How to find the degree of a field extension

I don't quite understand how to find the degree of a field extension. First, what does the notation [R:K] mean exactly?

If I had, for example, to find the degree of $\mathbb Q (\sqrt7)$ over $\mathbb Q$, how would I go about it? And how would it be different from, say, $\mathbb C (\sqrt7)$ over $\mathbb C$?

Would it involve finding the minimal polynomial? In the case of $\mathbb Q (\sqrt7)$, I find the minimal polynomial to be $x^2-7$ which is of degree 2, so would this be the value of $[\mathbb Q (\sqrt7):\mathbb Q]$?

In $\mathbb C$, this polynomial is reducible, so I assume somehow it would need to be reduced to find the minimal polynomial. (I am guessing since $x^2-7=(x^2+1)-8=i-8$ this is the polynomial in $\mathbb C$) Would the degree of this be the degree of$[\mathbb C (\sqrt7):\mathbb C]$?

• Your TeX is broken which makes your post harder to read. – Qudit Feb 9 '15 at 20:13
• In $\mathbb{Q}$, the minimal polinomial of $\sqrt{7}$ is $x^2-7$, which is irreducible over $\mathbb{Q}$ since its roots are $\sqrt{7},-\sqrt{7}$ (and they don't lie in $\mathbb{Q}$), so yes, you're right: $[\mathbb{Q}(\sqrt{7}):\mathbb{Q}]=2$. On the other hand, the minimal polynomial of $\sqrt{7}$ over $\mathbb{C}$ is $x-\sqrt{7}$ (since $\sqrt{7}$ lies in $\mathbb{C}$), so $[\mathbb{C}(\sqrt{7}):\mathbb{C}]=1$ (this implies that $\mathbb{C}(\sqrt{7})=\mathbb{C}$, which is verified directly). – Daniel Feb 9 '15 at 20:16

A field $K$ over a field $F$ is in particular a vector space over $F$, and $[K:F]$ is its dimension. For $F(\alpha)$ it's true that this dimension is the degree of the minimal polynomial of $\alpha$ over $F$, because $K$ then has a basis over $F$ given by $1,\alpha,...,\alpha^n$. $\mathbb{C}$ is algebraically closed, so all its algebraic extensions are trivial, that is, have degree $1$. But your computation of the minimal polynomial of $\mathbb{C}(\sqrt 7)$ is not correct. It's simply $x-\sqrt 7$, since $\mathbb{C}$ contains a square root of $7$. One more error: $x^2+1$ is not equal to $i$ in $\mathbb{C}[x]$. You're confusing the latter ring with the expression of $\mathbb{C}$ as $\mathbb{R}[t]/(t^2+1)$. In those terms $\mathbb{C}[x]$ is $(\mathbb{R}[t]/(t^2+1))[x]$: the two indeterminates $t$ and $x$ have nothing to do with each other.
• What if I was working with a cube root of a number? Would an irreducible polynomial in $\mathbb C$ be of degree 3 in this case, since cube roots are not in $\mathbb C$? – AccioHogwarts Feb 9 '15 at 23:23
• @AccioHogwarts Actually, cube roots are in $\mathbb{C}$. That's part of what it means to say $\mathbb{C}$ is algebraically closed, which is $\mathbb{C}$'s most important property. – Kevin Carlson Feb 10 '15 at 4:39