Understanding two facts about simple left modules

Question

On one hand, we have

Proposition 1. If $M$ is a left ideal of a ring $R$, satisfying $RM\neq 0$, then $M$ is a simple left $R$-module iff $M$ is a minimal left ideal.

On the other hand, we also have

Proposition 2. A left $R$-module $M$ is simple iff there is a $R$-module isomorphism from $M$ to $R/I$, where $I$ is a modular maximal left ideal.

Question: Does the above mean, if $M$ is a minimal left ideal of $R$, then there exists a maximal modular left ideal $N$ of $R$ such that $R=M\oplus N$? But if this is the case, since $N$ is modular, if $R$ has an identity, wouldn't it mean $M$ is supposed to contain the identity element, and so making it the whole ring? Thanks.

Update:

Based on the answer by @rschwieb, the correct statement should be

For a minimal left ideal $M$ of a ring $R$ such that $RM\neq 0$, let $N$ be the kernel of $R$-module epimorphism $r\mapsto rm$ from $R$ to $M$, for a fixed $m\in M\setminus \{0\}$. Then, there is a $R$-module decomposition $R=M\oplus N$ iff $N\cap M=(0)$.

Proof. By the two propositions, we have exact sequence of $R$-modules $0\to N\to R\to M\to 0$. The inclusion map $\iota: M\hookrightarrow R$ is a section iff $N\cap M=(0)$. Then, by splitting lemma, we have $R= M\oplus N$ as left $R$-modules. $\square$

In this case, if $R$ has an identity, then there exists $m_e\in M$ such that $m_e\in 1_R +N$ or equivalently $m_e-1_R\in N$. This causes no contradictions to the module isomorphism $M\simeq R/N\ni [1_R]$.

Answer 1

Does the above mean, if $$ is a minimal left ideal of $$, then there exists a maximal modular left ideal $$ of $$ such that $=⊕$?

No, it does not mean that. Consider $\mathbb Z/4\mathbb Z$. The ideal $M=2\mathbb Z/4\mathbb Z$ is the unique maximal ideal and $R/M$ is a simple module, in fact, the only simple module for $R$ up to isomorphism. There do not exist any pair $A,B$ of nonzero submodules such that $A\oplus B=R$.

But if this is the case, since is modular, if has an identity, wouldn't it mean is supposed to contain the identity element

No, there is no reason to say that. The definition of a modular right ideal $T$ is that there is an element $r\in R$ such that $rm-m\in T$ for every $m\in T$. When $R$ has identity, then $r=1$ always works (that's why modularity is automatic in rings with identity.) This does not imply at all that "$r\in T$".