How do we introduce subtraction from these field axioms? I am familiar with two different sets of field axioms. The first one is from "Mathematical Analysis" by Apostol. It has the first $3$ usual axioms, but the $4^{th}$ one is different:
Axiom 1:
Commutative Laws
$x+y=y+x$, $xy=yx$
Axiom 2: Associative Laws
$x+(y+z)=(x+y)+z$, $x(yz)=(xy)z$
Axiom 3: Distributive Law
$x(y+z)=xy+xz$
Axiom 4:
Given any two real numbers $x$ and $y$, there exists a real number $z$ such that $x+z=y$. This $z$ is denoted by $y-x$; the number $x-x$ is denoted by $0$ (it can be proved that $0$ is independent of $x$.) We write $-x$ for $0-x$ and call $-x$ the negative of $x$. 

Subtraction doesn't immediately follow as an operation from axiom 4, but we can prove that $y-x=y+(-x)$ and define subtraction this way. However, more commonly have seen axiom 4 being replaced by
$x+ (-x) = 0$
$x + 0 = x$
How do we introduce subtraction with this set of axioms? I have seen problems giving the second version of axiom 4 and having $y-x$, should I just assume it's meant as $y+(-x)$?
 A: Subtraction $y-x$ is defined as $y+(-x)$, where $-x$ is defined in axiom 4.
A: Edit: In order to prove unicity of the element $0$ defined in Axiom 4, it seems necessary to upgrade Axiom 4 to a stronger statement, namely:
Axiom 4:
Given any two real numbers $x$ and $y$, there exists a unique real number $z$ such that $x+z=y$. This $z$ is denoted by $y-x$; the number $x-x$ is denoted by $0$ (it can be proved that $0$ is independent of $x$.) We write $-x$ for $0-x$ and call $-x$ the negative of $x$. 

Hopefully this clears up the remaining confusion.
Axiom 5: There exists a real number $0$ such that $z + 0= z$ for any real number $z$. For any real number $x$, there exists a real number $-x$ satisfying $x + (-x) = 0$.
Note that $0$ is unique, for if $0'$ is another such number then $0 = 0 + 0' = 0'$. Also note that the inverse element $-x$ is uniquely determined by $x$, for if $z$ is any other number such that $x + z = 0$, we must have $$z = z + 0 = z + (x + (-x)) = (z + x) + -x = 0 + (-x) = -x.$$ (Note that we need associativity here.)
Fact: Axiom 4 $\iff$ Axiom 5.
Assume Axiom 4 holds. Define $0 := z -z$, where $z$ is any real number. Axiom 4 assures us that this is a well-defined real number and that $z + 0 = z$ for any real number $z$. Given a real number $x$, we set $-x := 0 - x$, which is a real number whose existence is assured by Axiom 4. By the definition of $0 - x$ given in Axiom 4, this means that $x + (-x) = 0$, so Axiom 5 is satisfied.
Conversely, assume Axiom 5 holds. Let $x,y$ be real numbers. To prove Axiom 4, we need to define a real number $y - x$ satisfying $x + (y-x) = y$, and prove that it is unique. In fact, if this relation were to hold, we could add the number $-x$ provided by Axiom 5 to both sides to get (using commutativity and associativity)
$$ y + (-x) = x + (y-x) + (-x) = x + (-x) + (y-x) = 0 + (y-x) = y - x.$$
Hence we see that we have to define $y - x := y + (-x)$, where $-x$ is the real number whose existence is assured by Axiom 5. This gives us the unicity part. Then, by the definitions of $-x$ and $0$ given in Axiom 5 and the other axioms we have that $x+ (y-x) = x + y + (-x) = x + (-x) + y = 0 + y = y$, which proves the existence part and hence Axiom 4.
