Characteristic of integral domain if $(x+1)^2=x^2+1$

Question

Let $A$ be an integral domain. Prove that if $(x+1)^2=x^2+1$ in $A[x]$, then $A$ must have characteristic $2$.

We have $x^2+2x+1=x^2+1$, so $2=0$, and hence the characteristic must be $1$ or $2$. Now, I don't see anything wrong with the characteristic being $1$. So, is the problem correct?

Answer 1

The characteristic of an integral domain is the smallest positive integer $n$ such that $n \cdot 1_A= 0_A$. So if $n=1$, then $1_A = 0_A$ in your integral domain $A$. But, we usually (i.e. pretty much always) assume that an integral domain must have unity different from the $0$. Note that a characteristic of an integral domain must always be prime.

Answer 2

What does characteristic equal to $1$ mean? It means that $1 x =x = 0$ for all $x.$ So, your $A$ is the zero ring, in particular it has no $1,$ in particular not an integral domain.