How are the cipher, the key and the initial message (that is not encrypted) are releted?

Question

Suppose that $m$ is a message that someone player $i$ wants to send to a network of other players $j\neq -i$. The player to prevent his message from cheating by others uses an encyrpstion scheme. Say $$g:M\times Y \to X$$ denotes a cipher where $Y$ is the key and $X$ a code that makes the message to look random. The standard assumptions to be made are that $|Y|\geq |M|$ and $g(\cdot,y)$ is a bijection namely every pair of $(m,y)$ is associated with only one $x$. My question is how are the key $y$, the code $x$, and the message $m$ are associated? for example if we could make some operations among $g$, $y$ and $m$, what would that be? could we claim that $x\oplus y \underbrace{=}_{?}m$? or somehting like this?

Answer 1

Taking into account the book. I write here an example. Suppose, that we have a mechanism of communication $\mathcal{M}=(g,h)$ such that $\mathcal{M}$ is defined over $(Y,M,X)$, where $Y$ is the key, $M$ the message and $X$ the cipher spaces respectively. To simplify the problem even more I assume that $Y=M=L=\{0,1\}^l=G$ instead of an arbitrarily finite field $\mathbb{F}^n$ and write below

$$g(y,m)=x,\quad\text{is the encrypted message, which by definition equals $x$}$$

$$h(y,x)=m,\quad\text{is the decrypted message, which by definition equals $m$}$$

So, indeed $(y,x)$ is defined to be associated with only one $m$ and hence $g(y,\cdot)$ is bijective by definition. To anser the question how are they associated, when someone knows both $x$ and $y$, then indeed $x\oplus_{G} y=m$

In order to decrypt the message we have that

$$h(y,x)=h(y,g(y,m))=y\oplus_G x=m$$

where $\oplus_{G}$ is the operation of $+$ as it is defined in the finite field $G$. And hence we have show that the calculation that you ask for, it holds by definition.