Splitting field over K of an infinite set of polynomial The 2019 Stack Overflow Developer Survey Results Are In$K$ is a splitting field $iff$ any irreducible polynomial with a root in $K$ splits completely over $K$.Splitting field over $K$ of a finite set of polynomialsFiniteness of a field that is a homomorphic image of a polynomial ringAlgebraic and Galois Extension is a Splitting Field of some set.Question on the proof of existence of splitting fields for a family of polynomialsIs every element in a finite splitting field K over F a root in a polynomial?Question regarding gcd in polynomial ring over a field$E/F$ is a normal extension iff $E$ is a splitting field for some polynomial $fin F[X]$.$mathbbC$ is not the splitting field of any polynomial over $mathbbQ$ (without cardinality)A splitting field over $BbbF_5$$K$ is a splitting field $iff$ any irreducible polynomial with a root in $K$ splits completely over $K$.
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Splitting field over K of an infinite set of polynomial
The 2019 Stack Overflow Developer Survey Results Are In$K$ is a splitting field $iff$ any irreducible polynomial with a root in $K$ splits completely over $K$.Splitting field over $K$ of a finite set of polynomialsFiniteness of a field that is a homomorphic image of a polynomial ringAlgebraic and Galois Extension is a Splitting Field of some set.Question on the proof of existence of splitting fields for a family of polynomialsIs every element in a finite splitting field K over F a root in a polynomial?Question regarding gcd in polynomial ring over a field$E/F$ is a normal extension iff $E$ is a splitting field for some polynomial $fin F[X]$.$mathbbC$ is not the splitting field of any polynomial over $mathbbQ$ (without cardinality)A splitting field over $BbbF_5$$K$ is a splitting field $iff$ any irreducible polynomial with a root in $K$ splits completely over $K$.
$begingroup$
Suppose $F$ is a finite splitting field over $K$ of $X=lbrace f_i(x)rbrace_iin I$, some infinite set. Is there necessarily a finite set $Ysubseteq X$ such that $F$ is a finite splitting field of $Y$?
I'm curious if there is a way to generalize: $F$ is a splitting field over $K$ of a finite set $lbrace f_1,…,f_nrbrace$ of polynomials in $K[x]$ if and only if $F$ is a splitting field over $K$ of the single polynomial $f=f_1f_2⋯f_n$.
The example I have in mind is $mathbbC$ over $mathbbR$. Clearly $mathbbC$ is the splitting field for $X=lbrace ax^2+bx+cmid a,b,cin mathbbR, b^2-4ac<0 rbrace$ and but also for just $x^2+1$.
abstract-algebra field-theory extension-field
asked Mar 24 at 3:55
DeneDene
61
$endgroup$
|
show 2 more comments
$begingroup$
Suppose $F$ is a finite splitting field over $K$ of $X=lbrace f_i(x)rbrace_iin I$, some infinite set. Is there necessarily a finite set $Ysubseteq X$ such that $F$ is a finite splitting field of $Y$?
I'm curious if there is a way to generalize: $F$ is a splitting field over $K$ of a finite set $lbrace f_1,…,f_nrbrace$ of polynomials in $K[x]$ if and only if $F$ is a splitting field over $K$ of the single polynomial $f=f_1f_2⋯f_n$.
The example I have in mind is $mathbbC$ over $mathbbR$. Clearly $mathbbC$ is the splitting field for $X=lbrace ax^2+bx+cmid a,b,cin mathbbR, b^2-4ac<0 rbrace$ and but also for just $x^2+1$.
abstract-algebra field-theory extension-field
asked Mar 24 at 3:55
DeneDene
61
$endgroup$
$begingroup$
No, in general it is not possible to find a finite subset. If you could, as you note, the finite subset could then be replaced by a single polynomial, and so the extension would necessarily be a finite extension. However, you can find infinite extensions, e.g., of $mathbbQ$: the extension given by the splitting field of the set of polynomials $f_n(x) = x^n-1$ for all $ngeq 1$ is not given as the splitting field of any finite set of polynomials, let alone a subset of $f_n$.
$endgroup$
– Arturo Magidin
Mar 24 at 6:00
$begingroup$
That was another example I was thinking about. Although $mathbbQ(sqrt2,sqrt3,...)$ is an infinite extension. If we assumed our restriction was finite then could we, for some reason, take some finite subset?
$endgroup$
– Dene
Mar 24 at 11:51
$begingroup$
If the extensions is finitely generated and algebraic, then it is finite, and then you can take a single polynomial.
$endgroup$
– Arturo Magidin
Mar 24 at 17:12
$begingroup$
Yes, this is actually what I'm trying to prove. This issue has arised because the question is worded in such a way that our only assumption is that the extension is finite and normal. Then I am to prove that it is the splitting field of a single polynomial. So I'm curious if the set of polynomials can always be finite.
$endgroup$
– Dene
Mar 24 at 19:55
$begingroup$
No, “a set of polynomials can always be finite” is not correct, as already noted. But a finite extension is necessarily finitely generated, and you only need one polynomial for each generator, and then you can take the product of all of these polynomials.
$endgroup$
– Arturo Magidin
Mar 24 at 21:19
|
show 2 more comments
$begingroup$
Suppose $F$ is a finite splitting field over $K$ of $X=lbrace f_i(x)rbrace_iin I$, some infinite set. Is there necessarily a finite set $Ysubseteq X$ such that $F$ is a finite splitting field of $Y$?
I'm curious if there is a way to generalize: $F$ is a splitting field over $K$ of a finite set $lbrace f_1,…,f_nrbrace$ of polynomials in $K[x]$ if and only if $F$ is a splitting field over $K$ of the single polynomial $f=f_1f_2⋯f_n$.
The example I have in mind is $mathbbC$ over $mathbbR$. Clearly $mathbbC$ is the splitting field for $X=lbrace ax^2+bx+cmid a,b,cin mathbbR, b^2-4ac<0 rbrace$ and but also for just $x^2+1$.
abstract-algebra field-theory extension-field
asked Mar 24 at 3:55
DeneDene
61
$endgroup$
Suppose $F$ is a finite splitting field over $K$ of $X=lbrace f_i(x)rbrace_iin I$, some infinite set. Is there necessarily a finite set $Ysubseteq X$ such that $F$ is a finite splitting field of $Y$?
I'm curious if there is a way to generalize: $F$ is a splitting field over $K$ of a finite set $lbrace f_1,…,f_nrbrace$ of polynomials in $K[x]$ if and only if $F$ is a splitting field over $K$ of the single polynomial $f=f_1f_2⋯f_n$.
The example I have in mind is $mathbbC$ over $mathbbR$. Clearly $mathbbC$ is the splitting field for $X=lbrace ax^2+bx+cmid a,b,cin mathbbR, b^2-4ac<0 rbrace$ and but also for just $x^2+1$.
abstract-algebra field-theory extension-field
abstract-algebra field-theory extension-field
asked Mar 24 at 3:55
DeneDene
61
asked Mar 24 at 3:55
DeneDene
61
asked Mar 24 at 3:55
DeneDene
61
asked Mar 24 at 3:55
DeneDene
61
asked Mar 24 at 3:55
DeneDene
61
61
$begingroup$
No, in general it is not possible to find a finite subset. If you could, as you note, the finite subset could then be replaced by a single polynomial, and so the extension would necessarily be a finite extension. However, you can find infinite extensions, e.g., of $mathbbQ$: the extension given by the splitting field of the set of polynomials $f_n(x) = x^n-1$ for all $ngeq 1$ is not given as the splitting field of any finite set of polynomials, let alone a subset of $f_n$.
$endgroup$
– Arturo Magidin
Mar 24 at 6:00
$begingroup$
That was another example I was thinking about. Although $mathbbQ(sqrt2,sqrt3,...)$ is an infinite extension. If we assumed our restriction was finite then could we, for some reason, take some finite subset?
$endgroup$
– Dene
Mar 24 at 11:51
$begingroup$
If the extensions is finitely generated and algebraic, then it is finite, and then you can take a single polynomial.
$endgroup$
– Arturo Magidin
Mar 24 at 17:12
$begingroup$
Yes, this is actually what I'm trying to prove. This issue has arised because the question is worded in such a way that our only assumption is that the extension is finite and normal. Then I am to prove that it is the splitting field of a single polynomial. So I'm curious if the set of polynomials can always be finite.
$endgroup$
– Dene
Mar 24 at 19:55
$begingroup$
No, “a set of polynomials can always be finite” is not correct, as already noted. But a finite extension is necessarily finitely generated, and you only need one polynomial for each generator, and then you can take the product of all of these polynomials.
$endgroup$
– Arturo Magidin
Mar 24 at 21:19
|
show 2 more comments
$begingroup$
No, in general it is not possible to find a finite subset. If you could, as you note, the finite subset could then be replaced by a single polynomial, and so the extension would necessarily be a finite extension. However, you can find infinite extensions, e.g., of $mathbbQ$: the extension given by the splitting field of the set of polynomials $f_n(x) = x^n-1$ for all $ngeq 1$ is not given as the splitting field of any finite set of polynomials, let alone a subset of $f_n$.
$endgroup$
– Arturo Magidin
Mar 24 at 6:00
$begingroup$
That was another example I was thinking about. Although $mathbbQ(sqrt2,sqrt3,...)$ is an infinite extension. If we assumed our restriction was finite then could we, for some reason, take some finite subset?
$endgroup$
– Dene
Mar 24 at 11:51
$begingroup$
If the extensions is finitely generated and algebraic, then it is finite, and then you can take a single polynomial.
$endgroup$
– Arturo Magidin
Mar 24 at 17:12
$begingroup$
Yes, this is actually what I'm trying to prove. This issue has arised because the question is worded in such a way that our only assumption is that the extension is finite and normal. Then I am to prove that it is the splitting field of a single polynomial. So I'm curious if the set of polynomials can always be finite.
$endgroup$
– Dene
Mar 24 at 19:55
$begingroup$
No, “a set of polynomials can always be finite” is not correct, as already noted. But a finite extension is necessarily finitely generated, and you only need one polynomial for each generator, and then you can take the product of all of these polynomials.
$endgroup$
– Arturo Magidin
Mar 24 at 21:19
$begingroup$
No, in general it is not possible to find a finite subset. If you could, as you note, the finite subset could then be replaced by a single polynomial, and so the extension would necessarily be a finite extension. However, you can find infinite extensions, e.g., of $mathbbQ$: the extension given by the splitting field of the set of polynomials $f_n(x) = x^n-1$ for all $ngeq 1$ is not given as the splitting field of any finite set of polynomials, let alone a subset of $f_n$.
$endgroup$
– Arturo Magidin
Mar 24 at 6:00
$begingroup$
No, in general it is not possible to find a finite subset. If you could, as you note, the finite subset could then be replaced by a single polynomial, and so the extension would necessarily be a finite extension. However, you can find infinite extensions, e.g., of $mathbbQ$: the extension given by the splitting field of the set of polynomials $f_n(x) = x^n-1$ for all $ngeq 1$ is not given as the splitting field of any finite set of polynomials, let alone a subset of $f_n$.
$endgroup$
– Arturo Magidin
Mar 24 at 6:00
$begingroup$
That was another example I was thinking about. Although $mathbbQ(sqrt2,sqrt3,...)$ is an infinite extension. If we assumed our restriction was finite then could we, for some reason, take some finite subset?
$endgroup$
– Dene
Mar 24 at 11:51
$begingroup$
That was another example I was thinking about. Although $mathbbQ(sqrt2,sqrt3,...)$ is an infinite extension. If we assumed our restriction was finite then could we, for some reason, take some finite subset?
$endgroup$
– Dene
Mar 24 at 11:51
$begingroup$
If the extensions is finitely generated and algebraic, then it is finite, and then you can take a single polynomial.
$endgroup$
– Arturo Magidin
Mar 24 at 17:12
$begingroup$
If the extensions is finitely generated and algebraic, then it is finite, and then you can take a single polynomial.
$endgroup$
– Arturo Magidin
Mar 24 at 17:12
$begingroup$
Yes, this is actually what I'm trying to prove. This issue has arised because the question is worded in such a way that our only assumption is that the extension is finite and normal. Then I am to prove that it is the splitting field of a single polynomial. So I'm curious if the set of polynomials can always be finite.
$endgroup$
– Dene
Mar 24 at 19:55
$begingroup$
Yes, this is actually what I'm trying to prove. This issue has arised because the question is worded in such a way that our only assumption is that the extension is finite and normal. Then I am to prove that it is the splitting field of a single polynomial. So I'm curious if the set of polynomials can always be finite.
$endgroup$
– Dene
Mar 24 at 19:55
$begingroup$
No, “a set of polynomials can always be finite” is not correct, as already noted. But a finite extension is necessarily finitely generated, and you only need one polynomial for each generator, and then you can take the product of all of these polynomials.
$endgroup$
– Arturo Magidin
Mar 24 at 21:19
$begingroup$
No, “a set of polynomials can always be finite” is not correct, as already noted. But a finite extension is necessarily finitely generated, and you only need one polynomial for each generator, and then you can take the product of all of these polynomials.
$endgroup$
– Arturo Magidin
Mar 24 at 21:19
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
It is a classic result in Galois theory that a finite extension $F/K$ is the splitting field of a single polynomial (i.e. generated by all the roots of a single polynomial) if and only if $F/K$ is normal, i.e. every polynomial in $K$ with a root in $F$ splits completely. See this post, for instance.
answered Mar 24 at 5:07
Joshua MundingerJoshua Mundinger
2,9321028
$endgroup$
$begingroup$
Actually, this is the question I was being asked. But was worded in such a way that we don't necessarily have a finite set of polynomials. When our set was finite I had already produced a proof. If the extension is finite and the set of polynomials is infinite then as of right now my proof doesn't apply.
$endgroup$
– Dene
Mar 24 at 11:59
add a comment |
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$begingroup$
It is a classic result in Galois theory that a finite extension $F/K$ is the splitting field of a single polynomial (i.e. generated by all the roots of a single polynomial) if and only if $F/K$ is normal, i.e. every polynomial in $K$ with a root in $F$ splits completely. See this post, for instance.
answered Mar 24 at 5:07
Joshua MundingerJoshua Mundinger
2,9321028
$endgroup$
$begingroup$
Actually, this is the question I was being asked. But was worded in such a way that we don't necessarily have a finite set of polynomials. When our set was finite I had already produced a proof. If the extension is finite and the set of polynomials is infinite then as of right now my proof doesn't apply.
$endgroup$
– Dene
Mar 24 at 11:59
add a comment |
$begingroup$
It is a classic result in Galois theory that a finite extension $F/K$ is the splitting field of a single polynomial (i.e. generated by all the roots of a single polynomial) if and only if $F/K$ is normal, i.e. every polynomial in $K$ with a root in $F$ splits completely. See this post, for instance.
answered Mar 24 at 5:07
Joshua MundingerJoshua Mundinger
2,9321028
$endgroup$
$begingroup$
Actually, this is the question I was being asked. But was worded in such a way that we don't necessarily have a finite set of polynomials. When our set was finite I had already produced a proof. If the extension is finite and the set of polynomials is infinite then as of right now my proof doesn't apply.
$endgroup$
– Dene
Mar 24 at 11:59
add a comment |
$begingroup$
It is a classic result in Galois theory that a finite extension $F/K$ is the splitting field of a single polynomial (i.e. generated by all the roots of a single polynomial) if and only if $F/K$ is normal, i.e. every polynomial in $K$ with a root in $F$ splits completely. See this post, for instance.
answered Mar 24 at 5:07
Joshua MundingerJoshua Mundinger
2,9321028
$endgroup$
It is a classic result in Galois theory that a finite extension $F/K$ is the splitting field of a single polynomial (i.e. generated by all the roots of a single polynomial) if and only if $F/K$ is normal, i.e. every polynomial in $K$ with a root in $F$ splits completely. See this post, for instance.
answered Mar 24 at 5:07
Joshua MundingerJoshua Mundinger
2,9321028
answered Mar 24 at 5:07
Joshua MundingerJoshua Mundinger
2,9321028
answered Mar 24 at 5:07
Joshua MundingerJoshua Mundinger
2,9321028
answered Mar 24 at 5:07
Joshua MundingerJoshua Mundinger
2,9321028
2,9321028
$begingroup$
Actually, this is the question I was being asked. But was worded in such a way that we don't necessarily have a finite set of polynomials. When our set was finite I had already produced a proof. If the extension is finite and the set of polynomials is infinite then as of right now my proof doesn't apply.
$endgroup$
– Dene
Mar 24 at 11:59
add a comment |
$begingroup$
Actually, this is the question I was being asked. But was worded in such a way that we don't necessarily have a finite set of polynomials. When our set was finite I had already produced a proof. If the extension is finite and the set of polynomials is infinite then as of right now my proof doesn't apply.
$endgroup$
– Dene
Mar 24 at 11:59
$begingroup$
Actually, this is the question I was being asked. But was worded in such a way that we don't necessarily have a finite set of polynomials. When our set was finite I had already produced a proof. If the extension is finite and the set of polynomials is infinite then as of right now my proof doesn't apply.
$endgroup$
– Dene
Mar 24 at 11:59
$begingroup$
Actually, this is the question I was being asked. But was worded in such a way that we don't necessarily have a finite set of polynomials. When our set was finite I had already produced a proof. If the extension is finite and the set of polynomials is infinite then as of right now my proof doesn't apply.
$endgroup$
– Dene
Mar 24 at 11:59
add a comment |
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$begingroup$
No, in general it is not possible to find a finite subset. If you could, as you note, the finite subset could then be replaced by a single polynomial, and so the extension would necessarily be a finite extension. However, you can find infinite extensions, e.g., of $mathbbQ$: the extension given by the splitting field of the set of polynomials $f_n(x) = x^n-1$ for all $ngeq 1$ is not given as the splitting field of any finite set of polynomials, let alone a subset of $f_n$.
$endgroup$
– Arturo Magidin
Mar 24 at 6:00
$begingroup$
That was another example I was thinking about. Although $mathbbQ(sqrt2,sqrt3,...)$ is an infinite extension. If we assumed our restriction was finite then could we, for some reason, take some finite subset?
$endgroup$
– Dene
Mar 24 at 11:51
$begingroup$
If the extensions is finitely generated and algebraic, then it is finite, and then you can take a single polynomial.
$endgroup$
– Arturo Magidin
Mar 24 at 17:12
$begingroup$
Yes, this is actually what I'm trying to prove. This issue has arised because the question is worded in such a way that our only assumption is that the extension is finite and normal. Then I am to prove that it is the splitting field of a single polynomial. So I'm curious if the set of polynomials can always be finite.
$endgroup$
– Dene
Mar 24 at 19:55
$begingroup$
No, “a set of polynomials can always be finite” is not correct, as already noted. But a finite extension is necessarily finitely generated, and you only need one polynomial for each generator, and then you can take the product of all of these polynomials.
$endgroup$
– Arturo Magidin
Mar 24 at 21:19