# What is the index of field extension $\mathbb{C}/\mathbb{R}$?

What is the index of field extension $$\mathbb{C}/\mathbb{R}$$?

I know that the answer is $$2$$, but if so, that means $$\mathbb{C}/\mathbb{R}=\{\mathbb{R}, i+\mathbb{R}\}$$, and how come $$5i+\mathbb{R}=i+\mathbb{R}$$?

• I think the index of $\mathbb R$ in $\mathbb C$ is not $2$, as you rightly noted. The dimension of $\mathbb C$, as a real vector-space, however, is $2$. – awllower Mar 23 at 11:15
• @awllower My actuall qustion is: My professor defined the degree of a field extension to be $\text{dim}_F(E):=[F:E]$. But $[F:E]$ is the index of $F/E$. How can it all settle down in the case of $E=\mathbb{C},F=\mathbb{R}$? – J. Doe Mar 23 at 11:19
• Note that $\Bbb C$ is of dimension $2$ over $\Bbb R.$ As every complex number can be written as a $\Bbb R$-linear combinations of $1$ and $i.$ – Dbchatto67 Mar 23 at 11:32
• No, $[F:E]$ is not the index. It is the thing it was just defined to be (which happens to disagree with what the notation would mean in a different context). – Tobias Kildetoft Mar 23 at 11:39

The degree of the field extension is 2: $$[\mathbb{C}:\mathbb{R}] = 2$$ because that is the dimension of a basis of $$\mathbb{C}$$ over $$\mathbb{R}$$.
As additive groups, $$\mathbb{R}$$ is normal in $$\mathbb{C}$$, so we get that $$\mathbb{C} / \mathbb{R}$$ is a group. The cardinality of this group is uncountably infinite (we have an answer for this here), which you should attempt to prove along the lines you suggested in the body of your question.
Take the map $$f : \Bbb C \longrightarrow \Bbb C$$ defined by $$a+bi \mapsto bi,\ a,b \in \Bbb R.$$ Observe that the kernel of the map $$f$$ is isomorphic to $$\Bbb R$$ and the image is the imaginary axis. So by the first group isomorphism theorem we can conclude that as a group $$\Bbb C/\Bbb R$$ is isomorphic to the imaginary axis which is uncountably infinite.