TITLE: Radical(R) multiplied by simple $R$ module is $0$
QUESTION [0 upvotes]: Let $R$ be a ring. We define the radical of $R$ bot be the left ideal $N$ which is the intersection of all maximal left ideals of $R$. Show that $NE = 0$ for every simple $R$-module $E$.
I want to say suppose there was an $x \neq 0$ such that $x \in NE$. Then $x = ne$ for some $n \in N, e \in E$ both not equal to $0$. From this I want to say that $nE$ is a proper submodule of $E$, which contradicts its simplicity. However, I haven't used the fact that $N$ is the radical of $R$.
For the second part

REPLY [1 votes]: If $x$ is a non-zero element of $NE$, this doesn’t mean that $x = ne$, where $n \in N$ and $e \in E$ are both non-zero.
If $M$ is a left $R$-module, and $I$ a left ideal of $R$, the elements of $IM$ are sums of elements of the form $im$, with $i \in I$ and $m \in M$. Note that the condition $IM \neq 0$ implies that there exists $i \in I$ and $m \in M$ such that $im \neq 0$ (otherwise, $im=0$ for every $i \in I$ and $m \in M$, and then $IM=0$).
Thus, what you wrote is not entirely incorrect, is just a small writting issue.

Honestly, I don't know how to finish this from what you've done, and I don't think it's that easy. The "standard" approach to this exercise is to use the following two well-known facts.

Every simple left $R$-module is isomorphic to the quotient of $R$ by some maximal left ideal of $R$.

The Jacobson radical of $R$ is a two-sided ideal of $R$.


Notice that by 1 we know that it suffices to show that, for any maximal left ideal $\mathfrak m$ of $R$, $N(R/\mathfrak m)=0$. Indeed, given $\mathfrak m$, for every $r \in R$ and $a \in N$ we have that (by 2) $ar \in N \subseteq \mathfrak m$, and then
$$
a(r+\mathfrak m) = ar+\mathfrak m = 0+\mathfrak m.
$$
Therefore $N(R/\mathfrak m)=0$.


Proof of 1: Given a simple left $R$-module $S$, consider a non-zero $x \in S$. Then the image of the map $f \colon R \to S,\, r \mapsto rx$ is a non-zero submodule of $S$, and hence it has to be $S$. Thus $R/(\ker f) \cong S$, and the simplicity of $S$ implies that $\ker f$ is a maximal left ideal of $R$.

Proof of 2: First observe that $f[N] \subseteq N$ for any $f \in \operatorname{End}_R(R)$. (In particular, if we choose $f$ to be right-multiplication by some element of $R$, it follows that $N$ is a right ideal of $R$).
Indeed, if $\mathfrak m$ is a maximal left ideal of $R$, $R/\mathfrak m$ is simple, and then the image of the composition of $f$ with the canonical projection $\pi \colon R \to R/\mathfrak m$ is either zero or $R/\mathfrak m$.
Since $f^{-1}[\mathfrak m] = \ker(\pi \circ f)$; in the first case, $f^{-1}[\mathfrak m]$ is $R$, and in the second one, $f^{-1}[\mathfrak m]$ is a maximal left ideal of $R$ (because $R/f^{-1}[\mathfrak m] \cong R/\mathfrak m$). In either case, $N \subseteq f^{-1}[\mathfrak m]$, so that $f[N] \subseteq \mathfrak m$.
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