Does one element of a field represent some other element of any arbitrary field?

Question

Three elements a,b,c of a field form an arithmetic progression if their successive difference are equal: b-a=c-b. Prove that b=a+c/2. when I start proving it: => b-a+b=c-b+b => b+b-a=c => b+b-a+a=c+a => b+b=c+a in next step, how I write b+b as 2b as it is not specified that b belongs to real number field. for example if F= {(a,b)| a,b belongs to R} where addition is defined as simple vector addition [(a,b)+(c,d)=(a+c,b+d)] and multiplication is defined as pairwise product[(a,b).(c,d)=(a.c,b.d)] Then, (2,3)+(2,3) = (4,6) = (2,2).(2,3) not equal to 2.(2,3) and 2.(2,3) is also not defined. Does 2 in the above question is just a representation of corresponding "2" of an arbitrary field F or it represent something else? Please clarify this.

Answer 1

Your example isn’t a field because there are non-zero elements without multiplicative inverse, e.g. $(1,0)$.

The literal $2$ is generally taken to refer to the element $1+1$ of the field, where $1$ is the multiplicative identity. More generally, an integer is naturally associated with its image under the unique group homomorphism from $\mathbb Z$ to the additive group of the field that maps $1$ to $1$. Thus, $b+b=1\cdot b+1\cdot b=(1+1)\cdot b=2b$.

The problem is implicitly making the assumption that the characteristic of the field is not $2$, that is, that $2\ne0$, so that you can divide through by $2$.