Show that the algebraic extension $K(alpha,beta)/K$ is simple if $alpha$ is separable over $K$
$begingroup$
Let $L=K(alpha,beta)$ be an algebraic field extension, with $alpha$ separable over K.
Show that $L/K$ is simple.
My attempt:
If we could show that $L/K$ is finite and separable then the claim would follow by the primitive element theorem.
Since $L/K$ is algebraic and finitely generated, it is a finite extension.
Since $alpha$ is separable the minimal polynomial of $alpha$ in $overline{K}$ is of the form:
$m_{alpha}(X)=(X-alpha_1)(X-alpha_2)cdot....cdot(X-alpha_n)$ with pairwise distinct $alpha_i$ and $alpha=alpha_i$ for one $i$.
Then $f_{alpha,beta}(X)=(X-alpha_1)(X-alpha_2)cdot....cdot(X-alpha_n)cdot(X-beta)$ is a separable polynomial with $f_{alpha,beta}(beta)=0$, hence $beta$ is also separable. The conclusion follows because then $K(alpha,beta)$ is separable, since it's generated by separable elements.
Can someone go through it? I also have some questions about the proof:
- Our definition of a separable element $alpha$ is: the minimal polynomial of $alpha$ is separable. In the proof above I used that $alpha$ is separable because there exists a polynomial $f$ with $f(alpha)=0$. Are these two characterizations equivalent?
$(Leftarrow)$ If $f$ is separable with $f(alpha)=0$ then $m_alphacdot g=f$, hence $m_alpha$ can't have a multiple zero, because of the degrees.
But how does the other direction work?
- The polynomial $f$ in the proof: is it already the minimal polynomial of $beta$? If yes, why?
Thanks a lot!
abstract-algebra proof-verification field-theory extension-field
edited Dec 13 '18 at 19:29
Brahadeesh
6,42442363
asked Jan 17 '15 at 16:49
RayBuriRayBuri
161
$endgroup$
add a comment |
$begingroup$
Let $L=K(alpha,beta)$ be an algebraic field extension, with $alpha$ separable over K.
Show that $L/K$ is simple.
My attempt:
If we could show that $L/K$ is finite and separable then the claim would follow by the primitive element theorem.
Since $L/K$ is algebraic and finitely generated, it is a finite extension.
Since $alpha$ is separable the minimal polynomial of $alpha$ in $overline{K}$ is of the form:
$m_{alpha}(X)=(X-alpha_1)(X-alpha_2)cdot....cdot(X-alpha_n)$ with pairwise distinct $alpha_i$ and $alpha=alpha_i$ for one $i$.
Then $f_{alpha,beta}(X)=(X-alpha_1)(X-alpha_2)cdot....cdot(X-alpha_n)cdot(X-beta)$ is a separable polynomial with $f_{alpha,beta}(beta)=0$, hence $beta$ is also separable. The conclusion follows because then $K(alpha,beta)$ is separable, since it's generated by separable elements.
Can someone go through it? I also have some questions about the proof:
- Our definition of a separable element $alpha$ is: the minimal polynomial of $alpha$ is separable. In the proof above I used that $alpha$ is separable because there exists a polynomial $f$ with $f(alpha)=0$. Are these two characterizations equivalent?
$(Leftarrow)$ If $f$ is separable with $f(alpha)=0$ then $m_alphacdot g=f$, hence $m_alpha$ can't have a multiple zero, because of the degrees.
But how does the other direction work?
- The polynomial $f$ in the proof: is it already the minimal polynomial of $beta$? If yes, why?
Thanks a lot!
abstract-algebra proof-verification field-theory extension-field
edited Dec 13 '18 at 19:29
Brahadeesh
6,42442363
asked Jan 17 '15 at 16:49
RayBuriRayBuri
161
$endgroup$
2
$begingroup$
$beta$ doesn't have to be separable. Your polynomial $f_{alpha,beta}(X)=m_{alpha,K}(X)(X-beta)$ is not in $K[X]$, if $betanotin K$. Hence it is not minimal polynomial of $beta$ over $K$. You can find the proof here: math.cornell.edu/~kbrown/6310/primitive.pdf
$endgroup$
– SMM
Jan 17 '15 at 18:53
add a comment |
$begingroup$
Let $L=K(alpha,beta)$ be an algebraic field extension, with $alpha$ separable over K.
Show that $L/K$ is simple.
My attempt:
If we could show that $L/K$ is finite and separable then the claim would follow by the primitive element theorem.
Since $L/K$ is algebraic and finitely generated, it is a finite extension.
Since $alpha$ is separable the minimal polynomial of $alpha$ in $overline{K}$ is of the form:
$m_{alpha}(X)=(X-alpha_1)(X-alpha_2)cdot....cdot(X-alpha_n)$ with pairwise distinct $alpha_i$ and $alpha=alpha_i$ for one $i$.
Then $f_{alpha,beta}(X)=(X-alpha_1)(X-alpha_2)cdot....cdot(X-alpha_n)cdot(X-beta)$ is a separable polynomial with $f_{alpha,beta}(beta)=0$, hence $beta$ is also separable. The conclusion follows because then $K(alpha,beta)$ is separable, since it's generated by separable elements.
Can someone go through it? I also have some questions about the proof:
- Our definition of a separable element $alpha$ is: the minimal polynomial of $alpha$ is separable. In the proof above I used that $alpha$ is separable because there exists a polynomial $f$ with $f(alpha)=0$. Are these two characterizations equivalent?
$(Leftarrow)$ If $f$ is separable with $f(alpha)=0$ then $m_alphacdot g=f$, hence $m_alpha$ can't have a multiple zero, because of the degrees.
But how does the other direction work?
- The polynomial $f$ in the proof: is it already the minimal polynomial of $beta$? If yes, why?
Thanks a lot!
abstract-algebra proof-verification field-theory extension-field
edited Dec 13 '18 at 19:29
Brahadeesh
6,42442363
asked Jan 17 '15 at 16:49
RayBuriRayBuri
161
$endgroup$
Let $L=K(alpha,beta)$ be an algebraic field extension, with $alpha$ separable over K.
Show that $L/K$ is simple.
My attempt:
If we could show that $L/K$ is finite and separable then the claim would follow by the primitive element theorem.
Since $L/K$ is algebraic and finitely generated, it is a finite extension.
Since $alpha$ is separable the minimal polynomial of $alpha$ in $overline{K}$ is of the form:
$m_{alpha}(X)=(X-alpha_1)(X-alpha_2)cdot....cdot(X-alpha_n)$ with pairwise distinct $alpha_i$ and $alpha=alpha_i$ for one $i$.
Then $f_{alpha,beta}(X)=(X-alpha_1)(X-alpha_2)cdot....cdot(X-alpha_n)cdot(X-beta)$ is a separable polynomial with $f_{alpha,beta}(beta)=0$, hence $beta$ is also separable. The conclusion follows because then $K(alpha,beta)$ is separable, since it's generated by separable elements.
Can someone go through it? I also have some questions about the proof:
- Our definition of a separable element $alpha$ is: the minimal polynomial of $alpha$ is separable. In the proof above I used that $alpha$ is separable because there exists a polynomial $f$ with $f(alpha)=0$. Are these two characterizations equivalent?
$(Leftarrow)$ If $f$ is separable with $f(alpha)=0$ then $m_alphacdot g=f$, hence $m_alpha$ can't have a multiple zero, because of the degrees.
But how does the other direction work?
- The polynomial $f$ in the proof: is it already the minimal polynomial of $beta$? If yes, why?
Thanks a lot!
abstract-algebra proof-verification field-theory extension-field
abstract-algebra proof-verification field-theory extension-field
edited Dec 13 '18 at 19:29
Brahadeesh
6,42442363
asked Jan 17 '15 at 16:49
RayBuriRayBuri
161
edited Dec 13 '18 at 19:29
Brahadeesh
6,42442363
asked Jan 17 '15 at 16:49
RayBuriRayBuri
161
edited Dec 13 '18 at 19:29
Brahadeesh
6,42442363
edited Dec 13 '18 at 19:29
Brahadeesh
6,42442363
edited Dec 13 '18 at 19:29
Brahadeesh
6,42442363
6,42442363
asked Jan 17 '15 at 16:49
RayBuriRayBuri
161
asked Jan 17 '15 at 16:49
RayBuriRayBuri
161
asked Jan 17 '15 at 16:49
RayBuriRayBuri
161
161
2
$begingroup$
$beta$ doesn't have to be separable. Your polynomial $f_{alpha,beta}(X)=m_{alpha,K}(X)(X-beta)$ is not in $K[X]$, if $betanotin K$. Hence it is not minimal polynomial of $beta$ over $K$. You can find the proof here: math.cornell.edu/~kbrown/6310/primitive.pdf
$endgroup$
– SMM
Jan 17 '15 at 18:53
add a comment |
2
$begingroup$
$beta$ doesn't have to be separable. Your polynomial $f_{alpha,beta}(X)=m_{alpha,K}(X)(X-beta)$ is not in $K[X]$, if $betanotin K$. Hence it is not minimal polynomial of $beta$ over $K$. You can find the proof here: math.cornell.edu/~kbrown/6310/primitive.pdf
$endgroup$
– SMM
Jan 17 '15 at 18:53
2
2
$begingroup$
$beta$ doesn't have to be separable. Your polynomial $f_{alpha,beta}(X)=m_{alpha,K}(X)(X-beta)$ is not in $K[X]$, if $betanotin K$. Hence it is not minimal polynomial of $beta$ over $K$. You can find the proof here: math.cornell.edu/~kbrown/6310/primitive.pdf
$endgroup$
– SMM
Jan 17 '15 at 18:53
$begingroup$
$beta$ doesn't have to be separable. Your polynomial $f_{alpha,beta}(X)=m_{alpha,K}(X)(X-beta)$ is not in $K[X]$, if $betanotin K$. Hence it is not minimal polynomial of $beta$ over $K$. You can find the proof here: math.cornell.edu/~kbrown/6310/primitive.pdf
$endgroup$
– SMM
Jan 17 '15 at 18:53
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Your proof is incorrect.
Notice that you have apparently proved that $beta$ is separable over $K$, when in fact you have not made any initial assumptions on $beta$. Surely this is a red flag: after all, you could have started with $beta notin K$ and $beta$ purely inseparable over $K$. Then it is quite impossible for $beta$ to also be separable over $K$.
Your proof fails because $f_{alpha,beta}$ need not be in $K[X]$. In fact, it definitely won't be when $beta notin K$. This also answers (2): it is not the minimal polynomial of $beta$ over $K$ because it does not lie in $K[X]$ (unless $beta in K$, which is a trivial scenario).
Regarding (1): I am not sure what your question is, because you have definitely used the fact that $alpha$ is separable because its minimal polynomial is separable.
For a full proof of the result, you can see here: math.cornell.edu/~kbrown/6310/primitive.pdf.
answered Dec 13 '18 at 18:54
BrahadeeshBrahadeesh
6,42442363
$endgroup$
add a comment |
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$begingroup$
Your proof is incorrect.
Notice that you have apparently proved that $beta$ is separable over $K$, when in fact you have not made any initial assumptions on $beta$. Surely this is a red flag: after all, you could have started with $beta notin K$ and $beta$ purely inseparable over $K$. Then it is quite impossible for $beta$ to also be separable over $K$.
Your proof fails because $f_{alpha,beta}$ need not be in $K[X]$. In fact, it definitely won't be when $beta notin K$. This also answers (2): it is not the minimal polynomial of $beta$ over $K$ because it does not lie in $K[X]$ (unless $beta in K$, which is a trivial scenario).
Regarding (1): I am not sure what your question is, because you have definitely used the fact that $alpha$ is separable because its minimal polynomial is separable.
For a full proof of the result, you can see here: math.cornell.edu/~kbrown/6310/primitive.pdf.
answered Dec 13 '18 at 18:54
BrahadeeshBrahadeesh
6,42442363
$endgroup$
add a comment |
$begingroup$
Your proof is incorrect.
Notice that you have apparently proved that $beta$ is separable over $K$, when in fact you have not made any initial assumptions on $beta$. Surely this is a red flag: after all, you could have started with $beta notin K$ and $beta$ purely inseparable over $K$. Then it is quite impossible for $beta$ to also be separable over $K$.
Your proof fails because $f_{alpha,beta}$ need not be in $K[X]$. In fact, it definitely won't be when $beta notin K$. This also answers (2): it is not the minimal polynomial of $beta$ over $K$ because it does not lie in $K[X]$ (unless $beta in K$, which is a trivial scenario).
Regarding (1): I am not sure what your question is, because you have definitely used the fact that $alpha$ is separable because its minimal polynomial is separable.
For a full proof of the result, you can see here: math.cornell.edu/~kbrown/6310/primitive.pdf.
answered Dec 13 '18 at 18:54
BrahadeeshBrahadeesh
6,42442363
$endgroup$
add a comment |
$begingroup$
Your proof is incorrect.
Notice that you have apparently proved that $beta$ is separable over $K$, when in fact you have not made any initial assumptions on $beta$. Surely this is a red flag: after all, you could have started with $beta notin K$ and $beta$ purely inseparable over $K$. Then it is quite impossible for $beta$ to also be separable over $K$.
Your proof fails because $f_{alpha,beta}$ need not be in $K[X]$. In fact, it definitely won't be when $beta notin K$. This also answers (2): it is not the minimal polynomial of $beta$ over $K$ because it does not lie in $K[X]$ (unless $beta in K$, which is a trivial scenario).
Regarding (1): I am not sure what your question is, because you have definitely used the fact that $alpha$ is separable because its minimal polynomial is separable.
For a full proof of the result, you can see here: math.cornell.edu/~kbrown/6310/primitive.pdf.
answered Dec 13 '18 at 18:54
BrahadeeshBrahadeesh
6,42442363
$endgroup$
Your proof is incorrect.
Notice that you have apparently proved that $beta$ is separable over $K$, when in fact you have not made any initial assumptions on $beta$. Surely this is a red flag: after all, you could have started with $beta notin K$ and $beta$ purely inseparable over $K$. Then it is quite impossible for $beta$ to also be separable over $K$.
Your proof fails because $f_{alpha,beta}$ need not be in $K[X]$. In fact, it definitely won't be when $beta notin K$. This also answers (2): it is not the minimal polynomial of $beta$ over $K$ because it does not lie in $K[X]$ (unless $beta in K$, which is a trivial scenario).
Regarding (1): I am not sure what your question is, because you have definitely used the fact that $alpha$ is separable because its minimal polynomial is separable.
For a full proof of the result, you can see here: math.cornell.edu/~kbrown/6310/primitive.pdf.
answered Dec 13 '18 at 18:54
BrahadeeshBrahadeesh
6,42442363
answered Dec 13 '18 at 18:54
BrahadeeshBrahadeesh
6,42442363
answered Dec 13 '18 at 18:54
BrahadeeshBrahadeesh
6,42442363
answered Dec 13 '18 at 18:54
BrahadeeshBrahadeesh
6,42442363
6,42442363
add a comment |
add a comment |
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$begingroup$
$beta$ doesn't have to be separable. Your polynomial $f_{alpha,beta}(X)=m_{alpha,K}(X)(X-beta)$ is not in $K[X]$, if $betanotin K$. Hence it is not minimal polynomial of $beta$ over $K$. You can find the proof here: math.cornell.edu/~kbrown/6310/primitive.pdf
$endgroup$
– SMM
Jan 17 '15 at 18:53