# If $(U^c,+,*)$ is maximal ideal ($U$ group formed by unit elements) then $A/U^c$ is a fields? [closed]

If $$(A,+,*)$$ is a ring, then $$(U,*)$$ is a group ( formed by unit elements). Is it true that if $$(U^c,+,*)$$ is maximal ideal, then $$A/U^c$$ a field?

• If $A$ is a commutative ring with unity, then your argument is true. – M. A. SARKAR Oct 18 at 7:56

Let $$R=D[[x, \sigma]]$$ be skew power series ring or Hilbert twisted power series ring (pege 10 of A first course in noncommutative ring, T. Y. Lam) , were $$D$$ be a division ring but not field and $$\sigma$$ is nontrivial automorphism on $$D$$.
Proposition 1: $$\alpha \in S[[x, \sigma]]$$ is unit iff $$a_{0}\in S$$ is unit.
Proposition 2: Let $$A=B[[x, \sigma]]$$ and $$J(R)$$ be Jacobson radical of $$R$$. Then $$A/J(A) \simeq B/J(B).($$B\$ is any ring). Proof : exercise 6 of Lam's book in page 82.
We know that this example is noncommutative local ring. So $$J(R)=R\backslash U(R) = U^{c}$$ and by proposition 2 at above, $$R/J(R)=R/U^{c}$$ is division ring but not field.