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So, let's define "modular ideal".

The modular ideal is the left ideal $I$ of the ring $R$ with the following property: in the ring $R$ there is at least one such element $x$. That for all $\forall r \in R$ the difference $rx-r$ belongs to $I$.

I have M - modular maximal left ideal of the algebra R. Also element $s \in R \setminus M.$

I should prove that:

$ (M : s)_l = Ms^{-1} = \{ r \in R \, | rs \in M \} $ - modular maximal left ideal of the algebra R.

I understand that, I must show, that this ideal is modular or equals R.

Is the mapping $ r \mapsto rs + M $ epimorphism of $R$-module $R$ on the submodule $Rs + M = R(s + M)$ irreducible $R$-module $R / M $? How can I prove this statement? Do you have any ideas?

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    $\begingroup$ $R\setminus M$ is not the notation for the quotient module $R/M$. It's the notation for the set difference. Are you sure that isn't what you mean? $\endgroup$ – rschwieb Jun 15 '17 at 23:48
  • $\begingroup$ @rschwieb Sorry, I make mistake. I have corrected. $\endgroup$ – Dimatywe1 Jun 16 '17 at 0:27
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Since $M$ is maximal, $M+Rs$ is either $M$ or $R$.

Now, you're on the right track when you notice that $Ms^{-1}$ is the kernel of the map from $R\to \frac{M+Rs}{M}$. Consequently, $\frac{R}{Ms^{-1}}\cong \frac{M+Rs}{M}$.

If $M=M+Rs$, we are obviously in the case where $Ms^{-1}=R$. Otherwise $\frac{R}{Ms^{-1}}\cong R/M$ and there exists an $x\in R$ such that $r\equiv rx\pmod M$. Obviously the image of $x$ in $\frac{R}{Ms^{-1}}$ has the same property, so $Ms^{-1}$ is modular.

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