Prime field of a characteristic zero field is isomorphic to $\mathbb{Q}$ I was attempting to prove that the prime field $P$ of a characteristic 0 field $K$ is isomorphic  to $\mathbb{Q}$. Here, I will use the notation
$nx=\left\{\begin{array}{cc}\underbrace{x+\cdots+x}_{n \text{ times}} & \text{if }n>0\\ 0 & \text{if } n=0 \\ \underbrace{(-x)+\cdots+(-x)}_{n \text{ times}} & \text{if }n<0 \end{array}\right.$
Firstly, I showed that $P=\{(m1)\cdot(n1)^{-1}; m,n\in\mathbb{Z},n\neq 0\}$, where 1 is the mulitplicative identity of $K$. Then, I showed that $\phi:\mathbb{Q}\to P$ such that $\dfrac{m}{n}\mapsto(m1)\cdot(n1)^{-1}$ is the isomorphism that I needed to build.
I accessed the ProofWiki to check my answer and there I found this page with the proof of the same thing but, in their proof, they show that $\phi$ is well-defined, i.e., given $\dfrac{a}{b}=\dfrac{c}{d}$, we have that $\phi\left(\dfrac{a}{b}\right)=\phi\left(\dfrac{c}{d}\right)$. Why is it necessary to show that $\phi$ is well-defined?

If you define a map $\phi : S\to T$, you must always check that $\phi$ is well defined; if it isn't well defined, it's not a function at all! More accurately, whenever you define $\phi$ based on a presentation of an element $x\in S$, you must check that if $x'$ is a different presentation of $x$, then $\phi(x) = \phi(x')$. If $\phi$ is to be a function, then $\phi(x)$ must have one and only one value. So, since $x = x'$, $\phi(x)$ must equal $\phi(x')$.
In this case, a rational number may be written in more than one way: $1/2 = 2/4 = 4/8 = \dots$. So, if you define a map $\phi : \Bbb Q\to S$, and you define $\phi(a/b)$ by some formula that you plug $a$ and $b$ into, then you need to check that if $a/b = c/d$, then the formula gives the same values when you put $c$ and $d$ into it instead of $a$ and $b$. As an example: $\phi : \Bbb Q\to\Bbb Z$ given by $\phi(a/b) = a - b$ is not well defined, since it would imply that $4 = 8 - 4 = \phi(8/4) = \phi(2/1) = 2 - 1 = 1$.
If an element $x$ of your $P$ can be written in more than one way, you must check that your $\phi : P\to\Bbb Q$ gives the same value $\phi(x)$ no matter which way you write $x$.
