Surjectivity of a composition of two module homomorphisms

Question

I am looking at the proof of the following Lemma: let $V$ be a simple $\mathfrak{m}$-module and $v_1,v_2 \in V$ s.t. $v_1$ is not proportional to $v_2$. Then the smallest submodule of $V \oplus V $ containing $(v_1,v_2)$ is $V \oplus V$. The proof starts as follows: let $U$ be the smallest submodule of $V \oplus V$ containing $(v_1,v_2)$. The inclusion map $i: U \hookrightarrow{} V \oplus V$ is a module homomorphism, as are two projection maps $p_1,p_2: V \oplus V \rightarrow V$. Therefore maps $p_1 \circ i, p_2 \circ i: U \rightarrow V$ are also homomorphisms. And here comes the bit I don't understand: since $V$ is simple, $U$ non-zero and $p_1 \circ i$ a non-zero map, it must be surjective. $U$ contains $(v_1,v_2)$ so the non-zero bit is clear. $p_1 \circ i$ as a composition of non-zero maps must non-zero so that's also ok. But how do I show that $p_1 \circ i$ is surjective using all these things?

Answer 1

First, let us note that both $v_1$ and $v_2$ are non-zero by hypothesis, because every element of $V$ is proportional to $0$. Let us look at $p_1\circ i$, and denote by $W_1$ its image. We know that $W_1$ is a non-zero submodule of $V$, because it contains $v_1=p_1\circ i(v_1,v_2) \not = 0$. But $V$ is simple: it follows that $W_1$ must automatically be the whole of $V$. The same argument goes for $p_2\circ i$.

Be careful about your statement « $p_1\circ i$, as a composition of non-zero maps, must be non-zero ». This is false in general. For instance, in $A=\mathbb Z/4\mathbb Z$ seen as a module over itself, considering $f(x)=2x$ the multiplication-by-$2$ endomophism, we have $f\circ f = 0$ even though $f\not = 0$.

Another remark : it is quite unusual to denote a ring by the letter $\mathfrak m$. Usually, this refers to some maximal ideal of a ring.