Show that there exists fields properly containing $mathbb{C}$. Does that field have the property that every...












2












$begingroup$


This question comes after showing that if $mathrm R$ is a domain, then $mathrm R[t]$ is also a domain. But I don't quite see the connections here. Since polynomials of complex coefficients isn't a field since there aren't inverses to every element.






abstract-algebra







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$endgroup$












  • $begingroup$
    The rational functions over $Bbb C$ form a field properly containing $Bbb C$.
    $endgroup$
    – TonyK
    Jan 25 at 2:06
















2












$begingroup$


This question comes after showing that if $mathrm R$ is a domain, then $mathrm R[t]$ is also a domain. But I don't quite see the connections here. Since polynomials of complex coefficients isn't a field since there aren't inverses to every element.






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    The rational functions over $Bbb C$ form a field properly containing $Bbb C$.
    $endgroup$
    – TonyK
    Jan 25 at 2:06














2












2








2





$begingroup$


This question comes after showing that if $mathrm R$ is a domain, then $mathrm R[t]$ is also a domain. But I don't quite see the connections here. Since polynomials of complex coefficients isn't a field since there aren't inverses to every element.






abstract-algebra







share|cite|improve this question











$endgroup$




This question comes after showing that if $mathrm R$ is a domain, then $mathrm R[t]$ is also a domain. But I don't quite see the connections here. Since polynomials of complex coefficients isn't a field since there aren't inverses to every element.






abstract-algebra




abstract-algebra






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share|cite|improve this question













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share|cite|improve this question








edited Jan 25 at 2:34









Thomas Shelby

4,1842726




4,1842726










asked Jan 25 at 1:44









davidhdavidh

3138




3138












  • $begingroup$
    The rational functions over $Bbb C$ form a field properly containing $Bbb C$.
    $endgroup$
    – TonyK
    Jan 25 at 2:06


















  • $begingroup$
    The rational functions over $Bbb C$ form a field properly containing $Bbb C$.
    $endgroup$
    – TonyK
    Jan 25 at 2:06
















$begingroup$
The rational functions over $Bbb C$ form a field properly containing $Bbb C$.
$endgroup$
– TonyK
Jan 25 at 2:06




$begingroup$
The rational functions over $Bbb C$ form a field properly containing $Bbb C$.
$endgroup$
– TonyK
Jan 25 at 2:06










1 Answer
1






active

oldest

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2












$begingroup$

Consider $Bbb{C}(x)$, where $Bbb{C}(x)$ is the field of fractions of the polynomial ring $Bbb{C}[x]$. Clearly $Bbb{C}subset Bbb{C} (x) $.




A field $F$ is algebraically closed if any polynomial of non-zero
degree over $F$ has at least one root in $F.$




$Bbb{C}(x)$ is not algebraically closed. To see this, let $F=Bbb{C}(x)$ and $F[y]$ be the polynomial ring over $F$.



Consider $$f(y)=y^2-frac1xin F[y].$$ If there exists a root in $F$, it must be of the form $frac{p(x)}{q(x)}$,$p (x),q(x)in Bbb{C}[x]$. Then $${left(frac{p(x)}{q(x)}right)}^2=frac1xiff x{p(x)}^2={q(x)}^2.$$
Can you see the contradiction?



Hence $Bbb{C}(x)$ is not algebraically closed.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think I get the first part now. But I think the question asks for whether polynomials over this field has root, not in this field. Do you mean $mathbb{C}(x)$ is also algebraically closed?
    $endgroup$
    – davidh
    Jan 25 at 2:38










  • $begingroup$
    I've edited my answer.
    $endgroup$
    – Thomas Shelby
    Jan 25 at 3:07






  • 1




    $begingroup$
    Is the contradiction that $q(x)$ must be $sqrt{x}p(x)$. and $sqrt{x}$ is not in $mathbb{C}(x)$
    $endgroup$
    – davidh
    Jan 25 at 6:27










  • $begingroup$
    Yes. Also note that degree of $x {p (x)}^2$ is odd whereas degree of ${q (x)}^2$ is even.
    $endgroup$
    – Thomas Shelby
    Jan 25 at 7:35











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









2












$begingroup$

Consider $Bbb{C}(x)$, where $Bbb{C}(x)$ is the field of fractions of the polynomial ring $Bbb{C}[x]$. Clearly $Bbb{C}subset Bbb{C} (x) $.




A field $F$ is algebraically closed if any polynomial of non-zero
degree over $F$ has at least one root in $F.$




$Bbb{C}(x)$ is not algebraically closed. To see this, let $F=Bbb{C}(x)$ and $F[y]$ be the polynomial ring over $F$.



Consider $$f(y)=y^2-frac1xin F[y].$$ If there exists a root in $F$, it must be of the form $frac{p(x)}{q(x)}$,$p (x),q(x)in Bbb{C}[x]$. Then $${left(frac{p(x)}{q(x)}right)}^2=frac1xiff x{p(x)}^2={q(x)}^2.$$
Can you see the contradiction?



Hence $Bbb{C}(x)$ is not algebraically closed.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think I get the first part now. But I think the question asks for whether polynomials over this field has root, not in this field. Do you mean $mathbb{C}(x)$ is also algebraically closed?
    $endgroup$
    – davidh
    Jan 25 at 2:38










  • $begingroup$
    I've edited my answer.
    $endgroup$
    – Thomas Shelby
    Jan 25 at 3:07






  • 1




    $begingroup$
    Is the contradiction that $q(x)$ must be $sqrt{x}p(x)$. and $sqrt{x}$ is not in $mathbb{C}(x)$
    $endgroup$
    – davidh
    Jan 25 at 6:27










  • $begingroup$
    Yes. Also note that degree of $x {p (x)}^2$ is odd whereas degree of ${q (x)}^2$ is even.
    $endgroup$
    – Thomas Shelby
    Jan 25 at 7:35
















2












$begingroup$

Consider $Bbb{C}(x)$, where $Bbb{C}(x)$ is the field of fractions of the polynomial ring $Bbb{C}[x]$. Clearly $Bbb{C}subset Bbb{C} (x) $.




A field $F$ is algebraically closed if any polynomial of non-zero
degree over $F$ has at least one root in $F.$




$Bbb{C}(x)$ is not algebraically closed. To see this, let $F=Bbb{C}(x)$ and $F[y]$ be the polynomial ring over $F$.



Consider $$f(y)=y^2-frac1xin F[y].$$ If there exists a root in $F$, it must be of the form $frac{p(x)}{q(x)}$,$p (x),q(x)in Bbb{C}[x]$. Then $${left(frac{p(x)}{q(x)}right)}^2=frac1xiff x{p(x)}^2={q(x)}^2.$$
Can you see the contradiction?



Hence $Bbb{C}(x)$ is not algebraically closed.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think I get the first part now. But I think the question asks for whether polynomials over this field has root, not in this field. Do you mean $mathbb{C}(x)$ is also algebraically closed?
    $endgroup$
    – davidh
    Jan 25 at 2:38










  • $begingroup$
    I've edited my answer.
    $endgroup$
    – Thomas Shelby
    Jan 25 at 3:07






  • 1




    $begingroup$
    Is the contradiction that $q(x)$ must be $sqrt{x}p(x)$. and $sqrt{x}$ is not in $mathbb{C}(x)$
    $endgroup$
    – davidh
    Jan 25 at 6:27










  • $begingroup$
    Yes. Also note that degree of $x {p (x)}^2$ is odd whereas degree of ${q (x)}^2$ is even.
    $endgroup$
    – Thomas Shelby
    Jan 25 at 7:35














2












2








2





$begingroup$

Consider $Bbb{C}(x)$, where $Bbb{C}(x)$ is the field of fractions of the polynomial ring $Bbb{C}[x]$. Clearly $Bbb{C}subset Bbb{C} (x) $.




A field $F$ is algebraically closed if any polynomial of non-zero
degree over $F$ has at least one root in $F.$




$Bbb{C}(x)$ is not algebraically closed. To see this, let $F=Bbb{C}(x)$ and $F[y]$ be the polynomial ring over $F$.



Consider $$f(y)=y^2-frac1xin F[y].$$ If there exists a root in $F$, it must be of the form $frac{p(x)}{q(x)}$,$p (x),q(x)in Bbb{C}[x]$. Then $${left(frac{p(x)}{q(x)}right)}^2=frac1xiff x{p(x)}^2={q(x)}^2.$$
Can you see the contradiction?



Hence $Bbb{C}(x)$ is not algebraically closed.






share|cite|improve this answer











$endgroup$



Consider $Bbb{C}(x)$, where $Bbb{C}(x)$ is the field of fractions of the polynomial ring $Bbb{C}[x]$. Clearly $Bbb{C}subset Bbb{C} (x) $.




A field $F$ is algebraically closed if any polynomial of non-zero
degree over $F$ has at least one root in $F.$




$Bbb{C}(x)$ is not algebraically closed. To see this, let $F=Bbb{C}(x)$ and $F[y]$ be the polynomial ring over $F$.



Consider $$f(y)=y^2-frac1xin F[y].$$ If there exists a root in $F$, it must be of the form $frac{p(x)}{q(x)}$,$p (x),q(x)in Bbb{C}[x]$. Then $${left(frac{p(x)}{q(x)}right)}^2=frac1xiff x{p(x)}^2={q(x)}^2.$$
Can you see the contradiction?



Hence $Bbb{C}(x)$ is not algebraically closed.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 25 at 6:08

























answered Jan 25 at 2:09









Thomas ShelbyThomas Shelby

4,1842726




4,1842726








  • 1




    $begingroup$
    I think I get the first part now. But I think the question asks for whether polynomials over this field has root, not in this field. Do you mean $mathbb{C}(x)$ is also algebraically closed?
    $endgroup$
    – davidh
    Jan 25 at 2:38










  • $begingroup$
    I've edited my answer.
    $endgroup$
    – Thomas Shelby
    Jan 25 at 3:07






  • 1




    $begingroup$
    Is the contradiction that $q(x)$ must be $sqrt{x}p(x)$. and $sqrt{x}$ is not in $mathbb{C}(x)$
    $endgroup$
    – davidh
    Jan 25 at 6:27










  • $begingroup$
    Yes. Also note that degree of $x {p (x)}^2$ is odd whereas degree of ${q (x)}^2$ is even.
    $endgroup$
    – Thomas Shelby
    Jan 25 at 7:35














  • 1




    $begingroup$
    I think I get the first part now. But I think the question asks for whether polynomials over this field has root, not in this field. Do you mean $mathbb{C}(x)$ is also algebraically closed?
    $endgroup$
    – davidh
    Jan 25 at 2:38










  • $begingroup$
    I've edited my answer.
    $endgroup$
    – Thomas Shelby
    Jan 25 at 3:07






  • 1




    $begingroup$
    Is the contradiction that $q(x)$ must be $sqrt{x}p(x)$. and $sqrt{x}$ is not in $mathbb{C}(x)$
    $endgroup$
    – davidh
    Jan 25 at 6:27










  • $begingroup$
    Yes. Also note that degree of $x {p (x)}^2$ is odd whereas degree of ${q (x)}^2$ is even.
    $endgroup$
    – Thomas Shelby
    Jan 25 at 7:35








1




1




$begingroup$
I think I get the first part now. But I think the question asks for whether polynomials over this field has root, not in this field. Do you mean $mathbb{C}(x)$ is also algebraically closed?
$endgroup$
– davidh
Jan 25 at 2:38




$begingroup$
I think I get the first part now. But I think the question asks for whether polynomials over this field has root, not in this field. Do you mean $mathbb{C}(x)$ is also algebraically closed?
$endgroup$
– davidh
Jan 25 at 2:38












$begingroup$
I've edited my answer.
$endgroup$
– Thomas Shelby
Jan 25 at 3:07




$begingroup$
I've edited my answer.
$endgroup$
– Thomas Shelby
Jan 25 at 3:07




1




1




$begingroup$
Is the contradiction that $q(x)$ must be $sqrt{x}p(x)$. and $sqrt{x}$ is not in $mathbb{C}(x)$
$endgroup$
– davidh
Jan 25 at 6:27




$begingroup$
Is the contradiction that $q(x)$ must be $sqrt{x}p(x)$. and $sqrt{x}$ is not in $mathbb{C}(x)$
$endgroup$
– davidh
Jan 25 at 6:27












$begingroup$
Yes. Also note that degree of $x {p (x)}^2$ is odd whereas degree of ${q (x)}^2$ is even.
$endgroup$
– Thomas Shelby
Jan 25 at 7:35




$begingroup$
Yes. Also note that degree of $x {p (x)}^2$ is odd whereas degree of ${q (x)}^2$ is even.
$endgroup$
– Thomas Shelby
Jan 25 at 7:35


















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