We have three claims , and the task is to decide if they are true or false.
Claim 1: If `bbb"F" subseteq bbb"E"` and `alpha in bbb"E"` is algebraic over `bbb"F"` , then `alpha^2` is algebraic over `bbb"F"`.
Claim 2: If `bbb"F" subseteq bbb"E"` and `alpha in bbb"E"` is algebraic over `bbb"F"` , then `alpha + alpha^2` is algebraic over `bbb"F"`.
Claim 3: Every vector space has a basis.

Write down the specific Theory used :
Theorem (Finite Extensions): If `alpha` is algebraic over `bbb"F"` , then the extension `bbb"F"(alpha)` is a vector space over `bbb"F"` with a finite dimension n , where n is the degree of the minimal polynomial of `alpha`.
Theorem (Elements of Finite Extensions): In a finite-dimensional vector space (extension field) , every element is algebraic over the base field F.
The basis Theorem: Every vector space `bbb"V"` has a basis. (Note: For infinite-dimensional spaces , this proof relies on Zorn's Lemma or Axiom of Choice).

Do the Exercise:

Claim 1: `alpha^2` is algebraic. True
Since `alpha in bbb"F"(alpha)` , and `bbb"F"(alpha)` is a field , it is closed under multiplication. Thus `alpha * alpha = alpha^2 in bbb"F"(alpha)`. Because `bbb"F"(alpha)` is a finite-dimensional vector space over `bbb"F"` , any element in it must be algebraic.

Explanation: `bbb"F"(alpha)` is infinite-dimensional precisely when `alpha` is transcendental over `bbb"F"` , but the converse fails: algebraic extensions can be infinite-dimensional , e.g. adjoining infinitely many algebraic elements like all algebraic numbers over `QQ`. Meaning a complex number `alpha` is algebraic over `QQ` if there exists a non-zero polynomial `f(x) in QQ[x]` with `f(alpha) = 0`. The set of all such `alpha` is denoted `overline QQ` and forms a field (closed under addition , subtraction , multiplication , and division). As an extension `overline QQ "/" QQ` , it is algebraic (every element is algebraic over `QQ`) but of infinite degree as a vector space over `QQ`. So `overline QQ subset CC` , not just a single number field like `QQ(sqrt 2)`.
This gives us that finite dimensionality implies algebraicity but not vice versa:
- transcendental extension `implies` infinite dimensional
- finite-dimension `implies` algebraic

Claim 2: `alpha + alpha^2` is algebraic. True
Since `alpha` and `alpha^2` are both in the field `bbb"F"(alpha)` , their sum `alpha + alpha^2` is also in `bbb"F"(alpha)`. Because `bbb"F"(alpha)` is a finite-dimensional extension of `bbb"F"` , the sum is algebraic.

Claim 3: Every vector space has a basis. True
This is a fundamental axiom/theorem of linear algebra. Whether the space is is `RR^n` or an infinite-dimensional space of functions , a basis always exists.

Final Answer:
- Claim 1: True
- Claim 2: True
- Claim 3: True
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