1
$\begingroup$

I have stumbled upon two different definitions of a submodule and I cant see why they are equivalent, let me state them ($M$ is a module and $R$ a commutative ring):

First Definition: A set $N \subset M$ is called a submodule of $M$ if $$an + bm \in N$$ for all $a,b \in R$ and all $m,n \in N$

Second definition: A set $N \subset M$ is called a submodule of $M$ if (i) $m+n \in N$ for all $m,n \in N$ and (ii) $an \in N$ for all $a \in R$ and $n \in N$.

However I cant see they the definitions are equivalent for instance; if the second definition holds then the first holds, but the converse is not true? If the first definitions holds then we can choose $a=b=1$ thus (i) in the second definition holds but how can I get (ii) to hold in the second definition? More precisely: The fact that $an+bm \in N$ does not imply that $an \in N$, is this correct or am I missing something?

$\endgroup$
3
  • $\begingroup$ Both definitions are missing a condition: we must also have $0 \in N$. $\endgroup$
    – Zhen Lin
    Mar 31, 2015 at 23:08
  • $\begingroup$ @ZhenLin sure you are right, but this does still not answer my question. $\endgroup$
    – jdaUU
    Mar 31, 2015 at 23:14
  • $\begingroup$ @ZhenLin why? Can't we get this from the fact that $0 \in R$? $\endgroup$
    – lanf
    Mar 29 at 17:25

1 Answer 1

Reset to default
3
$\begingroup$

There must be a $0$ in the ring $A$. Let $b = 0$ in $an + bm \in N$; then $an + 0m = an \in N$.

$\endgroup$
2
  • $\begingroup$ What I thought first aswell, but why $0m =0$? Part of the definition of a module, we have $1_A m = m$ but not $0m=m$? What am I missing? $\endgroup$
    – jdaUU
    Mar 31, 2015 at 23:12
  • 1
    $\begingroup$ As part of the definition of a module $M$, we need scalar multiplication to be a homomorphism so that $(a+b)m = am + bm$, for $a,b \in A$ and $m \in M$. If we let this $b=0$, $(a+0)m = am = am +bm$, we must have $bm = 0m = 0$. $\endgroup$
    – calavicci
    Mar 31, 2015 at 23:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .