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Rings

Let R R be an integral domain and let x,y∈R x, y \in R Show that (x)=(y) (x) = (y) if and only if ∃λ∈R∗ \exists \lambda \in R^* such that x=λy x = \lambda y
Where (x)={rx:r∈R},R∗={x:x is a unit} (x) = \{rx: r \in R \}, R^* = \{x: x \ is \ a \ unit \} .
Proving from left to right.
In part of the answer it says x∈(x)=(y)⇒x=λy x \in (x) = (y) \Rightarrow x = \lambda y for some λ∈R \lambda \in R and y∈(y)=(x)⇒y=μx y \in (y) = (x) \Rightarrow y = \mu x for some μ∈R \mu \in R . Thus x=λy=λμx⇒x−λμx=0R⇒x(1R−λμ)=0R⇒x=0R x = \lambda y = \lambda \mu x \Rightarrow x-\lambda \mu x = 0_R \Rightarrow x(1_R - \lambda \mu) = 0_R \Rightarrow x=0_R or (1R−λμ)=0R (1_R - \lambda \mu) = 0_R .
If x=0R⇒y=μ0R=0R⇒x=1Ry x = 0_R \Rightarrow y = \mu 0_R = 0_R \Rightarrow x = 1_R y . I don't understand the last implication, how does it imply λ=1R \lambda = 1_R from the case when x=0R x= 0_R ?
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nuodai
17
If x=0x=0 then it implies that y=0y=0, and so x=1.yx=1.y. In fact, for any λ\lambda it'd be true that x=λyx=\lambda y in this case; they just gave a concrete example of a value of λ\lambda that works.
Oh I see, thanks.:biggrin:

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