0
$\begingroup$

Let $R$ a ring and $M$ a $R-$module. Is $M$ an ideal of $R$ ? And conversely, if $I$ is an ideal of $R$ is it a $R-$module ? To me, I'd say yes (obviously), but a friend of mine told me that it's wrong... So I just want your opinion.

  • 1
    The first question does not make sense. An ideal of $R$ is in particular a subset of $R$, which an $R$-module need not be. – 2017-01-26
  • 0
    of, then is $M$ is a subset of $R$ which is an $R-$module. Is $M$ an ideal ? I know that an ideal is an $R-$module, but is the converse true (when $M$ is a subset of $R$) @TobiasKildetoft – 2017-01-26
  • 3
    If $M$ is a subset of $R$ and the module structure is given by the multiplication in $R$ then yes, $M$ is an ideal (the definitions end up precisely coinciding in fact, so there is not even anything to show). – 2017-01-26
  • 0
    However, $\Bbb Z$ is a subset of the ring $\Bbb Z[X]$, which is a $\Bbb Z[X]$-module by defining $P \cdot n = nP(0)$ (i.e. $\Bbb Z \cong \Bbb Z[X]/(X)$ induces a structure of $\Bbb Z[X]$-module on $\Bbb Z$), but $\Bbb Z$ is not an ideal of $\Bbb Z[X]$. – 2017-01-26
  • 0
    Let be $R$ is a ring and $I$ is a left ideal in $R$ then $I$ is a $R-$ left module and if $I$ is a right ideal in $R$ then $I$ is a $R-$ right module – 2017-01-26
  • 0
    @Mustafa : not quite, you need that the $R$-module structure is given by the multiplication of $R$. – 2017-01-26
  • 0
    if we define multiply in module the multiply in ring – 2017-01-26

1 Answers 1

5

Let $R$ a ring and $M$ a $R-$module. Is $M$ an ideal of $R$ ?

Certainly not, as $M$ might not even be a subset of $R$. For example, $\mathbb Z^2$ is a $\mathbb Z$-module.