Prove there is no field F satisfying R ⊊ F ⊊ C

Question

I've (obviously) just learnt the basics of groups and fields and I've come across this exercise: Prove there is no field $\mathbb{R} ⊊\mathbb{F} ⊊\mathbb{C}$. My reasoning is simply that since $\mathbb{F} \neq \mathbb{R} $ it must contain at least one complex number, say $x+yi$. So then it contains $(x+k) + yi$ for any real number $k$ and from there it contains $c(x+k) + cyi$ for some other real number $c$ so $\mathbb{F}$ would be equal to $\mathbb{C}$. However I am not sure if this qualifies as a "proof" or if this is right. Any help is appreciated!

Answer 1

You are on the correct path, but I'd personally tidy that up a bit! Let your field $\mathbb{F}$ satisfy $\mathbb{R}\subsetneq\mathbb{F}\subseteq\mathbb{C}$. We will now show it has to be equal to $\mathbb{C}$.

As you mentioned, there is a complex number $x+yi\in\mathbb{F}$ with $y\neq0$. Notice that $x+yi-r$ and hence $\frac{x+yi - r}{s}$ are also in $\mathbb{F}$ for any reals $r, s$, with $s\neq0$ (can you tell me why?). Choosing the correct $r, s$, we may show that $i\in\mathbb{F}$, and then it will follow (why?) that $\mathbb{F}=\mathbb{C}$.

Answer 2

Yes, You are on a right track... As someone said above you need to assume $y \neq 0$ then what you want to do is take any complex number z=a+ib and show that there exists some u and v such that u=c(x+k) and v=cy for some c and k in Reals.. can you do that? If you did that then your proof is complete ... do you understand why? let me know in the comments