Existence of ring homomorphism and embedding
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I don't know how to prove the following:
Let $Ksubset E$ be a field extension.
i) Let $x$ be an element of $E$. Show that $x$ is transcendent over $K$ if and only if the map $ev_{x}: K[X] to E$ is one to one.
An element $xin E$ is called transcendent over $K$ if for every polynomial $f(X)in K[X]$, with $f(X)neq 0$, $f(x)neq 0$ holds.
ii) Let $K(X)$ be the field of fraction of the polynomial ring $K[X]$ and $phi : K[X] to E$ an one to one ring homomorphism with $phi(a)=a$ for all $ain K$. Show that there exists a ring homomorphism $overline{phi}: K(X) to E$ with $overline{phi}(f(X))=phi(f(X))$ for all $f(X) in K[X]$.
$K(X) := {f(X)/g(X) | f(X), g(X) in K[X], g(X)neq 0 }/ sim$ with $f(X)/g(X) sim a(X)/b(X)$ if and only if $b(X)f(X) = a(X) g(X)$.
iii) Show that an embedding $tau: mathbb{Q}(X) to mathbb{R}$ over $mathbb{Q}$ exists.
I tried to find similar questions in the forum but couldn't find a helpful post. Any help would be appreciated, thanks in advance.
abstract-algebra extension-field
asked Dec 8 '18 at 20:38
ManwellManwell
114
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add a comment |
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I don't know how to prove the following:
Let $Ksubset E$ be a field extension.
i) Let $x$ be an element of $E$. Show that $x$ is transcendent over $K$ if and only if the map $ev_{x}: K[X] to E$ is one to one.
An element $xin E$ is called transcendent over $K$ if for every polynomial $f(X)in K[X]$, with $f(X)neq 0$, $f(x)neq 0$ holds.
ii) Let $K(X)$ be the field of fraction of the polynomial ring $K[X]$ and $phi : K[X] to E$ an one to one ring homomorphism with $phi(a)=a$ for all $ain K$. Show that there exists a ring homomorphism $overline{phi}: K(X) to E$ with $overline{phi}(f(X))=phi(f(X))$ for all $f(X) in K[X]$.
$K(X) := {f(X)/g(X) | f(X), g(X) in K[X], g(X)neq 0 }/ sim$ with $f(X)/g(X) sim a(X)/b(X)$ if and only if $b(X)f(X) = a(X) g(X)$.
iii) Show that an embedding $tau: mathbb{Q}(X) to mathbb{R}$ over $mathbb{Q}$ exists.
I tried to find similar questions in the forum but couldn't find a helpful post. Any help would be appreciated, thanks in advance.
abstract-algebra extension-field
asked Dec 8 '18 at 20:38
ManwellManwell
114
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$begingroup$
Similar posts are this one, this one, ...
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– Dietrich Burde
Dec 8 '18 at 20:44
add a comment |
$begingroup$
I don't know how to prove the following:
Let $Ksubset E$ be a field extension.
i) Let $x$ be an element of $E$. Show that $x$ is transcendent over $K$ if and only if the map $ev_{x}: K[X] to E$ is one to one.
An element $xin E$ is called transcendent over $K$ if for every polynomial $f(X)in K[X]$, with $f(X)neq 0$, $f(x)neq 0$ holds.
ii) Let $K(X)$ be the field of fraction of the polynomial ring $K[X]$ and $phi : K[X] to E$ an one to one ring homomorphism with $phi(a)=a$ for all $ain K$. Show that there exists a ring homomorphism $overline{phi}: K(X) to E$ with $overline{phi}(f(X))=phi(f(X))$ for all $f(X) in K[X]$.
$K(X) := {f(X)/g(X) | f(X), g(X) in K[X], g(X)neq 0 }/ sim$ with $f(X)/g(X) sim a(X)/b(X)$ if and only if $b(X)f(X) = a(X) g(X)$.
iii) Show that an embedding $tau: mathbb{Q}(X) to mathbb{R}$ over $mathbb{Q}$ exists.
I tried to find similar questions in the forum but couldn't find a helpful post. Any help would be appreciated, thanks in advance.
abstract-algebra extension-field
asked Dec 8 '18 at 20:38
ManwellManwell
114
$endgroup$
I don't know how to prove the following:
Let $Ksubset E$ be a field extension.
i) Let $x$ be an element of $E$. Show that $x$ is transcendent over $K$ if and only if the map $ev_{x}: K[X] to E$ is one to one.
An element $xin E$ is called transcendent over $K$ if for every polynomial $f(X)in K[X]$, with $f(X)neq 0$, $f(x)neq 0$ holds.
ii) Let $K(X)$ be the field of fraction of the polynomial ring $K[X]$ and $phi : K[X] to E$ an one to one ring homomorphism with $phi(a)=a$ for all $ain K$. Show that there exists a ring homomorphism $overline{phi}: K(X) to E$ with $overline{phi}(f(X))=phi(f(X))$ for all $f(X) in K[X]$.
$K(X) := {f(X)/g(X) | f(X), g(X) in K[X], g(X)neq 0 }/ sim$ with $f(X)/g(X) sim a(X)/b(X)$ if and only if $b(X)f(X) = a(X) g(X)$.
iii) Show that an embedding $tau: mathbb{Q}(X) to mathbb{R}$ over $mathbb{Q}$ exists.
I tried to find similar questions in the forum but couldn't find a helpful post. Any help would be appreciated, thanks in advance.
abstract-algebra extension-field
abstract-algebra extension-field
asked Dec 8 '18 at 20:38
ManwellManwell
114
asked Dec 8 '18 at 20:38
ManwellManwell
114
asked Dec 8 '18 at 20:38
ManwellManwell
114
asked Dec 8 '18 at 20:38
ManwellManwell
114
asked Dec 8 '18 at 20:38
ManwellManwell
114
114
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Similar posts are this one, this one, ...
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– Dietrich Burde
Dec 8 '18 at 20:44
add a comment |
$begingroup$
Similar posts are this one, this one, ...
$endgroup$
– Dietrich Burde
Dec 8 '18 at 20:44
$begingroup$
Similar posts are this one, this one, ...
$endgroup$
– Dietrich Burde
Dec 8 '18 at 20:44
$begingroup$
Similar posts are this one, this one, ...
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– Dietrich Burde
Dec 8 '18 at 20:44
add a comment |
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– Dietrich Burde
Dec 8 '18 at 20:44