If $E\cong F$ and $E\subset F$, can we say that $E=F$?

Question

Let $R $ be a commutative ring and $E $ and $F $ be two $R$-modules such that : $$ E\cong_{R} F~~and~~E\subset F. $$ Can we say that $$E=F\ ?$$

Answer 1

Absolutely not. For example, take $F=\mathbb Z\times \mathbb Z$ and take the submodule $E$ generated by $(2,0)$ and $(0,2)$. Then your conditions are satisfied but $E\neq F$. More simply, we could even take $F=\mathbb Z$ and $E=2\mathbb Z$.

This is clearly true by cardinality if $F$ is finite, however.

Answer 2

About the simplest example you can think of will easily contradict this. $\Bbb Z$-modules are just Abelian groups, and the additive group $\Bbb Z$ has infinitely many submodules $n\Bbb Z$, all of which but one are isomorphic to $\Bbb Z$, yet all of which but one are unequal to $\Bbb Z$.

More generally whenever you have an integral domain $R$ that is not a field, so it contains nonzero elements that are neither invertible nor zero divisors, then the multiplication map $R\to R$ by such an element will be injective without being surjective, and its image is a submodule of $R$ isomorphic to but distinct from $R$.