I'm trying to solve the following problem:
I have shown that the map is definable with 0. In trying to prove that the map is not definable without parameters, this is my work so far:
Let $G= \{ (a,b,c) : (a,b)\in A\times A \} $ be the graph of the addition map. Suppose that $G$ is definable over $A$ without parameters. Then $\exists \phi(u_1,u_2,u_3)$ formula such that $$(b_1,b_2,b_3)\in G \iff A\vDash \phi[b_1,b_2,b_3]$$
But, a 3-tuple is in the graph iff the third element is the sum of the first two. So, $$(b_1,b_2,b_3)\in G \iff b_1+b_2=b_3 \iff b_1+b_2=b_3+0\iff (b_1,b_2,b_3,0)\in R^A$$
Thus we get the equivalence $$A\vDash \phi[b_1,b_2,b_3]\iff A\vDash R[b_1,b_2,b_3,0]$$
This is where I'm unsure if I can continue further, or if this is the wrong direction entirely. I want to conclude that the formula and relation must be logically equivalent in the structure, but this would only be true if the equivalence held for all assignments of variables. While it does hold for any assignment of $b_1,b_2,b_3$, the 0 cannot be varied, so I do not think this would mean they are equivalent. If this is the right direction, how might I reach a contradiction? If it is not the right direction, how should I have started instead?
If the formula $\phi$ for addition map is definable over $(A,R^A)$ without parameters, then you can recover $0$ from $\phi$ (as the unique element such that $A,R^A\vDash\phi[0,0,0]$). So $(A,R^A)$ alone is sufficient to determine $0$, which you should know is not true because you could make any $a\in A$ the zero element of the group structure using $\phi[x,y,z]\iff R^A[x,y,z,a]$.