# I Simple Modules and Right Annihilator Ideals ...

1. Feb 6, 2017

### 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:

In the above text from Bland ... in the proof of (1) we read the following:

" ... ... we see that $\text{ann}_r( R / \mathfrak{m} ) = \text{ann}_r(S)$. But $a \in \text{ann}_r( R / \mathfrak{m} )$ implies that $a + \mathfrak{m} = ( 1 + \mathfrak{m} ) a = 0$ , so $a \in \mathfrak{m}$. "

Can someone please explain exactly why $a \in \text{ann}_r( R / \mathfrak{m} )$ implies that $a + \mathfrak{m} = ( 1 + \mathfrak{m} ) a = 0$ ... ... ?

Can someone also explain how this then implies that $a \in \mathfrak{m}$ ... ?

Hope someone can help ...

Peter

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2. Feb 6, 2017

### andrewkirk

Assume $a \in \text{ann}_r( R / \mathfrak{m} )$. That means that, for any element of $R/\mathfrak m$, say $s+\mathfrak m$, we have
$$(s+\mathfrak m)a=0_{R/\mathfrak m}\ \ \ \ \ (1)$$
Hence, in particular, we have
$$(1+\mathfrak m)a=0_{R/\mathfrak m}\ \ \ \ (2)$$
(I assume, since the proof refers to $1$, that $R$ is assumed to contain a multiplicative identity, ie it is a 'unitary ring')

Now by definition of multiplication in a quotient:
$$(1+\mathfrak m)a=1\times a+\mathfrak m=a+\mathfrak m\ \ \ \ (3)$$
and by (2) this is equal to $0_{R/\mathfrak m}$, which is $\mathfrak m$.

Thus
$$a+\mathfrak m=\mathfrak m\ \ \ \ (4)$$
which means that $a\in\mathfrak m$.

3. Feb 7, 2017

### Math Amateur

Thanks Andrew ... most clear and very helpful ...