How to show that this simple extension avoids adding more elements in $L$?
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$Fsubset L subset K$ is a field extension, $α in K$ is algebraic element over $F$ , whose minimal polynomial $p(x)$ over $F$ is irreducible over $L$, show that $F(α)cap L = F$
I don't know how to do..thanks for any help..
field-theory extension-field
asked Jan 16 at 0:29
likemathlikemath
16519
$endgroup$
add a comment |
$begingroup$
$Fsubset L subset K$ is a field extension, $α in K$ is algebraic element over $F$ , whose minimal polynomial $p(x)$ over $F$ is irreducible over $L$, show that $F(α)cap L = F$
I don't know how to do..thanks for any help..
field-theory extension-field
asked Jan 16 at 0:29
likemathlikemath
16519
$endgroup$
add a comment |
$begingroup$
$Fsubset L subset K$ is a field extension, $α in K$ is algebraic element over $F$ , whose minimal polynomial $p(x)$ over $F$ is irreducible over $L$, show that $F(α)cap L = F$
I don't know how to do..thanks for any help..
field-theory extension-field
asked Jan 16 at 0:29
likemathlikemath
16519
$endgroup$
$Fsubset L subset K$ is a field extension, $α in K$ is algebraic element over $F$ , whose minimal polynomial $p(x)$ over $F$ is irreducible over $L$, show that $F(α)cap L = F$
I don't know how to do..thanks for any help..
field-theory extension-field
field-theory extension-field
asked Jan 16 at 0:29
likemathlikemath
16519
asked Jan 16 at 0:29
likemathlikemath
16519
asked Jan 16 at 0:29
likemathlikemath
16519
asked Jan 16 at 0:29
likemathlikemath
16519
asked Jan 16 at 0:29
likemathlikemath
16519
16519
add a comment |
add a comment |
1 Answer
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$newcommand{Size}[1]{leftlvert #1 rightrvert}$Recall that every element of $F(alpha)$ can be written uniquely as $q(alpha)$, where $q(x) in F[x]$ is either zero, or has degree less than the degree of $p(x)$.
Also, since $p(x)$ is irreducible over $L$, it is the minimal polynomial of $alpha$ not only over $F$, but also over $L$
Suppose by way of contradiction that $F(alpha) cap L supsetneq F$. Then there is a non-constant polynomial $q(x) in F[x]$, of degree smaller than the degree of $p(x)$, such that $q(alpha) = b in L$. But then $alpha$ is a root of $0 ne q(x) - b in L[x]$. This is a contradiction, as $q(x) - b$ has degree less than $p(x)$, which was assumed to be the minimal polynomial of $alpha$ over $L$.
answered Jan 16 at 1:10
Andreas CarantiAndreas Caranti
56.9k34397
$endgroup$
$begingroup$
Thanks, I got it.
$endgroup$
– likemath
Jan 16 at 1:26
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@likemath, you're welcome.
$endgroup$
– Andreas Caranti
Jan 16 at 1:27
add a comment |
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1 Answer
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$begingroup$
$newcommand{Size}[1]{leftlvert #1 rightrvert}$Recall that every element of $F(alpha)$ can be written uniquely as $q(alpha)$, where $q(x) in F[x]$ is either zero, or has degree less than the degree of $p(x)$.
Also, since $p(x)$ is irreducible over $L$, it is the minimal polynomial of $alpha$ not only over $F$, but also over $L$
Suppose by way of contradiction that $F(alpha) cap L supsetneq F$. Then there is a non-constant polynomial $q(x) in F[x]$, of degree smaller than the degree of $p(x)$, such that $q(alpha) = b in L$. But then $alpha$ is a root of $0 ne q(x) - b in L[x]$. This is a contradiction, as $q(x) - b$ has degree less than $p(x)$, which was assumed to be the minimal polynomial of $alpha$ over $L$.
answered Jan 16 at 1:10
Andreas CarantiAndreas Caranti
56.9k34397
$endgroup$
$begingroup$
Thanks, I got it.
$endgroup$
– likemath
Jan 16 at 1:26
$begingroup$
@likemath, you're welcome.
$endgroup$
– Andreas Caranti
Jan 16 at 1:27
add a comment |
$begingroup$
$newcommand{Size}[1]{leftlvert #1 rightrvert}$Recall that every element of $F(alpha)$ can be written uniquely as $q(alpha)$, where $q(x) in F[x]$ is either zero, or has degree less than the degree of $p(x)$.
Also, since $p(x)$ is irreducible over $L$, it is the minimal polynomial of $alpha$ not only over $F$, but also over $L$
Suppose by way of contradiction that $F(alpha) cap L supsetneq F$. Then there is a non-constant polynomial $q(x) in F[x]$, of degree smaller than the degree of $p(x)$, such that $q(alpha) = b in L$. But then $alpha$ is a root of $0 ne q(x) - b in L[x]$. This is a contradiction, as $q(x) - b$ has degree less than $p(x)$, which was assumed to be the minimal polynomial of $alpha$ over $L$.
answered Jan 16 at 1:10
Andreas CarantiAndreas Caranti
56.9k34397
$endgroup$
$begingroup$
Thanks, I got it.
$endgroup$
– likemath
Jan 16 at 1:26
$begingroup$
@likemath, you're welcome.
$endgroup$
– Andreas Caranti
Jan 16 at 1:27
add a comment |
$begingroup$
$newcommand{Size}[1]{leftlvert #1 rightrvert}$Recall that every element of $F(alpha)$ can be written uniquely as $q(alpha)$, where $q(x) in F[x]$ is either zero, or has degree less than the degree of $p(x)$.
Also, since $p(x)$ is irreducible over $L$, it is the minimal polynomial of $alpha$ not only over $F$, but also over $L$
Suppose by way of contradiction that $F(alpha) cap L supsetneq F$. Then there is a non-constant polynomial $q(x) in F[x]$, of degree smaller than the degree of $p(x)$, such that $q(alpha) = b in L$. But then $alpha$ is a root of $0 ne q(x) - b in L[x]$. This is a contradiction, as $q(x) - b$ has degree less than $p(x)$, which was assumed to be the minimal polynomial of $alpha$ over $L$.
answered Jan 16 at 1:10
Andreas CarantiAndreas Caranti
56.9k34397
$endgroup$
$newcommand{Size}[1]{leftlvert #1 rightrvert}$Recall that every element of $F(alpha)$ can be written uniquely as $q(alpha)$, where $q(x) in F[x]$ is either zero, or has degree less than the degree of $p(x)$.
Also, since $p(x)$ is irreducible over $L$, it is the minimal polynomial of $alpha$ not only over $F$, but also over $L$
Suppose by way of contradiction that $F(alpha) cap L supsetneq F$. Then there is a non-constant polynomial $q(x) in F[x]$, of degree smaller than the degree of $p(x)$, such that $q(alpha) = b in L$. But then $alpha$ is a root of $0 ne q(x) - b in L[x]$. This is a contradiction, as $q(x) - b$ has degree less than $p(x)$, which was assumed to be the minimal polynomial of $alpha$ over $L$.
answered Jan 16 at 1:10
Andreas CarantiAndreas Caranti
56.9k34397
edited Jan 16 at 19:21
answered Jan 16 at 1:10
Andreas CarantiAndreas Caranti
56.9k34397
answered Jan 16 at 1:10
Andreas CarantiAndreas Caranti
56.9k34397
answered Jan 16 at 1:10
Andreas CarantiAndreas Caranti
56.9k34397
56.9k34397
$begingroup$
Thanks, I got it.
$endgroup$
– likemath
Jan 16 at 1:26
$begingroup$
@likemath, you're welcome.
$endgroup$
– Andreas Caranti
Jan 16 at 1:27
add a comment |
$begingroup$
Thanks, I got it.
$endgroup$
– likemath
Jan 16 at 1:26
$begingroup$
@likemath, you're welcome.
$endgroup$
– Andreas Caranti
Jan 16 at 1:27
$begingroup$
Thanks, I got it.
$endgroup$
– likemath
Jan 16 at 1:26
$begingroup$
Thanks, I got it.
$endgroup$
– likemath
Jan 16 at 1:26
$begingroup$
@likemath, you're welcome.
$endgroup$
– Andreas Caranti
Jan 16 at 1:27
$begingroup$
@likemath, you're welcome.
$endgroup$
– Andreas Caranti
Jan 16 at 1:27
add a comment |
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