$(A,+)$ and $(U(A), \cdot)$ are not isomorphic

Question

Let $A$ be a ring with $0\ne1$ and without divisors of $0$. Prove that the groups $(A,+)$ and $(U(A), \cdot)$ are not isomorphic.

I supposed that there is an isomorphism and I got $f(x)=1,\forall x\in A$. Can somebody tell me what can I do now?

Answer 1

$f(0+0)=f(0)=f(0)^2$ implies that $f(0)=1$.

Let $p$ be the characteristtic suppose that $p>0$, $f(p-(p-1))=f(p)f(-(p-1))=f(-(p-1))=f(1)$ implies that $1-p=1$ and $p=2$. This implies that $f(1+1)=f(1)^2=f(0)=1$ $(f(1)-1)(f(1)+1)=0$ since $R$ does not have divisors of zero, and $p=2$, $f(1)=1$ contradiction since $f$ is injective, thus the characteristic is zero.

$f(a)=2,f(b)=3, f(b-a)=f(1)=3/2$, $f(c)=4,f(d)=5$, $f(d-c)=f(1)=5/4$ implies that $5/4=3/2$ and $10=12$ contradiction since no zero divisors.

Answer 2

Assume there existed an isomorphism $f \in \mathrm{Hom}_{\mathrm{Gr}} (A, \mathrm{U}(A))$. For arbitrary integer $n \in \mathbb{Z}$ denote $n_{A}=n1_{A}$. Since $f$ is surjective and $-1_A \in \mathrm{U}(A)$ there exists $a \in A$ such that $f(a)=-1_A$. It follows that $1_A=f(0_A)=\left(-1_A\right)^2=f(a)^2=f(2a)$, whence by injectivity of $f$ we infer $2a=2_Aa=0_A$.

Since $A$ has no zero-divisors, we have either $2_A=0_A$ or $a=0_A$ which also leads to $1_A=f(0_A)=f(a)=-1_A$ and hence $2_A=0_A$. This however entails for any $x \in A$ that $f(x)^2=f(2x)=f(0_A)=1_A$, which by virtue of the surjectivity of $f$ means that for any $y \in \mathrm{U}(A)$ the relation $y^2=1_A$ holds.

Since any element of $A$ commutes with the unity $1_A$, we have the binomial development $(y-1_A)^2=y^2-2y+1_A=y^2-1_A$ (taking $2y=0_A$ and $1_A=-1_A$ again into account), which combined with the previous relation signifies that $(y-1_A)^2=0_A$ for every unit $y \in \mathrm{U}(A)$. By once again appealing to the absence of zero-divisors we infer that $\mathrm{U}(A)=\{1_A\}$, in other words the group of units is trivial. Since this multiplicative group is isomorphic to the additive group $(A, +)$ we gather that the latter is also trivial, in other words $A=\{0_A\}$. This however means that the ring $A$ is null (degenerate) and contradicts our starting premise.