This is a bit of a wall of text and so to clarify things I've put my questions in bold. I've always summarized my question in the final section so one could look at those and then look above for the context I've given, rather than having to read through this whole thing. My apologies this question is so long and a bit scattered but I suppose it represents my level of understanding (or really, lack thereof) on this topic. Any help is GREATLY appreciated. Thank you.
I'm currently working through the tail end of an abstract algebra $1$ course, so things like fields from rings, extension fields, polynomials and Galois Theory and one (likely simple) definition that I'm struggling with gaining the intuition for and seeing how to determine the value of is the notion of $[\mathbb{K}:\mathbb{F}]$ where $\mathbb{K}, \mathbb{F}$ are field with $\mathbb{F} \subset \mathbb{K}$ an extension field.
Now the book says to view this as the dimension of $\mathbb{K}$ when viewing $\mathbb{K}$ as a vector space over $\mathbb{F}$. So, to my mind that means determining the number of basis vectors for $\mathbb{K}$ over $\mathbb{F}$, correct? Well, if that's correct then I seem to have quite a bit of difficulty doing just that for fields which are a bit more abstract than just your usual $\mathbb{C}, \mathbb{R},$ etc...
Consider, for example, the field $\mathbb{Q}(\sqrt[3](3), \zeta_3) \subset \mathbb{C}$. In the book this field is just given to us with the only reference to an object like this being the statement a few sections prior that
Suppose we already have some extension field $\mathbb{F} \subset \mathbb{L}$ and $\exists \beta \in \mathbb{L}$ which is a root of the irreducible polynomial $p \in \mathbb{F}(x)$ in $\mathbb{L}$. Let $\mathbb{F}(\beta)$ denote the smallest subfield of $\mathbb{L}$ containing both $\mathbb{F}$ and $\beta$.
Now given that then I would interpret $\mathbb{Q}(\sqrt[3](3), \zeta_3)$ as the smallest subfield of $\mathbb{C}$ that contains $\mathbb{Q}, \sqrt[3](3), \zeta_3$ where the latter $2$ are roots in $\mathbb{C}$, correct? Based on some previous experience with the field $\mathbb{Q}(\sqrt 2) = \{a + b\sqrt 2 \mid a,b \in \mathbb{Q} \}$ I would interpret the elements of this field as $\mathbb{Q}(\sqrt[3](3), \zeta_3) = \{ a + b\sqrt[3](3) + c\zeta_3 \mid a,b,c \in \mathbb{Q} \}$ since we can see $\mathbb{Q}$ is contained in the field for $b=c=0$ and similarly for the other $2$ roots when their respective coefficients are the only nonzero ones. Is this correct thinking for describing elements of these kinds of groups?
The text, in the example referring to this group, casually says that "$\mathbb{Q}(\sqrt[3](3), \zeta_3)$ is a degree $2$ extension of $\mathbb{Q}(\sqrt[3](3))$ which is itself a degree $3$ extension over $\mathbb{Q}$" Which I imagine is referencing the number of basis vectors of the extension field when viewed as a vector space over the subfield. And then proceeds to say that $$[\mathbb{Q}(\sqrt[3](3), \zeta_3): \mathbb{Q}] = [\mathbb{Q}(\sqrt[3](3), \zeta_3):\mathbb{Q}(\sqrt[3](3))][\mathbb{Q}(\sqrt[3](3)): \mathbb{Q}] = 2 \cdot 3 = 6$$ Now I understand why the end value is $6$ when the intermediate values are $2$ and $3$ respectively, but I don't understand how to see that $[\mathbb{Q}(\sqrt[3](3), \zeta_3):\mathbb{Q}(\sqrt[3](3))] = 2$ and $[\mathbb{Q}(\sqrt[3](3)): \mathbb{Q}] = 2$
Finally, the example goes on to say that the basis of $\mathbb{Q}(\sqrt[3](3), \zeta_3)$ over $\mathbb{Q}$ is $$\mathcal{B} = \{1, \sqrt[3](3), \sqrt[3](9), \zeta_3, \sqrt[3](3)\zeta_3, \sqrt[3](9)\zeta_3 \}$$ Which contradicts what I imagine the element of $\mathbb{Q}(\sqrt[3](3), \zeta_3)$ to be since in my case the basis would just be $\{ 1, \sqrt[3](3), \zeta_3 \}$. So How do I find the basis of fields of the form $\mathbb{F}(\alpha, \beta, \gamma, ... )$?
Summarizing Questions
- How do I interpret fields of the form $\mathbb{F}(\alpha, \beta, \gamma, ...)$?
- Like in the example above $\mathbb{Q}(\sqrt[3](3), \zeta_3)$
- Are these just the smallest subfield $\mathbb{F} \subset \mathbb{L}$ that contains $\mathbb{F}, \alpha, \beta, \gamma, ...$ Where the greek letters denote roots of an irreducible polynomial $p \in \mathbb{F}[x]$ in $\mathbb{L}$? So that $\mathbb{Q}(\sqrt[3](3), \zeta_3)$ is the smallest subfield of $\mathbb{C}$ that contains $\mathbb{Q}, \sqrt[3](3), \text{and}\ \zeta_3$ where $\sqrt[3](3), \text{and}\ \zeta_3$ are the roots of some polynomial $p \in \mathbb{Q}[x]$ in $\mathbb{C}$?
- How do I interpret the elements of such a field?
- I always thought it was, say $\mathbb{Q}(\sqrt 2) = \{a + b\sqrt 2 \mid a,b \in \mathbb{Q} \}$ for $\mathbb{Q}(\sqrt[3](3), \zeta_3)$ over $\mathbb{Q}$, but that doesn't seem to track. Let alone for something like $\mathbb{Q}(\sqrt[3](3), \zeta_3)$ over $\mathbb{Q}(\sqrt[3](3))$
- How do I find the basis for these kinds of fields?
- As above, the basis for $\mathbb{Q}(\sqrt[3](3), \zeta_3)$ over $\mathbb{Q}$ is $\mathcal{B} = \{1, \sqrt[3](3), \sqrt[3](9), \zeta_3, \sqrt[3](3)\zeta_3, \sqrt[3](9)\zeta_3 \}$ but interpreting this extension field in the way that I have I would have never got this as a basis.
- Finally, is the value of $[\mathbb{K}:\mathbb{F}]$ simply just the number of basis vectors when viewing $\mathbb{K}$ as a vector space over $\mathbb{F}$
I assume some familiarity with university level algebra. In particular that given a field $\mathbb{F}$ you know the polynomial ring $$R_n:=\mathbb{F}[x_1,\ldots,x_n]=\left\{\sum_{I\in\mathbb{N}_0^n}a_Ix^I: \text{for all but finitely many $I$ we have }0=a_I\in\mathbb{F}\right\},$$ where for $I=(i_1,\ldots,i_n)\in\mathbb{N}_0^n$ we abbreviate the monomial $x^I=x_1^{i_1}\cdot\ldots\cdot x_n^{i_n}$.
I will also only deal with finite, algebraic extension and provide only examples over $\mathbb{F}=\mathbb{Q}$, as this seems to be the specific context of the question.
$\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\alpha,\beta)$ etc. are simply the "homomorphic images" of $R_n$ when you replace $x_1=\alpha$, resp. $x_2=\beta$ etc. Let's start slowely with $\mathbb{Q}(\sqrt{2})$. We have $R_1=\left\{\sum\limits_{i=0}^da_ix_1^i: a_i\in\mathbb{Q}, d\in\mathbb{N}_0\right\}$. Replacing $x_1=\sqrt{2}$ we get $\mathbb{Q}(\sqrt{2})=\left\{\sum\limits_{i=0}^da_i\sqrt{2}^i: a_i\in\mathbb{Q}, d\in\mathbb{N}_0\right\}$. However we can write, e.g., $a_2\sqrt{2}^2=a_2\cdot 2=2a_2\cdot \sqrt{2}^0\quad$ or $\quad a_3\sqrt{2}^3=a_3\cdot 2\sqrt{2}=2a_3\cdot \sqrt{2}^1\quad$ or $\quad a_4\sqrt{2}^4=a_4\cdot 4=4a_4\cdot \sqrt{2}^0$ and so on and in a combined way, e.g., $$1\sqrt{2}^0+2\sqrt{2}^1+3\sqrt{2}^2+4\sqrt{2}^3+5\sqrt{2}^4=1\sqrt{2}^0+2\sqrt{2}^1+3\cdot 2\sqrt{2}^0+4\cdot 2\sqrt{2}^1+5\cdot 4\sqrt{2}^0=(1+3\cdot 2+5\cdot 4)\sqrt{2}^0+(2+4\cdot 2)\sqrt{2}^1=27+10\sqrt{2},$$ thus we do not get anything new, by letting $d\geq 2$. Thus it turns out, after some work, that $$\mathbb{Q}(\sqrt{2})=\left\{\sum\limits_{i=0}^da_i\sqrt{2}^i: a_i\in\mathbb{Q}, d\in\mathbb{N}_0\right\}=\left\{\sum\limits_{i=0}^1b_i\sqrt{2}^i: b_i\in\mathbb{Q}\right\}=\{b_0+b_1\sqrt{2}:b_0,b_1\in\mathbb{Q}\}.$$ Let's play the game with $\mathbb{Q}(\sqrt[3]{2})$. Replacing $x_1=\sqrt[3]{2}$ we get $\mathbb{Q}(\sqrt[3]{2})=\left\{\sum\limits_{i=0}^da_i\sqrt[3]{2}^i: a_i\in\mathbb{Q}, d\in\mathbb{N}_0\right\}$. However we can write, e.g., $a_3\sqrt[3]{2}^3=a_3\cdot 2=2a_3\cdot \sqrt[3]{2}^0\quad$ or $\quad a_4\sqrt{2}^4=a_4\cdot 2\sqrt[3]{2}=2a_4\cdot \sqrt[3]{2}^1\quad$ or $\quad a_8\sqrt[3]{2}^8=a_8\cdot 4\sqrt[3]{2}^2=4a_8\cdot \sqrt[3]{2}^2$ and so on and in a combined way, e.g., $$1\sqrt[3]{2}^0+2\sqrt[3]{2}^1+3\sqrt[3]{2}^2+4\sqrt[3]{2}^3+5\sqrt[3]{2}^4=1\sqrt[3]{2}^0+2\sqrt[3]{2}^1+3\sqrt[3]{2}^2+4\cdot 2\sqrt[3]{2}^0+5\cdot 2\sqrt[3]{2}^1=(1+4\cdot 2)\sqrt[3]{2}^0+(2+5\cdot 2)\sqrt[3]{2}^1+3\sqrt[3]{2}^2=9+12\sqrt[3]{2}+3\sqrt[3]{2}^2,$$ thus we do not get anything new, by letting $d\geq 3$. Thus it turns out, after some work, that $$\mathbb{Q}(\sqrt[3]{2})=\left\{\sum\limits_{i=0}^da_i\sqrt[3]{2}^i: a_i\in\mathbb{Q}, d\in\mathbb{N}_0\right\}=\left\{\sum\limits_{i=0}^2b_i\sqrt[3]{2}^i: b_i\in\mathbb{Q}\right\}=\{b_0+b_1\sqrt[3]{2}+b_2\sqrt[3]{2}^3:b_i\in\mathbb{Q}\}.$$ This game gets quite tough pretty soon. Set $\alpha=\sqrt{2}+\sqrt{3}$ and look, e.g., at $\mathbb{Q}(\alpha)$. In this case it turns out, that you get old stuff for $d\geq 4$. Explicitly: $$ \alpha^4= ((\sqrt{2}+\sqrt{3})^2)^2= (2+2\sqrt{6}+3)^2= (5+2\sqrt{6})^2= 25+20\sqrt{6}+24= 49+20\sqrt{6}= (50+20\sqrt{6})-1= 10(5+2\sqrt{6})-1= 10(2+2\sqrt{6}+3)-1= 10(\sqrt{2}+\sqrt{3})^2-1= 10\alpha^2-1= 10\alpha^2-1\alpha^0. $$ You should be able to verify this calculation as correct, but you probably do not have any idea, how one comes up with it: the key word is minimal polynomial and we leave it at that for the moment. It turns out (I've shown that you can discard any $d\geq 4$ and ask you to believe me that you cannot discard any $d\leq 3$) that $$\mathbb{Q}(\alpha)=\left\{\sum\limits_{i=0}^da_i\alpha^i: a_i\in\mathbb{Q}, d\in\mathbb{N}_0\right\}=\left\{\sum\limits_{i=0}^3b_i\alpha^i: b_i\in\mathbb{Q}\right\}=\{b_0+b_1\alpha+b_2\alpha^2+b_3\alpha^3:b_i\in\mathbb{Q}\}.$$ So far we only "adjoined a single element", let's do an example where we adjoin two elements, e.g. $\sqrt{2}$ and $\sqrt{3}$. We have $R_2=\left\{\sum\limits_{i,j=0}^da_{i,j}x_1^ix_2^j: a_{i,j}\in\mathbb{Q}, d\in\mathbb{N}_0\right\}$. Replacing $x_1=\sqrt{2}$ and $x_2=\sqrt{3}$ we get $\mathbb{Q}(\sqrt{2},\sqrt{3})=\left\{\sum\limits_{i,j=0}^da_{i,j}\sqrt{2}^i\sqrt{3}^j: a_{i,j}\in\mathbb{Q}, d\in\mathbb{N}_0\right\}$. By now it should be no surprise that we can cut down on $d$, anything with $d\geq 2$ can be simplified, e.g.: $\sqrt{2}^2=2\sqrt{2}^0,\quad$ $\sqrt{3}^5=9\sqrt{3}^1,\quad$ $\sqrt{2}^6\cdot\sqrt{3}^3=24\sqrt{2}^0\sqrt{3}^1$. It turns out that $$\mathbb{Q}(\sqrt{2},\sqrt{3})=\left\{\sum\limits_{i,j=0}^da_{i,j}\sqrt{2}^i\sqrt{3}^j: a_{i,j}\in\mathbb{Q}, d\in\mathbb{N}_0\right\}=\left\{\sum\limits_{i,j=0}^1b_{i,j}\sqrt{2}^i\sqrt{3}^j: b_{i,j}\in\mathbb{Q}\right\}=\{b_{0,0}+b_{1,0}\sqrt{2}+b_{0,1}\sqrt{3}+b_{1,1}\sqrt{6}:b_{i,j}\in\mathbb{Q}\}.$$ With a bit of work, one can show that $$\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\alpha).$$ So we see that the number of adjoined roots is not an invariant of a number field. Obviously, it might be simpler to compute in the representation $\mathbb{Q}(\sqrt{2},\sqrt{3})$ as opposed to $\mathbb{Q}(\sqrt{2}+\sqrt{3})$, but in principle there is no difference whether we choose one or the other: it's like in linear algebra, you can "choose different bases", the choice is irrelevant from an abstract point of view, but some bases might be easier to compute in. As a final example we look at $\mathbb{Q}(\sqrt[3]{3},\zeta_3)$. Again this is a homomorphic image of $R_2$ replacing $x_1=\sqrt[3]{3}$ and $x_2=\zeta_3$. Clearly $\sqrt[3]{3}^3=3\sqrt[3]{3}^0$ so we can cut down all degrees $d_1\geq 3$ when being confronted with $\sqrt[3]{3}^{d_1}$. The story for $\zeta_3$ is a bit different. $\zeta_3$ is, esentially by definition, a solution of the equation $x^3=1\Leftrightarrow x^3-1=0$. However there is a factorization of $p=x^3-1=(x-1)(x^2+x+1)$ involving only polynomials with coefficients in the base field $\mathbb{Q}$, i.e. $p$ is not irreducible. Thus by the zero divisor rule either $\zeta_3$ solves $p_1=x-1=0$ or it solve $p_2=x^2+x+1=0$. Clearly the solution to $p_1$ is $\xi=1\in\mathbb{Q}$ and since $\zeta_3\not\in\mathbb{Q}$ we have $\zeta_3\neq\xi=1$. We are left with the second possibility: $\zeta_3$ solves $p_2=0$ or, after substitution, $\zeta_3^2+\zeta_3+1=0$. This leads to $\zeta_3^2=-1\cdot\zeta_3^0-1\cdot\zeta_3^1$. As a consequence we can cut down on any degrees $d_2\geq 2$ in $\zeta_3$. So the cut-off degrees for $\sqrt[3]{2}$ and $\zeta_3$ are $d_1=3>d_2=2$ and different. Combining everything, we have: $$\mathbb{Q}(\sqrt[3]{3},\zeta_3)=\left\{\sum\limits_{i,j=0}^da_{i,j}\sqrt{3}^i\zeta_3^j: a_{i,j}\in\mathbb{Q}, d\in\mathbb{N}_0\right\}=\left\{\sum\limits_{i=0}^2\sum\limits_{j=0}^1b_{i,j}\sqrt[3]{3}^i\zeta_3^j: b_{i,j}\in\mathbb{Q}\right\}=\{b_{0,0}+b_{1,0}\sqrt[3]{3}+b_{2,0}\sqrt[3]{3}^2+b_{0,1}\zeta_3+b_{1,1}\sqrt[3]{3}\zeta_3+b_{2,1}\sqrt[3]{3}^2\zeta_3:b_{i,j}\in\mathbb{Q}\}.$$ By now you should have a host of examples to see what it means to have $\mathbb{F}(\alpha),\mathbb{F}(\alpha,\beta)$, etc. What I have shown is essentially a generative process. This does coincide with a subtractive process in spirit of "start with $\mathbb{C}$ and cut away everything, that is not necessary to accomodate the generators without loosing the field property". For now, I will not explain, why these two perspectives coincide. I think that the "generative approach" is much more enlightening in the beginning as the "subtractive approach".
I hope that the answer for 1. also answered 2. passing by.
Dito 2.
Yes, $[\mathbb{K}:\mathbb{F}]=\dim_{\mathbb{F}\text{-vector space}}(\mathbb{K})$. From the different presentations above in the answer to 1. you should see: $$ [\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2, \quad [\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3, \quad [\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=4, \quad [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=4, \quad [\mathbb{Q}(\sqrt[3]{3},\zeta_3):\mathbb{Q}]=6. $$
I hope that answers all your questions.
Final remark: You actually have $[\mathbb{Q}(\sqrt[3]{3},\zeta_3):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{3}):\mathbb{Q}]\cdot[\mathbb{Q}(\zeta_3):\mathbb{Q}]=3\cdot 2=6$. This works sometimes, but this is not true in general. Here is a "counter example": $[\mathbb{Q}(\sqrt[4]{2},i\sqrt[4]{2}):\mathbb{Q}]=8$, but $[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]\cdot[\mathbb{Q}(i\sqrt[4]{2}):\mathbb{Q}]=4\cdot 4=16\neq 8$. The reason why such a product formula does not work out is that there is some "overlap" beyond $\mathbb{Q}$: $-1\cdot \sqrt[4]{2}^2=-\sqrt{2}=\left(i\sqrt[4]{2}\right)^2$.