description/proof of that field is canonically \(1\)-dimensional vectors space
Topics
About: field
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of dimension of vectors space.
Target Context
- The reader will have a description and a proof of the proposition that any field is canonically a \(1\)-dimensional vectors space.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\), with the addition, \(+: F \times F \to F, (r_1, r_2) \mapsto r_1 + r_2\), and the scalar multiplication, \(.: F \times F \to F, (r_1, r_2) \mapsto r_1 r_2\)
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Statements:
\(F \in \{\text{ the } F \text{ vectors spaces }\}\)
\(\land\)
\(\{1\} \in \{\text{ the bases for } F\}\)
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2: Proof
Whole Strategy: Step 1: see that \(F\) satisfies the conditions to be an \(F\) vectors space; Step 2: see that \(\{1\}\) is a basis for the vectors space, \(F\).
Step 1:
Let us see that \(F\) satisfies the conditions to be an \(F\) vectors space.
1) \(\forall r_1, r_2 \in F (r_1 + r_2 \in F)\) (closed-ness under addition): because \(F\) is a field, which is a ring, which is an Abelian group under addition.
2) \(\forall r_1, r_2 \in F (r_1 + r_2 = r_2 + r_1)\) (commutativity of addition): because \(F\) is a field, which is a ring, which is an Abelian group under addition.
3) \(\forall r_1, r_2, r_3 \in F ((r_1 + r_2) + r_3 = r_1 + (r_2 + r_3))\) (associativity of additions): because \(F\) is a field, which is a ring, which is an Abelian group under addition.
4) \(\exists 0 \in F (\forall r \in F (r + 0 = r))\) (existence of 0 element): because \(F\) is a field, which is a ring, which is an Abelian group under addition.
5) \(\forall r \in F (\exists r' \in F (r' + r = 0))\) (existence of inverse element): because \(F\) is a field, which is a ring, which is an Abelian group under addition.
6) \(\forall r \in F, \forall r' \in F (r . r' \in F)\) (closed-ness under scalar multiplication): because \(F\) is a field, which is a ring, which is a monoid under multiplication.
7) \(\forall r \in F, \forall r_1, r_2 \in F ((r_1 + r_2) . r = r_1 . r + r_2 . r)\) (scalar multiplication distributability for scalars addition): because \(F\) is a field, which is a ring.
8) \(\forall r_1, r_2 \in F, \forall r \in F (r . (r_1 + r_2) = r . r_1 + r . r_2)\) (scalar multiplication distributability for vectors addition): because \(F\) is a field, which is a ring.
9) \(\forall r \in F, \forall r_1, r_2 \in F ((r_1 r_2) . r = r_1 . (r_2 . r))\) (associativity of scalar multiplications): because \(F\) is a field, which is a ring, which is a monoid under multiplication.
10) \(\forall r \in F (1 . r = r)\) (identity of 1 multiplication): because \(F\) is a field, which is a ring, which is a monoid under multiplication.
So, \(F\) is an \(F\) vectors space.
Step 2:
\(\{1\} \subseteq F\) is linearly independent, because for \(r 1 = 0\), \(r 1 = r\), so, \(r = 0\).
For each \(r \in F\), \(r = r 1\).
So, \(\{1\}\) is a basis for the \(F\) vectors space, \(F\).
So, \(F\) is a \(1\)-dimensional vectors space.
