Why Does y ∈ xR Imply xR = yR in Theorem 3.2.19?

[Math Amateur]

I am reading The Basics of Abstract Algebra by Paul E. Bland ...

I am focused on Section 3.2 Subrings, Ideals and Factor Rings ... ...

I need help with the proof of Theorem 3.2.19 ... ... Theorem 3.2.19 and its proof reads as follows:
View attachment 8269In the above proof by Bland we read the following:"... ... If $$y \in xR$$ it immediately follows that $$xR = yR$$ ... ... Can someone please explain exactly why $$y \in xR$$ implies that $$xR = yR$$ ... ... Peter

 

[Math Amateur]

I am reading The Basics of Abstract Algebra by Paul E. Bland ...

I am focused on Section 3.2 Subrings, Ideals and Factor Rings ... ...

I need help with the proof of Theorem 3.2.19 ... ... Theorem 3.2.19 and its proof reads as follows:
In the above proof by Bland we read the following:"... ... If $$y \in xR$$ it immediately follows that $$xR = yR$$ ... ... Can someone please explain exactly why $$y \in xR$$ implies that $$xR = yR$$ ... ... Peter

I've been thinking about my question ... and think i have an answer ... as follows ...... to show $$y \in xR \Longrightarrow xR = yR$$ ...Assume $$y \in xR$$ ...

But then we also have $$y \in yR$$ ...

$$\Longrightarrow xR \subseteq yR$$ ... ... ... ... (1)Now ... let $$a \in yR$$ ... in addition to our assumption that $$y \in xR$$ ...

then $$ a = yb$$ for some $$b \in R$$ ... ... ... ... (2)

But $$y \in xR$$ so $$y = xc$$ for some $$c \in R$$ ... ... ... ... (3)

Now ... (2) (3) $$\Longrightarrow a = xcb = xd$$ where $$d \in R$$

So ... $$a = xd \in xR$$ ...

$$\Longrightarrow yR \subseteq xR$$ ... ... ... ... (4)Therefore (1) (4) $$\Longrightarrow xR = yR$$ ...
Is the above proof correct?Peter

 

[steenis]

(4) is correct, but I am afraid that (1) is not correct.

In the proof it is supposed that $yR$ is an ideal such that $xR \subseteq yR \subseteq R$.

Furthermore, $xR$ is a nonzero ideal and $y \in xR$.

$xR$ is an ideal, so for for all $r \in R$ we have $yr \in xR$, thus $yR \subseteq xR$.