Mixing
Field extensions and Intermediate Rings
$begingroup$
Let $E/K$ be a field extension. Show the following equivalence:
(i) E/K is algebraic
(ii) Every Intermediate Ring of $E/K$ is a Field.
Here is what I tried:
(i)$Rightarrow$(ii) Let $E/K$ be algebraic. We need to show that for two Elements $a,b in R$ the multiplicative Inverse also is in $R$. Let $R$ be a Intermediate Ring of $E/K$. Then there are is Set $A$ with $K(A)=R$. Let $a,bin A$ with $aneq0$. Let's consider $K(a,b)subseteq R$. Since $a,b$ are algebraic elements, the extension $K(a,b)/K$ is finite $Rightarrow$ all elements of $K(a,b)$ are algebaric over K, so $a+b, a-b, ab, 1/a in K(a,b)subseteq R$.
(ii)$Rightarrow$(i)
Let $R$ be a Intermediate Ring of $E/K$ and $ain E$. We know that $K[a]$ is a field, so $a^{-1}in K[a]$. Now $K[a]:= {sumlimits_{i=1}^n x_ia^{i_1} : x_iin K, i_jin mathbb{N}}$, so there is a $sumlimits_{i=1}^n y_ia^{i_1} = a^{-1} Leftrightarrow sumlimits_{i=1}^n y_ia^{i_1+1} = 1 Leftrightarrow sumlimits_{i=1}^n y_ia^{i_1+1} - 1 = 0 Rightarrow$ a is algebraic.
Is this proof ok?
abstract-algebra ring-theory field-theory
asked Jan 13 at 10:34
KingDingelingKingDingeling
1517
$endgroup$
add a comment |
$begingroup$
Let $E/K$ be a field extension. Show the following equivalence:
(i) E/K is algebraic
(ii) Every Intermediate Ring of $E/K$ is a Field.
Here is what I tried:
(i)$Rightarrow$(ii) Let $E/K$ be algebraic. We need to show that for two Elements $a,b in R$ the multiplicative Inverse also is in $R$. Let $R$ be a Intermediate Ring of $E/K$. Then there are is Set $A$ with $K(A)=R$. Let $a,bin A$ with $aneq0$. Let's consider $K(a,b)subseteq R$. Since $a,b$ are algebraic elements, the extension $K(a,b)/K$ is finite $Rightarrow$ all elements of $K(a,b)$ are algebaric over K, so $a+b, a-b, ab, 1/a in K(a,b)subseteq R$.
(ii)$Rightarrow$(i)
Let $R$ be a Intermediate Ring of $E/K$ and $ain E$. We know that $K[a]$ is a field, so $a^{-1}in K[a]$. Now $K[a]:= {sumlimits_{i=1}^n x_ia^{i_1} : x_iin K, i_jin mathbb{N}}$, so there is a $sumlimits_{i=1}^n y_ia^{i_1} = a^{-1} Leftrightarrow sumlimits_{i=1}^n y_ia^{i_1+1} = 1 Leftrightarrow sumlimits_{i=1}^n y_ia^{i_1+1} - 1 = 0 Rightarrow$ a is algebraic.
Is this proof ok?
abstract-algebra ring-theory field-theory
asked Jan 13 at 10:34
KingDingelingKingDingeling
1517
$endgroup$
add a comment |
$begingroup$
Let $E/K$ be a field extension. Show the following equivalence:
(i) E/K is algebraic
(ii) Every Intermediate Ring of $E/K$ is a Field.
Here is what I tried:
(i)$Rightarrow$(ii) Let $E/K$ be algebraic. We need to show that for two Elements $a,b in R$ the multiplicative Inverse also is in $R$. Let $R$ be a Intermediate Ring of $E/K$. Then there are is Set $A$ with $K(A)=R$. Let $a,bin A$ with $aneq0$. Let's consider $K(a,b)subseteq R$. Since $a,b$ are algebraic elements, the extension $K(a,b)/K$ is finite $Rightarrow$ all elements of $K(a,b)$ are algebaric over K, so $a+b, a-b, ab, 1/a in K(a,b)subseteq R$.
(ii)$Rightarrow$(i)
Let $R$ be a Intermediate Ring of $E/K$ and $ain E$. We know that $K[a]$ is a field, so $a^{-1}in K[a]$. Now $K[a]:= {sumlimits_{i=1}^n x_ia^{i_1} : x_iin K, i_jin mathbb{N}}$, so there is a $sumlimits_{i=1}^n y_ia^{i_1} = a^{-1} Leftrightarrow sumlimits_{i=1}^n y_ia^{i_1+1} = 1 Leftrightarrow sumlimits_{i=1}^n y_ia^{i_1+1} - 1 = 0 Rightarrow$ a is algebraic.
Is this proof ok?
abstract-algebra ring-theory field-theory
asked Jan 13 at 10:34
KingDingelingKingDingeling
1517
$endgroup$
Let $E/K$ be a field extension. Show the following equivalence:
(i) E/K is algebraic
(ii) Every Intermediate Ring of $E/K$ is a Field.
Here is what I tried:
(i)$Rightarrow$(ii) Let $E/K$ be algebraic. We need to show that for two Elements $a,b in R$ the multiplicative Inverse also is in $R$. Let $R$ be a Intermediate Ring of $E/K$. Then there are is Set $A$ with $K(A)=R$. Let $a,bin A$ with $aneq0$. Let's consider $K(a,b)subseteq R$. Since $a,b$ are algebraic elements, the extension $K(a,b)/K$ is finite $Rightarrow$ all elements of $K(a,b)$ are algebaric over K, so $a+b, a-b, ab, 1/a in K(a,b)subseteq R$.
(ii)$Rightarrow$(i)
Let $R$ be a Intermediate Ring of $E/K$ and $ain E$. We know that $K[a]$ is a field, so $a^{-1}in K[a]$. Now $K[a]:= {sumlimits_{i=1}^n x_ia^{i_1} : x_iin K, i_jin mathbb{N}}$, so there is a $sumlimits_{i=1}^n y_ia^{i_1} = a^{-1} Leftrightarrow sumlimits_{i=1}^n y_ia^{i_1+1} = 1 Leftrightarrow sumlimits_{i=1}^n y_ia^{i_1+1} - 1 = 0 Rightarrow$ a is algebraic.
Is this proof ok?
abstract-algebra ring-theory field-theory
abstract-algebra ring-theory field-theory
asked Jan 13 at 10:34
KingDingelingKingDingeling
1517
asked Jan 13 at 10:34
KingDingelingKingDingeling
1517
asked Jan 13 at 10:34
KingDingelingKingDingeling
1517
asked Jan 13 at 10:34
KingDingelingKingDingeling
1517
asked Jan 13 at 10:34
KingDingelingKingDingeling
1517
1517
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The second part of the proof is correct, though you don't need to consider an arbitrary intermediate ring $R$, only the specific $K[a]$ for a given nonzero element $ain E$.
However, in the first part when you write $R=K(A)$, (at least the notation seems to suggest that) you implicitly assumed the conslusion, that $R$ is a field.
We only need that $Rsubseteq E$ and so every nonzero element $ain R$ is algebraic over $K$, implying $a^{-1}in K[a]subseteq R$.
answered Jan 13 at 10:59
BerciBerci
60.8k23673
$endgroup$
$begingroup$
Thank you. I think it would be correct if I would write $R=K[A]$. Can I leave the rest for the first part?
$endgroup$
– KingDingeling
Jan 13 at 11:01
$begingroup$
I'm not entirely sure we can always write $R=K[A]$. Nevertheless, for any $a,bin R$, their sum and difference and product is obviously in $R$, as it is a subring, and for the inverse of an $ain R$ we can purely use that $a$ is algebraic.
$endgroup$
– Berci
Jan 13 at 11:04
$begingroup$
Thanks again, how can we use that $a$ is algebraic?
$endgroup$
– KingDingeling
Jan 13 at 11:05
1
$begingroup$
As I wrote in my answer, if $0ne ain R$, then $K[a]subseteq R$, and as $a$ is algebraic, we have $a^{-1}in K[a]$.
$endgroup$
– Berci
Jan 13 at 11:15
add a comment |
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votes
$begingroup$
The second part of the proof is correct, though you don't need to consider an arbitrary intermediate ring $R$, only the specific $K[a]$ for a given nonzero element $ain E$.
However, in the first part when you write $R=K(A)$, (at least the notation seems to suggest that) you implicitly assumed the conslusion, that $R$ is a field.
We only need that $Rsubseteq E$ and so every nonzero element $ain R$ is algebraic over $K$, implying $a^{-1}in K[a]subseteq R$.
answered Jan 13 at 10:59
BerciBerci
60.8k23673
$endgroup$
$begingroup$
Thank you. I think it would be correct if I would write $R=K[A]$. Can I leave the rest for the first part?
$endgroup$
– KingDingeling
Jan 13 at 11:01
$begingroup$
I'm not entirely sure we can always write $R=K[A]$. Nevertheless, for any $a,bin R$, their sum and difference and product is obviously in $R$, as it is a subring, and for the inverse of an $ain R$ we can purely use that $a$ is algebraic.
$endgroup$
– Berci
Jan 13 at 11:04
$begingroup$
Thanks again, how can we use that $a$ is algebraic?
$endgroup$
– KingDingeling
Jan 13 at 11:05
1
$begingroup$
As I wrote in my answer, if $0ne ain R$, then $K[a]subseteq R$, and as $a$ is algebraic, we have $a^{-1}in K[a]$.
$endgroup$
– Berci
Jan 13 at 11:15
add a comment |
$begingroup$
The second part of the proof is correct, though you don't need to consider an arbitrary intermediate ring $R$, only the specific $K[a]$ for a given nonzero element $ain E$.
However, in the first part when you write $R=K(A)$, (at least the notation seems to suggest that) you implicitly assumed the conslusion, that $R$ is a field.
We only need that $Rsubseteq E$ and so every nonzero element $ain R$ is algebraic over $K$, implying $a^{-1}in K[a]subseteq R$.
answered Jan 13 at 10:59
BerciBerci
60.8k23673
$endgroup$
$begingroup$
Thank you. I think it would be correct if I would write $R=K[A]$. Can I leave the rest for the first part?
$endgroup$
– KingDingeling
Jan 13 at 11:01
$begingroup$
I'm not entirely sure we can always write $R=K[A]$. Nevertheless, for any $a,bin R$, their sum and difference and product is obviously in $R$, as it is a subring, and for the inverse of an $ain R$ we can purely use that $a$ is algebraic.
$endgroup$
– Berci
Jan 13 at 11:04
$begingroup$
Thanks again, how can we use that $a$ is algebraic?
$endgroup$
– KingDingeling
Jan 13 at 11:05
1
$begingroup$
As I wrote in my answer, if $0ne ain R$, then $K[a]subseteq R$, and as $a$ is algebraic, we have $a^{-1}in K[a]$.
$endgroup$
– Berci
Jan 13 at 11:15
add a comment |
$begingroup$
The second part of the proof is correct, though you don't need to consider an arbitrary intermediate ring $R$, only the specific $K[a]$ for a given nonzero element $ain E$.
However, in the first part when you write $R=K(A)$, (at least the notation seems to suggest that) you implicitly assumed the conslusion, that $R$ is a field.
We only need that $Rsubseteq E$ and so every nonzero element $ain R$ is algebraic over $K$, implying $a^{-1}in K[a]subseteq R$.
answered Jan 13 at 10:59
BerciBerci
60.8k23673
$endgroup$
The second part of the proof is correct, though you don't need to consider an arbitrary intermediate ring $R$, only the specific $K[a]$ for a given nonzero element $ain E$.
However, in the first part when you write $R=K(A)$, (at least the notation seems to suggest that) you implicitly assumed the conslusion, that $R$ is a field.
We only need that $Rsubseteq E$ and so every nonzero element $ain R$ is algebraic over $K$, implying $a^{-1}in K[a]subseteq R$.
answered Jan 13 at 10:59
BerciBerci
60.8k23673
edited Jan 13 at 11:01
answered Jan 13 at 10:59
BerciBerci
60.8k23673
answered Jan 13 at 10:59
BerciBerci
60.8k23673
answered Jan 13 at 10:59
BerciBerci
60.8k23673
60.8k23673
$begingroup$
Thank you. I think it would be correct if I would write $R=K[A]$. Can I leave the rest for the first part?
$endgroup$
– KingDingeling
Jan 13 at 11:01
$begingroup$
I'm not entirely sure we can always write $R=K[A]$. Nevertheless, for any $a,bin R$, their sum and difference and product is obviously in $R$, as it is a subring, and for the inverse of an $ain R$ we can purely use that $a$ is algebraic.
$endgroup$
– Berci
Jan 13 at 11:04
$begingroup$
Thanks again, how can we use that $a$ is algebraic?
$endgroup$
– KingDingeling
Jan 13 at 11:05
1
$begingroup$
As I wrote in my answer, if $0ne ain R$, then $K[a]subseteq R$, and as $a$ is algebraic, we have $a^{-1}in K[a]$.
$endgroup$
– Berci
Jan 13 at 11:15
add a comment |
$begingroup$
Thank you. I think it would be correct if I would write $R=K[A]$. Can I leave the rest for the first part?
$endgroup$
– KingDingeling
Jan 13 at 11:01
$begingroup$
I'm not entirely sure we can always write $R=K[A]$. Nevertheless, for any $a,bin R$, their sum and difference and product is obviously in $R$, as it is a subring, and for the inverse of an $ain R$ we can purely use that $a$ is algebraic.
$endgroup$
– Berci
Jan 13 at 11:04
$begingroup$
Thanks again, how can we use that $a$ is algebraic?
$endgroup$
– KingDingeling
Jan 13 at 11:05
1
$begingroup$
As I wrote in my answer, if $0ne ain R$, then $K[a]subseteq R$, and as $a$ is algebraic, we have $a^{-1}in K[a]$.
$endgroup$
– Berci
Jan 13 at 11:15
$begingroup$
Thank you. I think it would be correct if I would write $R=K[A]$. Can I leave the rest for the first part?
$endgroup$
– KingDingeling
Jan 13 at 11:01
$begingroup$
Thank you. I think it would be correct if I would write $R=K[A]$. Can I leave the rest for the first part?
$endgroup$
– KingDingeling
Jan 13 at 11:01
$begingroup$
I'm not entirely sure we can always write $R=K[A]$. Nevertheless, for any $a,bin R$, their sum and difference and product is obviously in $R$, as it is a subring, and for the inverse of an $ain R$ we can purely use that $a$ is algebraic.
$endgroup$
– Berci
Jan 13 at 11:04
$begingroup$
I'm not entirely sure we can always write $R=K[A]$. Nevertheless, for any $a,bin R$, their sum and difference and product is obviously in $R$, as it is a subring, and for the inverse of an $ain R$ we can purely use that $a$ is algebraic.
$endgroup$
– Berci
Jan 13 at 11:04
$begingroup$
Thanks again, how can we use that $a$ is algebraic?
$endgroup$
– KingDingeling
Jan 13 at 11:05
$begingroup$
Thanks again, how can we use that $a$ is algebraic?
$endgroup$
– KingDingeling
Jan 13 at 11:05
1
1
$begingroup$
As I wrote in my answer, if $0ne ain R$, then $K[a]subseteq R$, and as $a$ is algebraic, we have $a^{-1}in K[a]$.
$endgroup$
– Berci
Jan 13 at 11:15
$begingroup$
As I wrote in my answer, if $0ne ain R$, then $K[a]subseteq R$, and as $a$ is algebraic, we have $a^{-1}in K[a]$.
$endgroup$
– Berci
Jan 13 at 11:15
add a comment |
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