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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.

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  • $\begingroup$ The rational functions over $\Bbb C$ form a field properly containing $\Bbb C$. $\endgroup$ – TonyK Jan 25 at 2:06
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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 $$f(y)=y^2-\frac1x\in F[y].$$ 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 $${\left(\frac{p(x)}{q(x)}\right)}^2=\frac1x\iff x{p(x)}^2={q(x)}^2.$$ Can you see the contradiction?

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

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    $\begingroup$ 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
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    $\begingroup$ 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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