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 \(\displaystyle y \in xR\) it immediately follows that \(\displaystyle xR = yR\) ... ... Can someone please explain exactly why \(\displaystyle y \in xR\) implies that \(\displaystyle xR = yR\) ... ... 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 \(\displaystyle y \in xR\) it immediately follows that \(\displaystyle xR = yR\) ... ... Can someone please explain exactly why \(\displaystyle y \in xR\) implies that \(\displaystyle xR = yR\) ... ... Peter

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

But then we also have \(\displaystyle y \in yR\) ...

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

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

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

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

So ... \(\displaystyle a = xd \in xR\) ...

\(\displaystyle \Longrightarrow yR \subseteq xR\) ... ... ... ... (4)Therefore (1) (4) \(\displaystyle \Longrightarrow xR = yR\) ...
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 $yR \subseteq xR$.