
The induced map from $K [x]/(f) $ to $K (a) $ will be an isomorphism. Note that $K [x]/(f) $ is a field(why?).
$endgroup$
– Thomas Shelby
Mar 18 at 1:20
$begingroup$
It's a field because $f$ is irreducible
$endgroup$
– Jack Doherty
Mar 18 at 1:33
$begingroup$
You know that $phi : K[x] to K[a]$ being a surjective homomorphism implies $K[x]/ker(phi) cong K[a]$
$endgroup$
– reuns
Mar 18 at 1:44