# Show that there exists ﬁelds properly containing $\mathbb{C}$. Does that ﬁeld have the property that every non-constant polynomial over it has a root?

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.

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

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.

• 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? – davidh Jan 25 at 2:38
• I've edited my answer. – Thomas Shelby Jan 25 at 3:07
• Is the contradiction that $q(x)$ must be $\sqrt{x}p(x)$. and $\sqrt{x}$ is not in $\mathbb{C}(x)$ – davidh Jan 25 at 6:27
• Yes. Also note that degree of $x {p (x)}^2$ is odd whereas degree of ${q (x)}^2$ is even. – Thomas Shelby Jan 25 at 7:35