Let a, b, c be such that ab + c = 1 in a ring R. If there exists x ∈ R such that a + cx ∈ U(R), show that there exists y ∈ R such that b + yc ∈ U(R).
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Expert's answer
2012-10-25T09:54:59-0400
Write u = a + cx ∈U(R). We claim that the element y : = (1 − bx)u−1 works, i.e. v : = b + (1 − bx)u−1c ∈U(R). To see this, note that vx = bx + (1 − bx)u−1(u − a) = 1 − (1 − bx)u−1a, vx(1 − ba) = 1 − ba − (1 − bx)u−1a(1 − ba) = 1− ba − (1 − bx)u−1(1 − ab)a = 1− [b + (1 − bx)u−1c]a = 1− va. Therefore, for w : = a + x(1 − ba), we have vw = 1. We finish by showing that wv = 1 (for then v ∈U(R)). Note that wb = ab + xb(1 − ab) = ab + xbc, w(1 − bx) = a + x(1 − ba) − abx − xbcx = a + (1 − ab)x − xb(a + cx) = a + cx − xbu, w(1 − bx)u−1c = c − xbc. Adding the first and the last equation yields wv = ab + c = 1, as desired.
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