Simple Modules and Right Annihilator Ideals .... Bland, Proposition 6.1.7 ....

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 6.1 The Jacobson Radical ... ...

I need help with the proof of Proposition 6.1.7 ... Proposition 6.1.7 and its proof read as follows:View attachment 6394
View attachment 6395
In the above text from Bland ... in the proof of (1) we read the following:" ... ... we see that \(\displaystyle \text{ann}_r( R / \mathfrak{m} ) = \text{ann}_r(S)\). But \(\displaystyle a \in \text{ann}_r( R / \mathfrak{m} )\) implies that \(\displaystyle a + \mathfrak{m} = ( 1 + \mathfrak{m} ) a = 0\) , so \(\displaystyle a \in \mathfrak{m}\). "Can someone please explain exactly why \(\displaystyle a \in \text{ann}_r( R / \mathfrak{m} )\) implies that \(\displaystyle a + \mathfrak{m} = ( 1 + \mathfrak{m} ) a = 0 \) ... ... ?

Can someone also explain how this then implies that \(\displaystyle a \in \mathfrak{m}\) ... ?Hope someone can help ...

Peter

 

[steenis]

Q2: If $I$ is a two-sided ideal in a ring $R$, than $R/I$ is also a ring,
with addition $(a+I)+(b+I)=(a+b)+I$,
and multiplication $(a+I)(b+I)=ab+I$.
The zero element of $R/I$ is $0+I=I$, because $(a+I) + (0+I) = (a+I)$.
We have $a+I=b+I \Leftrightarrow a-b \in I$,
so if $a+I=I=0+I$ then $a=a-0 \in I$, i.e., $a+I=\bar{0}$ then $a \in I$

Q1: Let $M$ be a maximal two-sided ideal in a ring $R$ (Bland only speaks of right ideals, but I think $M$ has to be a two-sided ideal in order to $R/M$ be a ring (is this English?)).

$ann_r(R/M) =$
$\{ x \in R \mbox{ } | \mbox{ } \forall \bar{y} \in R/M \mbox{ } \bar{y}x = \bar{0} \} = $
$\{ x \in R \mbox{ } | \mbox{ } \forall \bar{y} \in R/M \mbox{ } (y+M)x = \bar{0} \} $

$Ma = M$

$a \in ann_r(R/M)$ thus
$a+M= a+Ma = (1+M)a = \bar{0}$ thus $a \in M$