Topic: `"Adjoining Multiple Elements"`.
We will give an example of how the "Aquarium" expands. When you add `sqrt 2` , you get a bigger tank.When you then add `sqrt 3` , you aren't just adding one more fish , you are multiplying the possibilities of the entire environment.

Introduction: `"Building Fields Piece by Piece"`.
Up until now , we have looked at adjoining a single element `alpha` to a field `bbb"F"`. We can add `alpha_1` to get `bbb"F"(alpha_1)` , and then add `alpha_2` to that result to get `bbb"F"(alpha_1 , alpha_2)`.
This process is called adjoining. The resulting field is the smallest possible "neighborhood" that contains `bbb"F"` and all your new elements. It doesen't matter which one you add first , the final destination is the same.

Theory: `"The Method of Intersection and Basis"`.
1. `"Characterisation"`:`bbb"F" (alpha_1 , ... \ , alpha_n)` can be defined as the intersection of all subfields of `bbb"E"` that contain both `bbb"F"` and the set `{alpha_1 , ... \ , alpha_n}`. It is the "tightest fit" possible.
2. `"Successive Extensions"`: We treat `bbb"F"(alpha_1 , alpha_2)` as `(bbb"F"(alpha_1))(alpha_2)`. This means we can find a basis for the first extension and then find the basis for the second extension over the first.
3. `"The Tower Result"`: If `{u_1 , ... \ , u_n}` is a basis for `bbb"K/E"` and `{v_1 , ... \ , v_n}` is a basis for `bbb"E/F"` , then the set of all products `{u_i , v_j}` is a basis for the total extension `bbb"K/F"`.

Exercise: `"The Mystery of" \ sqrt 2 + sqrt 3`.
`"Problem"`: Prove that `sqrt 3` is not in `QQ(sqrt 2)` and find a basis for `QQ(sqrt 2 , sqrt 3)` over `QQ`.

`"Solution"`:
1. `"Check the Degree"`: `gamma = sqrt 2 + sqrt 3` is a root of `x^4 - 10x^2 + 1`. This polynomial is irreducible over `QQ` , so `[QQ(sqrt 2 + sqrt 3) : QQ] = 4`.
2. `"Compare Spaces"`: We know `[QQ(sqrt 2) : QQ] = 2`. Since `4` is greater than `2` , the element `sqrt 2 + sqrt 3` cannot live in `QQ(sqrt 2)`.
3. `"Build the Tower"`:
- `QQ(sqrt 2)` has basis `{1 , sqrt 2}` over `QQ`.
- `QQ(sqrt 2 , sqrt 3)` has basis `{1 , sqrt 3}` over `QQ(sqrt 2)`.
4. `"Final Basis"`: By multiplying the bases together : `1 * 1 , 1 * sqrt 2 , sqrt 3 * 1 , sqrt 3 * sqrt 2 = sqrt 6`.

`"Result"`: The basis for `QQ(sqrt 2 , sqrt 3)` over `QQ` is `{1 , sqrt 2 , sqrt 3 , sqrt 6}`. QED

Additional Explanations: To show that `x^4 - 10x^2 + 1` is irreducible and to prove the "residence" of `sqrt 2 + sqrt 3` by contradiction , we need to look at how these fields "sift" through possible values.

`"Irreducibility of" \ \ x^4 - 10x^2 + 1 \ "over" \ QQ`: To prove that the polynomial `P(x) = x^4 -10x^2 + 1` is irreducible , we check for two types of failure: linear factors (roots) and quadratic factors.
Step 1: `"Rational Root Theorem"`.
Any rational root `"p/q"` must divide the contant term `(1)` and the leading coefficient `(1)`. The only possibilities are `+-1`.
- `P(1) = 1 - 10 + 1 = -8`
- `P(1) = 1 - 10 + 1 = -8`
`"Result"`: No rational roots , thus , no linear factors.

Step 2: `"Quadratic Factors"`: Assume `x^4 - 10x^2 + 1 = (x^2 + ax + b)(x^2 + cx + d)` for integers `a , b , c , d`. Foiling:
`x^4 + cx^3 + dx^2 +ax^3 + acx^2 + adx + bx^2 + bcx + bd =`
`x^4 + (a + c)x^3 +(ac + b + d)x^2 + (ad + bc)x + bd`.
By matching coefficients:
1. `bb x^3 \ term : a + c = 0`.
2. `bb x^2 \ term : ac + b + d = -10 rArr -a^2 + b + d = -10`.
3. `bb x \ term : ad + bc = 0 rArr ad + b(-a) = a(d - b) = 0`.
4. `"Constant term" : bd = 1`.
From `(4)` either `(b , d) = (1 , 1)` or `(-1 , -1)`. If we test `(b , d) = (1 , 1)` , then from `(2): -a^2 + 2 = -10 rArr a^2 = 12` (no integer solution).
If we test `(b , d) (-1 , -1)` , then from `(2) : -a^2 - 2 = -10 rArr a^2 = 8` (no integer solution).
Conclusion: Since `P(x)` cannot be split into smaller integer polynomials , it is irreducible over `QQ`.

`"Why"` `sqrt 2 + sqrt 3 in QQ(sqrt 2)`: `"Proof by Contradiction"`.
We can prove this by assuming the "fish" can live in the smaller tank and showing the math collapses.
`"The Assumption"`: Assume `sqrt 2 + sqrt 3 in QQ(sqrt 2)`.
Since `QQ(sqrt 2)` is a field and `sqrt 2` is already in it , the difference must also be in the field: `(sqrt 2 + sqrt 3) - sqrt 2 = sqrt 3`.
Therefore our assumption implies that `sqrt 3 in QQ(sqrt 2)`. If `sqrt 3 in QQ(sqrt 2)` there must exist rational numbers `a` and `b` such that `sqrt 3 = a + b sqrt 2 rArr 3 = (a + b sqrt 2)^2 rArr 3 = a^2 + 2ab sqrt 2 + 2b^2`.
Now we analyze the pieces:
1. If `a` and `b` are nonzero: We can solve for `sqrt 2`:
`sqrt 2 = {3 - a^2 - 2b^2}/{2 ab}`. The right side is a rational number (a fraction of integers). But we know `sqrt 2` is irrational. This is a contradiction.

2. If `b = 0`: Then `3 = a^2` , meaning `sqrt 3 = a`. But `sqrt 3` is irrational , so this is a contradiction.
3. If `a = 0` , then `3 = 2b^2 rArr b^2 = 3 "/" 2`. There is no rational number whose square is `1.5` , therefore this is a contradiction.

Final Verdict: Because every possible path leads to a logical impossibility , our original assumption was false. The element `sqrt 2 + sqrt 3` cannot live in `QQ(sqrt 2)`. It requires a "larger tank" - specifically the `4`-dimensional field `QQ(sqrt2 , sqrt 3)`.

`"What is a basis ?"`
Building a basis is like finding the "irreducible building blocks" of our space. If you have a set of blocks that can reach any point in the field , and non of them can be built from the others , you have your basis.

`"Basis for" \ QQ(sqrt 2) \ "over" \ QQ`.
We are building the field by adjoining `alpha = sqrt 2` to the rational numbers `QQ`.

`"The Logic of Reduction"`.
Any element in this field looks like a polynomial in `sqrt 2` with rational coefficients:
`c_0 + c_1 (sqrt 2) + c_2(sqrt 2)^2 + c_3(sqrt 2)^3 + ...`
Because `sqrt 2` is a root of `x^2 - 2 = 0` , we know that `(sqrt 2)^2 = 2`. This rule allows us to "collapse" all higher powers.
- `(sqrt 2)^2 = 2` (which is just a rational number , so it merges with `c_0`).
- `(sqrt 2)^3 = 2 sqrt 2` (which merges with the `c_1 sqrt 2` term).
- `(sqrt 2)^4 = 4` (merges with the rational term `c_0`).

The result: Every single power of `sqrt 2` eventually collapses into either a rational number or a multiple of `sqrt 2`. Therefore , any element in the field can be written as: `a + b sqrt 2` (where `a , b in QQ`). Since `1` and `sqrt 2` are linearly independent (you cannot write `sqrt 2` as a rational multiple of `1`) , the set `{1 , sqrt 2}` is a basis.

`"Basis of" \ QQ(sqrt 2 , sqrt 3)` over `QQ(sqrt 2)`.
Now , our base is no longer `QQ`. Our base is the field `bbb"K" = QQ(sqrt 2)`. We are adjoining `beta = sqrt 3` to this new base.

`"The Logic of Relative Adjunction"`: We treat the entire field `QQ(sqrt 2)` as if it where our new "rational" field. When we adjoin `sqrt 3` , we are essentially looking for a basis of the form: `x + y sqrt 3` (where `x , y in bbb"K"`).
- `x` represents any element from `QQ(sqrt 2)` , which we know as `a + b sqrt 2`.
- `y` also represents any element from `QQ(sqrt 2)` , which is `c + d sqrt 2`.

`"Why" \ {1 , sqrt 3} \ "is the basis"`.
Spanning: Any element in the new , larger field can be written as a linear combination of `1` and `sqrt 3` , with coefficients drawn from the base field `bbb"K"`.
Independence: We proved previously that `sqrt 3 notin QQ(sqrt 2)`. This is the crucial step. If `sqrt 3` were in `QQ(sqrt 2)` , it would be a multiple of 1 (a rational combination of `1` and `sqrt 2 \ : 1 * sqrt 3 + 0 * sqrt 2`) , and we would not have a `"degree"-2` extension. Because `sqrt 3` is "new" and "independent" relative to our base field , `{1 , sqrt 3}` works perfectly.

`"The Job Description of a Scalar"`.
In high school or early college days , "scalars" were effectivily defined as "the real numbers `RR`" or "the rational numbers `QQ`". But in the context of Field Theory and Lie Algebra "scalar" is not a type of number , it is a job description.
A scalar is any element of the field `bbb"F"` that satisfies the field axioms (you can add , subtract , multiply , and divide by them). When we perform the construction of a vector space over a field `bbb"K"` , every element in `bbb"K"` is eligible to be a scalar.
In the vector space `QQ(sqrt 2)` with basis vectors `{1 , sqrt 2}` the elements look like `a + b sqrt 2` , where the scalars are `a , b in QQ`.
The next level is represented by the vector space `QQ(sqrt 2 , sqrt 3)` with basis `{1 , sqrt 3}`. Here the elements look like `x + y sqrt 3` , where the scalars are `x , y in QQ(sqrt 2)`.

Transition Note: `"The Operator Connection"`.
In our Lie Algebra work , we often see that an operator `ad_x` might have multiple eigenvalues (like `sqrt2` and `sqrt 3`). To fully "see" what that operator is doing , we have to adjoin all those eigenvalues to our field. This creates a multi-dimensional field extension where the basis elements (like `sqrt 6`) represent the interaction between the different "currents" in the tank.