Matt Myers
Problem 6, 6.1 #17
November 11, 2009
Let R be a ring with identity and let I be an ideal in R.
a) If 1R ∈ I, prove that I = R.
Proof. Let r be an element of R. If 1R is an element of I, then by the
definition of an ideal, we have r · 1R = r ∈ I. Therefore every element of R
is an element of I, so R ⊆ I. But I is an ideal, so I is a subring of R, and
I ⊆ R. So we have I = R.
b) If I contains a unit, prove that I = R.
Proof. Let u ∈ I be a unit in R, and let r be any element of R. Since R
is closed under multiplication, we have ru−1 ∈ R, and u−1 r ∈ R. Then
(ru−1 )u = r = u(u−1 r), so r ∈ I. Therefore every element of R is an element
of I, and I is a subring of R, so I = R.
Alternatively, if u\in I, then 1_R = u*u^(-1) \in I, so I=R by part (a).
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