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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?

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    $\begingroup$ If $A$ is a commutative ring with unity, then your argument is true. $\endgroup$ – M. A. SARKAR Oct 18 at 7:56
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No. The following example is counterexample.

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 $D$.

We have following statements:

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.

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