Linear Algebra Field Definition and its Axioms

Question

I have some troubles understanding the definition of a field, and I tried to provide questions that might clear up my understanding of the definition. Thanks.

A field is a set $F$ with two binary operations:

1. $F \times F \to F, (a,b) \longmapsto a+b$ addition (Equation A)
2. $F \times F \to F, (a,b) \longmapsto a \cdot b$ multiplication (Equation B)

and two distinguished elements $0,1 \in F$ where $0 \neq 1$ such that the following axioms are satisfied for all, $a,b,c \in F$: Won't state them here.

Question:

1. Is $0,1$ referring to the real number $0,1?$

2. If we have an element $x \in F$, is $x \cdot x = x^2$ because of equation B?

3. If we have an element $x \in F$, is $x + x = 2x$ because of equation A?

4. What is the additive inverse of $1$? If this were to be real numbers I would say it's $-1$ but does $-1$ even exist in $F$? Since we are given those axioms one of them states $\forall a \in F \exists b \in F, s.t. a + b = 0$. So in this case we have $a= 1$ and we know that $a = 1 \in F$ thus, we know that there exists a $c \in F$ such that $1 + c=0$ it would be natural to say that $c = -1$ but how do we know a value of $-1$ even exists..?

Answer 1

The elements $0$ and $1$ are exactly what it says: distinguished elements. Because they are "special" elements they get special names, and we choose $0$ and $1$ as those special names because they satisfy the same properties we are used to when dealing with integers, rationals, reals, etc.

We typically use the notation $x^2 = x \cdot x$; it is not an equation that is being satisfied, rather a definition of what we mean by an exponent, so $x^k$ means we multiply $x$ my itself $k$ times.

Likewise with $2x$ being a shorthand for $x+x$. We can consider "2" to be an element of the field as a shorthand of $1+1$, but it is possible that $2=1+1=0$ in some fields (you will get more explanation of this when you learn about the characteristic of a field).

Again with the additive inverse of $1$, we define $-a$ to mean the additive inverse of $a$, so $-1$ is the additive inverse of $1$. Note that if we have $1+1=0$ in some field, this implies that $-1=1$.