Mixing
Are these extensions separable.
$begingroup$
Let $K[x]$ be a polynomial ring over the field $K$ and $F$ a subring of $K$. If $F$ is a perfect field and $f(x) in F[x]$ has no repeated irreducible factors in $F[x]$, prove that $f(x)$ has no repeated irreducible factors in $K[x]$.
My proof:
Let $f(x) in F[x]$. Then as $F$ is a perfect field then we have that $$f(x) = g(x)^p$$
for some $g(x) in F[x]$. But $f$ cannot have any repeated irreducible factors so that if $p geq 2$ then we would have a repeated irreducible factor. Therefore we conclude that the field must have characteristic 0. Thus $(f,f') =1$ since all of its irreducible factors must be separable in $F[x]$ and so in $K[x]$ as well. But this implies that $f$ is separable over $K[x]$ and thus cannot have any repeated irreducible factor.
Is my argument correct? If so, are there any ways I could improve this proof. Thanks in advance!
abstract-algebra proof-verification
edited Jan 23 at 21:53
Bernard
123k741116
asked Jan 23 at 21:27
Justin StevensonJustin Stevenson
957519
$endgroup$
add a comment |
$begingroup$
Let $K[x]$ be a polynomial ring over the field $K$ and $F$ a subring of $K$. If $F$ is a perfect field and $f(x) in F[x]$ has no repeated irreducible factors in $F[x]$, prove that $f(x)$ has no repeated irreducible factors in $K[x]$.
My proof:
Let $f(x) in F[x]$. Then as $F$ is a perfect field then we have that $$f(x) = g(x)^p$$
for some $g(x) in F[x]$. But $f$ cannot have any repeated irreducible factors so that if $p geq 2$ then we would have a repeated irreducible factor. Therefore we conclude that the field must have characteristic 0. Thus $(f,f') =1$ since all of its irreducible factors must be separable in $F[x]$ and so in $K[x]$ as well. But this implies that $f$ is separable over $K[x]$ and thus cannot have any repeated irreducible factor.
Is my argument correct? If so, are there any ways I could improve this proof. Thanks in advance!
abstract-algebra proof-verification
edited Jan 23 at 21:53
Bernard
123k741116
asked Jan 23 at 21:27
Justin StevensonJustin Stevenson
957519
$endgroup$
$begingroup$
This has an answer at this site; start searching from this question with related links.
$endgroup$
– Dietrich Burde
Jan 23 at 21:38
add a comment |
$begingroup$
Let $K[x]$ be a polynomial ring over the field $K$ and $F$ a subring of $K$. If $F$ is a perfect field and $f(x) in F[x]$ has no repeated irreducible factors in $F[x]$, prove that $f(x)$ has no repeated irreducible factors in $K[x]$.
My proof:
Let $f(x) in F[x]$. Then as $F$ is a perfect field then we have that $$f(x) = g(x)^p$$
for some $g(x) in F[x]$. But $f$ cannot have any repeated irreducible factors so that if $p geq 2$ then we would have a repeated irreducible factor. Therefore we conclude that the field must have characteristic 0. Thus $(f,f') =1$ since all of its irreducible factors must be separable in $F[x]$ and so in $K[x]$ as well. But this implies that $f$ is separable over $K[x]$ and thus cannot have any repeated irreducible factor.
Is my argument correct? If so, are there any ways I could improve this proof. Thanks in advance!
abstract-algebra proof-verification
edited Jan 23 at 21:53
Bernard
123k741116
asked Jan 23 at 21:27
Justin StevensonJustin Stevenson
957519
$endgroup$
Let $K[x]$ be a polynomial ring over the field $K$ and $F$ a subring of $K$. If $F$ is a perfect field and $f(x) in F[x]$ has no repeated irreducible factors in $F[x]$, prove that $f(x)$ has no repeated irreducible factors in $K[x]$.
My proof:
Let $f(x) in F[x]$. Then as $F$ is a perfect field then we have that $$f(x) = g(x)^p$$
for some $g(x) in F[x]$. But $f$ cannot have any repeated irreducible factors so that if $p geq 2$ then we would have a repeated irreducible factor. Therefore we conclude that the field must have characteristic 0. Thus $(f,f') =1$ since all of its irreducible factors must be separable in $F[x]$ and so in $K[x]$ as well. But this implies that $f$ is separable over $K[x]$ and thus cannot have any repeated irreducible factor.
Is my argument correct? If so, are there any ways I could improve this proof. Thanks in advance!
abstract-algebra proof-verification
abstract-algebra proof-verification
edited Jan 23 at 21:53
Bernard
123k741116
asked Jan 23 at 21:27
Justin StevensonJustin Stevenson
957519
edited Jan 23 at 21:53
Bernard
123k741116
asked Jan 23 at 21:27
Justin StevensonJustin Stevenson
957519
edited Jan 23 at 21:53
Bernard
123k741116
edited Jan 23 at 21:53
Bernard
123k741116
edited Jan 23 at 21:53
Bernard
123k741116
123k741116
asked Jan 23 at 21:27
Justin StevensonJustin Stevenson
957519
asked Jan 23 at 21:27
Justin StevensonJustin Stevenson
957519
asked Jan 23 at 21:27
Justin StevensonJustin Stevenson
957519
957519
$begingroup$
This has an answer at this site; start searching from this question with related links.
$endgroup$
– Dietrich Burde
Jan 23 at 21:38
add a comment |
$begingroup$
This has an answer at this site; start searching from this question with related links.
$endgroup$
– Dietrich Burde
Jan 23 at 21:38
$begingroup$
This has an answer at this site; start searching from this question with related links.
$endgroup$
– Dietrich Burde
Jan 23 at 21:38
$begingroup$
This has an answer at this site; start searching from this question with related links.
$endgroup$
– Dietrich Burde
Jan 23 at 21:38
add a comment |
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$begingroup$
This has an answer at this site; start searching from this question with related links.
$endgroup$
– Dietrich Burde
Jan 23 at 21:38