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# Answer to Question #16738 in Algebra for sanches

Question #16738
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).
Write u = a + cx &isin; U(R). We claim that the element y : = (1 &minus; bx)u&minus;1 works, i.e. v : = b + (1 &minus; bx)u&minus;1c &isin; U(R). To see this, note that
vx = bx + (1 &minus; bx)u&minus;1(u &minus; a) = 1 &minus; (1 &minus; bx)u&minus;1a,
vx(1 &minus; ba) = 1 &minus; ba &minus; (1 &minus; bx)u&minus;1a(1 &minus; ba) = 1&minus; ba &minus; (1 &minus; bx)u&minus;1(1 &minus; ab)a = 1&minus; [b + (1 &minus; bx)u&minus;1c]a = 1&minus; va.
Therefore, for w : = a + x(1 &minus; ba), we have vw = 1. We finish by showing that wv = 1 (for then v &isin; U(R)). Note that
wb = ab + xb(1 &minus; ab) = ab + xbc,
w(1 &minus; bx) = a + x(1 &minus; ba) &minus; abx &minus; xbcx
= a + (1 &minus; ab)x &minus; xb(a + cx)
= a + cx &minus; xbu,
w(1 &minus; bx)u&minus;1c = c &minus; xbc.
Adding the first and the last equation yields wv = ab + c = 1, as desired.

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