# surjective homomorphism between unitary rings

Let $f$ be a surjective homomorphism from a unitary ring $R$ to another unitary ring $R\,^{\prime}$.  Then

• $f(1)\;=\;1^{\prime},$

• $f(a^{-1})\;=\;(f(a))^{-1}$  for all elements $a$ belonging to the group of units of $R$.

Proof.$1^{\circ}$.  In a ring, the identity element is unique, whence it suffices to show that $f(1)$ has the properties required for the unity of the ring $R\,^{\prime}$.  When $a^{\prime}$ is an arbitrary element of this ring, there is by the surjectivity an element $a$ of $R$ such that  $f(a)=a^{\prime}$.  Thus we have

 $f(1)a^{\prime}\;=\;f(1)f(a)\;=\;f(1a)\;=\;f(a)\;=\;a^{\prime},\quad a^{\prime}% f(1)\;=\;f(a)f(1)\;=\;f(a1)\;=\;f(a)\;=\;a^{\prime}.$

$2^{\circ}$.  Let $a$ be a unit of $R$.  Then

 $f(a)f(a^{-1})\;=\;f(aa^{-1})\;=\;f(1)\;=\;1^{\prime},\quad f(a^{-1})f(a)\;=\;f% (a^{-1}a)\;=\;f(1)\;=\;1^{\prime},$

whence $f(a^{-1})$ is a multiplicative inverse of $f(a)$.

Title surjective homomorphism between unitary rings SurjectiveHomomorphismBetweenUnitaryRings 2013-03-22 19:10:22 2013-03-22 19:10:22 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 16B99 msc 13B10 IsomorphismSwappingZeroAndUnity