Algebraically closed concept
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Suppose $L$ is an extension field of $K$ and $L$ is algebraically closed. Then is it trivially true that $forall f(x)in K[x],~f(x)$ splits over $L$? I think the answer is yes since $L$ is algebraically closed means that every polynomial with coefficients in $L$ have a root in $L$ (and hence have all roots in $L$)
abstract-algebra
asked Dec 15 '18 at 12:32
EricEric
1,805615
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add a comment |
$begingroup$
Suppose $L$ is an extension field of $K$ and $L$ is algebraically closed. Then is it trivially true that $forall f(x)in K[x],~f(x)$ splits over $L$? I think the answer is yes since $L$ is algebraically closed means that every polynomial with coefficients in $L$ have a root in $L$ (and hence have all roots in $L$)
abstract-algebra
asked Dec 15 '18 at 12:32
EricEric
1,805615
$endgroup$
1
$begingroup$
Yes, this follows by induction on the degree of $f$, starting with one root. So there is nothing to ask in your question.
$endgroup$
– Dietrich Burde
Dec 15 '18 at 12:37
add a comment |
$begingroup$
Suppose $L$ is an extension field of $K$ and $L$ is algebraically closed. Then is it trivially true that $forall f(x)in K[x],~f(x)$ splits over $L$? I think the answer is yes since $L$ is algebraically closed means that every polynomial with coefficients in $L$ have a root in $L$ (and hence have all roots in $L$)
abstract-algebra
asked Dec 15 '18 at 12:32
EricEric
1,805615
$endgroup$
Suppose $L$ is an extension field of $K$ and $L$ is algebraically closed. Then is it trivially true that $forall f(x)in K[x],~f(x)$ splits over $L$? I think the answer is yes since $L$ is algebraically closed means that every polynomial with coefficients in $L$ have a root in $L$ (and hence have all roots in $L$)
abstract-algebra
abstract-algebra
asked Dec 15 '18 at 12:32
EricEric
1,805615
asked Dec 15 '18 at 12:32
EricEric
1,805615
asked Dec 15 '18 at 12:32
EricEric
1,805615
asked Dec 15 '18 at 12:32
EricEric
1,805615
asked Dec 15 '18 at 12:32
EricEric
1,805615
1,805615
1
$begingroup$
Yes, this follows by induction on the degree of $f$, starting with one root. So there is nothing to ask in your question.
$endgroup$
– Dietrich Burde
Dec 15 '18 at 12:37
add a comment |
1
$begingroup$
Yes, this follows by induction on the degree of $f$, starting with one root. So there is nothing to ask in your question.
$endgroup$
– Dietrich Burde
Dec 15 '18 at 12:37
1
1
$begingroup$
Yes, this follows by induction on the degree of $f$, starting with one root. So there is nothing to ask in your question.
$endgroup$
– Dietrich Burde
Dec 15 '18 at 12:37
$begingroup$
Yes, this follows by induction on the degree of $f$, starting with one root. So there is nothing to ask in your question.
$endgroup$
– Dietrich Burde
Dec 15 '18 at 12:37
add a comment |
1 Answer
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Yes, the splitting field of $f(x)in K[x]$ must be contained in $L$ since $L$ is algebraically closed.
answered Dec 15 '18 at 12:35
positrón0802positrón0802
4,313520
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add a comment |
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$begingroup$
Yes, the splitting field of $f(x)in K[x]$ must be contained in $L$ since $L$ is algebraically closed.
answered Dec 15 '18 at 12:35
positrón0802positrón0802
4,313520
$endgroup$
add a comment |
$begingroup$
Yes, the splitting field of $f(x)in K[x]$ must be contained in $L$ since $L$ is algebraically closed.
answered Dec 15 '18 at 12:35
positrón0802positrón0802
4,313520
$endgroup$
add a comment |
$begingroup$
Yes, the splitting field of $f(x)in K[x]$ must be contained in $L$ since $L$ is algebraically closed.
answered Dec 15 '18 at 12:35
positrón0802positrón0802
4,313520
$endgroup$
Yes, the splitting field of $f(x)in K[x]$ must be contained in $L$ since $L$ is algebraically closed.
answered Dec 15 '18 at 12:35
positrón0802positrón0802
4,313520
answered Dec 15 '18 at 12:35
positrón0802positrón0802
4,313520
answered Dec 15 '18 at 12:35
positrón0802positrón0802
4,313520
answered Dec 15 '18 at 12:35
positrón0802positrón0802
4,313520
4,313520
add a comment |
add a comment |
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Yes, this follows by induction on the degree of $f$, starting with one root. So there is nothing to ask in your question.
$endgroup$
– Dietrich Burde
Dec 15 '18 at 12:37