Topic: `"The Hierarchy of Algebraic Extensions"`.
Continuing our study of field extensions , we encounter two pivotal results that describe how finite dimensions guarantee algebraic behaviour and how these extensions "stack" on top of each other.

Theory:
Theorem: `"The Finiteness - Algebraic Link"`.
If an extension field `bbb"E"` of `bbb"F"` is a finite extension (meaning its dimension as a vector space `[bbb"E" : bbb"F"]` is a finite number `n`) , then `bbb"E"` is necessarily an `"algebraic"` extension of `bbb"F"`.

Explanation: Take a field extension `bbb"F" subseteq \ bbb"E" ` with `[bbb"E" : bbb"F"] = n` , and pick some `alpha in bbb"E"`. View `bbb"E"` as an
`n`-dimensional vector space over `bbb"F"`.

Step 1: `"The n + 1 powers of" \ alpha`.
Consider `n + 1` elements `1 , alpha , alpha ^2 , ... , alpha^n in bbb"E"`.
These are `n + 1` vectors in the `n`-dimensional `bbb"F"`-vector space `bbb"E"`. By linear algebra , any set of more than `n` vectors in an `n`-dimensional vector space is linearly dependent. So there exists scalars `a_0 , ... \ , a_n in bbb"F"` , not all zero , such that `a_0 * 1 + a_1 alpha + a_2 alpha^2 + ... + a_n alpha^n = 0`. (*)

Step 2: `"Turn the dependence into a polynomial"`.
Regard the left-hand side of (*) as a polynomial in an indeterminate `x` with coefficients in `bbb"F"`:
`p(x) = a_0 + a_1 x + ... + a_n x^n in bbb"F"[x]`.
By construction , `p(x)` is not the zero polynomial because not all `a_i` are zero.
Now evaluate this polynomial at `x = alpha`. Substituting `x = alpha` gives exactly the linear dependence relation:
`p(alpha) = a_0 + a_1 alpha + a_2 alpha^2 + ... + a_n alpha^n = 0`.
So we have produced a nonzero polynomial `p(x) in bbb"F"[x]` with `p(alpha) = 0`.

Step 3: `"Conclude that" \ alpha \ "is algebraic"`.
Since `alpha` was arbitrary , every element of `bbb"E"` is algebraic over `bbb"F"` , so the finite extension `bbb"E/F"` is an algebraic extension. QED

`"The Multiplicativity of Degrees and The Tower Law (or Theorem of The Product Formula)"`.

Theorem: If we have a "tower" of fields `bbb"F" subseteq bbb"E" subseteq bbb"K"` , where each step is a finite extension , then the total degree of the extension from the bottom to the top is the product of the intermediate degrees: `[bbb"K" : bbb"F"] = [bbb"K" : bbb"E"] * [bbb"E" : bbb"F"]`.

Exercise: `"Calculating the Tower Degree"`.
Let's look at a concrete tower of extensions:
1. Base Field: `QQ` (Rationals).
2. Intermediate Field: `bbb"E" = QQ(sqrt 2)`.
3. Top Field : `bbb"K" = QQ(sqrt 2 , sqrt 3)` (the field containing both roots).

Solution:
Step 1: We know from our previous lesson that the minimal polynomial of `sqrt 2` over `QQ` is `x^2 - 2`. Thus `[bbb"E" : QQ] = 2`.
Step 2: Now we look at `sqrt 3` over the field `bbb"E"`. We have that `sqrt 3` cannot be written in the form `a + b sqrt 2` , its minimal polynomial over `bbb"E"` is `x^2 - 3`. Thus , `[bbb"K : E"] = 2`.
Step 3: Apply the Tower law : `[bbb"K : Q"] = [bbb"K : E"] * [bbb"E : Q"] = 2 * 2 = 4`

Result: The total dimension of `bbb"K"` over `QQ` is 4.
A basis for this `4D` vector space would be `{1 , sqrt 2 , sqrt 3 , sqrt 6}`.

Explanation of the result: The construction of a basis for the field extension `QQ(sqrt 2 , sqrt 3)` over `QQ` (which is `4`-dimensional as a vector space) follows directly from the tower law for field extensions. We build the tower
`QQ subset QQ(sqrt 2) subset QQ(sqrt 2 , sqrt 3)`.
(The same holds if we adjoin `sqrt 3` first , the order does not matter).

1. `"The bottom step" \ QQ(sqrt 2)"/"QQ` has basis `{1 , sqrt 2}`.
This is the standard power basis for the quadratic extension generated by a root of `x^2 - 2`. By definition: `QQ(sqrt 2) = {a + b sqrt 2 \ : a , b in QQ}`. So every element is a `QQ`-linear combination of `1` and `sqrt 2` , i.e. the set `{1 , sqrt 2}` spans `QQ(sqrt 2)`.
To see linear independence , suppose `a * 1 + b * sqrt 2 = 0 , \ a , b in QQ`. Then `b sqrt 2 = -a` , so if `b ne 0` we get `sqrt 2 = -a"/"b in QQ` , which is false. Hence `b = 0` , and then `a = 0`. So only the trivial combination gives zero , and `{1 , sqrt 2}` is linearly independent. A spanning , linearly independent `2`-element set is a basis of a `2`-dimensional space.

2. `"The top step" \ QQ(sqrt 2 , sqrt 3)"/"QQ(sqrt 2) \ "has basis" \ {1 , sqrt 3}`.
The top step is obtained by adjoining `sqrt 3`. Its minimal polynomial over `QQ(sqrt 2)` is still `x^2 - 3` , since it remains irreducible (`3` is not a square in `QQ(sqrt 2)`). Thus the quadratic extension has basis `{1 , sqrt 3}`. To see this , we work over the field `bbb"F" = QQ(sqrt 2)`. First , `sqrt 3` satisfies the polynomial `x^2 - 3 in bbb"F"[x]` , and this polynomial is irreducible over `bbb"F"`: if it factored linearly , `sqrt 3` would lie in `QQ(sqrt 2)` , which it does not (one can show that no `a + b sqrt 2` squares to `sqrt3` with `a , b in QQ`). Hence the minimal polynomial of `sqrt 3` has degree `2` , so `[bbb"F"(sqrt 3) : bbb"F"] = 2`. But `bbb"F"(sqrt 3) = QQ(sqrt 2 , sqrt 3)` , so `[QQ(sqrt 2 , sqrt 3) : QQ(sqrt 2)] = 2`. For a quadratic extension `bbb"F"(alpha)"/" bbb"F"` , every element can be written as `u + v alpha \ , u , v in bbb"F"` , and `{1 , alpha}` is linearly independent for the same reason as above: If `u + v alpha = 0` and `v ne 0` , then `alpha = -u"/"v in bbb"F"` , contradicting `alpha notin bbb"F"`. So here an element of `QQ(sqrt 2 , sqrt 3)` is of the form `u + v sqrt 3 , \ u , v in QQ(sqrt 2)` , and `{1 , sqrt 3}` is linearly independent over `QQ(sqrt 2)`. Thus `{1 , sqrt 3}` is a basis of `QQ(sqrt 2 , sqrt 3)` as a vector space over `QQ(sqrt 2)`.

3. `"The Tower Law for composing bases in a tower of extensions"`.
By the tower law , a basis for the full extension `QQ(sqrt 2 , sqrt 3)"/" QQ` is obtained by taking all products of an element from the first basis with an element from the second basis:
`{1 , sqrt 2} times {1 , sqrt 3} = {b_1 * b_2 : b_1 in {1 , sqrt 2} , b_2 in {1 , sqrt 3}}`.
Explicitly multiplying gives exactly the four elements:
`1 * 1 , 1 * sqrt 3 , sqrt 2 * 1 , sqrt 2 * sqrt 3`. So the basis is `{1 , sqrt 2 , sqrt 3 , sqrt 6}`.

4. `"Showing that the four elements are linearly independent"`.
These four elements are linearly independent over `QQ` and span `QQ(sqrt 2 , sqrt 3)` because

- The tower law guarantees `[QQ(sqrt 2 , sqrt 3) : QQ] = [QQ(sqrt 2 , sqrt 3) : QQ(sqrt 2)] * [QQ(sqrt 2) : QQ] = 2 * 2 = 4`.

- Any vector space of dimension `4` has exactly four elements in a basis , so the four products automatically form one. The "multiply the bases" rule works for any tower of finite extensions (not just quadratics). This is the standard way to construct a basis for the composite field , when adjoining several algebraic numbers whose minimal polynomials are "independent" in the appropiate sense. The same construction also arises from the tensor-product description `QQ(sqrt 2) otimes_{QQ} QQ(sqrt 3)` , where the pure tensors `b_1 otimes b_2` correspond precisely to the field products `b_1 * b_2`. QED