Mixing
Transitivity of Algebraic Field Extensions
$begingroup$
Consider the fields $F, E,$ and $K$, where $F subseteq E subseteq K$. If $E$ is algebraic over $F$, and $K$ is algebraic over $E$, show that $K$ must be algebraic over $F$.
I know this is a well-known, proved theorem, but I'm trying to understand it on my own.
If $E$ is algebraic over $F$, then that means there is some $alpha_1$ in $E$ s.t. $f(alpha_1)$ = $0$ for some $f(x) in F[x]$. Similarly for $K$ and $E$, there is some $alpha_2$ in $K$ s.t. $e(alpha_2$) = $0$ for some $e(x) in E[x]$.
What I'm stuck on is connecting $alpha_1$ and $alpha_2$ together. If $alpha_2$ is algebraic over $E$, how does that translate to it also being algebraic over $F$? Thank you for your help.
abstract-algebra field-theory extension-field
edited Dec 12 '16 at 15:28
Adam Hughes
32.4k83770
asked Dec 12 '16 at 12:45
MaxMax
697317
$endgroup$
add a comment |
$begingroup$
Consider the fields $F, E,$ and $K$, where $F subseteq E subseteq K$. If $E$ is algebraic over $F$, and $K$ is algebraic over $E$, show that $K$ must be algebraic over $F$.
I know this is a well-known, proved theorem, but I'm trying to understand it on my own.
If $E$ is algebraic over $F$, then that means there is some $alpha_1$ in $E$ s.t. $f(alpha_1)$ = $0$ for some $f(x) in F[x]$. Similarly for $K$ and $E$, there is some $alpha_2$ in $K$ s.t. $e(alpha_2$) = $0$ for some $e(x) in E[x]$.
What I'm stuck on is connecting $alpha_1$ and $alpha_2$ together. If $alpha_2$ is algebraic over $E$, how does that translate to it also being algebraic over $F$? Thank you for your help.
abstract-algebra field-theory extension-field
edited Dec 12 '16 at 15:28
Adam Hughes
32.4k83770
asked Dec 12 '16 at 12:45
MaxMax
697317
$endgroup$
add a comment |
$begingroup$
Consider the fields $F, E,$ and $K$, where $F subseteq E subseteq K$. If $E$ is algebraic over $F$, and $K$ is algebraic over $E$, show that $K$ must be algebraic over $F$.
I know this is a well-known, proved theorem, but I'm trying to understand it on my own.
If $E$ is algebraic over $F$, then that means there is some $alpha_1$ in $E$ s.t. $f(alpha_1)$ = $0$ for some $f(x) in F[x]$. Similarly for $K$ and $E$, there is some $alpha_2$ in $K$ s.t. $e(alpha_2$) = $0$ for some $e(x) in E[x]$.
What I'm stuck on is connecting $alpha_1$ and $alpha_2$ together. If $alpha_2$ is algebraic over $E$, how does that translate to it also being algebraic over $F$? Thank you for your help.
abstract-algebra field-theory extension-field
edited Dec 12 '16 at 15:28
Adam Hughes
32.4k83770
asked Dec 12 '16 at 12:45
MaxMax
697317
$endgroup$
Consider the fields $F, E,$ and $K$, where $F subseteq E subseteq K$. If $E$ is algebraic over $F$, and $K$ is algebraic over $E$, show that $K$ must be algebraic over $F$.
I know this is a well-known, proved theorem, but I'm trying to understand it on my own.
If $E$ is algebraic over $F$, then that means there is some $alpha_1$ in $E$ s.t. $f(alpha_1)$ = $0$ for some $f(x) in F[x]$. Similarly for $K$ and $E$, there is some $alpha_2$ in $K$ s.t. $e(alpha_2$) = $0$ for some $e(x) in E[x]$.
What I'm stuck on is connecting $alpha_1$ and $alpha_2$ together. If $alpha_2$ is algebraic over $E$, how does that translate to it also being algebraic over $F$? Thank you for your help.
abstract-algebra field-theory extension-field
abstract-algebra field-theory extension-field
edited Dec 12 '16 at 15:28
Adam Hughes
32.4k83770
asked Dec 12 '16 at 12:45
MaxMax
697317
edited Dec 12 '16 at 15:28
Adam Hughes
32.4k83770
asked Dec 12 '16 at 12:45
MaxMax
697317
edited Dec 12 '16 at 15:28
Adam Hughes
32.4k83770
edited Dec 12 '16 at 15:28
Adam Hughes
32.4k83770
edited Dec 12 '16 at 15:28
Adam Hughes
32.4k83770
32.4k83770
asked Dec 12 '16 at 12:45
MaxMax
697317
asked Dec 12 '16 at 12:45
MaxMax
697317
asked Dec 12 '16 at 12:45
MaxMax
697317
697317
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We use the characterization that $alpha in k'/k$ is an algebraic element of $k'$ over $k$ iff $[k(alpha):k]<infty$. Then let $ain K$. By definition there is $p(x)in E[x]$ so that $p(a)=0$. Let
$$p(x) = sum_{i=1}^n a_ix^i$$
and then note that as each $a_iin E$, we have that
$$[F(a_0, a_1, a_2,ldots, a_n): F]le prod_i [F(a_i):F]<infty.$$
But then $[F(a_0, a_1,ldots, a_n, a):F] < infty$, so $a$ is algebraic over $F$.
answered Dec 12 '16 at 15:27
Adam HughesAdam Hughes
32.4k83770
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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active
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votes
$begingroup$
We use the characterization that $alpha in k'/k$ is an algebraic element of $k'$ over $k$ iff $[k(alpha):k]<infty$. Then let $ain K$. By definition there is $p(x)in E[x]$ so that $p(a)=0$. Let
$$p(x) = sum_{i=1}^n a_ix^i$$
and then note that as each $a_iin E$, we have that
$$[F(a_0, a_1, a_2,ldots, a_n): F]le prod_i [F(a_i):F]<infty.$$
But then $[F(a_0, a_1,ldots, a_n, a):F] < infty$, so $a$ is algebraic over $F$.
answered Dec 12 '16 at 15:27
Adam HughesAdam Hughes
32.4k83770
$endgroup$
add a comment |
$begingroup$
We use the characterization that $alpha in k'/k$ is an algebraic element of $k'$ over $k$ iff $[k(alpha):k]<infty$. Then let $ain K$. By definition there is $p(x)in E[x]$ so that $p(a)=0$. Let
$$p(x) = sum_{i=1}^n a_ix^i$$
and then note that as each $a_iin E$, we have that
$$[F(a_0, a_1, a_2,ldots, a_n): F]le prod_i [F(a_i):F]<infty.$$
But then $[F(a_0, a_1,ldots, a_n, a):F] < infty$, so $a$ is algebraic over $F$.
answered Dec 12 '16 at 15:27
Adam HughesAdam Hughes
32.4k83770
$endgroup$
add a comment |
$begingroup$
We use the characterization that $alpha in k'/k$ is an algebraic element of $k'$ over $k$ iff $[k(alpha):k]<infty$. Then let $ain K$. By definition there is $p(x)in E[x]$ so that $p(a)=0$. Let
$$p(x) = sum_{i=1}^n a_ix^i$$
and then note that as each $a_iin E$, we have that
$$[F(a_0, a_1, a_2,ldots, a_n): F]le prod_i [F(a_i):F]<infty.$$
But then $[F(a_0, a_1,ldots, a_n, a):F] < infty$, so $a$ is algebraic over $F$.
answered Dec 12 '16 at 15:27
Adam HughesAdam Hughes
32.4k83770
$endgroup$
We use the characterization that $alpha in k'/k$ is an algebraic element of $k'$ over $k$ iff $[k(alpha):k]<infty$. Then let $ain K$. By definition there is $p(x)in E[x]$ so that $p(a)=0$. Let
$$p(x) = sum_{i=1}^n a_ix^i$$
and then note that as each $a_iin E$, we have that
$$[F(a_0, a_1, a_2,ldots, a_n): F]le prod_i [F(a_i):F]<infty.$$
But then $[F(a_0, a_1,ldots, a_n, a):F] < infty$, so $a$ is algebraic over $F$.
answered Dec 12 '16 at 15:27
Adam HughesAdam Hughes
32.4k83770
edited Oct 11 '17 at 11:40
answered Dec 12 '16 at 15:27
Adam HughesAdam Hughes
32.4k83770
answered Dec 12 '16 at 15:27
Adam HughesAdam Hughes
32.4k83770
answered Dec 12 '16 at 15:27
Adam HughesAdam Hughes
32.4k83770
32.4k83770
add a comment |
add a comment |
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