Fields and cubic extensions

i have a question that i do not know how to solve. I am still a beginner in field theory and i want to answer this :

Does every field $\mathbb{K}$ have a cubic expansion ?

I was thinking about infinite fields and was wondering if they could posses a finite extension but i know that $ \mathbb{C}/ \mathbb{R}$ is an finite extension of degree $2$. But does $ \mathbb{C}$ have a degree $3$ extension ? Since it is algebraically closed i do not know if it is possible to extend it more.

Even more, to extend $ \mathbb{R}$ into $\mathbb{L}$ with this extension being of degree $3$ i have to find an irreducible polynomial $P$ of degree $3$ that cancels for an algebraic element $x \in \mathbb{L}$. Thus $P$ would be the minimal polynomial of $x$ over $\mathbb{R}$ and then $ \mathbb{R}/<P>= \mathbb{L} $ is an extension of degree $3$ over $\mathbb{R}$.

And i know that such a polynomial exists in characteristic $p$. As we showed in class that there exists at least $1$ irreducible monic polynomial fo any degree $r \geq 1$ over any finite fields $\mathbb{F}_p$ for $p$ a prime number. So if this work even in characteristic $0$ then the statement would be true !

Thank you for your time.

Solution 1:

No, $\mathbb{C}$ (or any algebraically closed field) cannot have an extension of finite degree, since any such extension must necessarily be algebraic. This is proved in many places, including on this site.

We can use this to see what goes wrong with your example involving $\mathbb{R}$, but we can also see directly: if you like it is a theorem in real analysis---a basic application of the intermediate value theorem---that every odd degree polynomial over the reals has a real root. So there are no degree $3$ irreducible polynomials over $\mathbb{R}$. We can also see this algebraically: if a real polynomial has a complex root $z$ then it must also have the root $\overline{z}$: this means that a real polynomial cannot have an odd number of complex roots, so any hypothetical degree $3$ real polynomial must again have at least one real root. (By "complex roots" here I mean "not purely real", i.e. so that $z \not = \overline{z}$.)