Mixing
If $Ma = (0:_Mb)$, where $(0:_Mb)=lbrace min M | mb=0_Mrbrace$, is it true that $Na=(0:_Nb)$ for any...
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Suppose that for a fixed $ain R$ there exists a $bin R$ such that $Ma = (0:_Mb)$, where $(0:_Mb)=lbrace min M | mb=0_Mrbrace$. Is it possible that for any $R$-submodule, $Nleq_RM$, of $M$ $Na=(0:_Nb)=lbrace nin N | nb=0_Nrbrace$? If not, is there any counter example?
In my proof I am able to show that if $Nleq_RM$, then $Nasubseteq Ma=(0:_Mb)$ and $Nasubseteq N$ so that $Nab=0_N$. Thus, $Nasubseteq (0:_Nb)$. To prove the reverse inclusion, let $xin (0:_Nb)$ so that $xb=0_N$ for some $xin N$. Then $x=na$ for some $nin N$.
This is where I have issues:
(1) Is it right to choose $x=na$? Aren't there any other elements $yin N$ for which $yb=0_N$ but $yneq ta, tin N$?
(1) Is the equality $Na=(0:_Nb)$ correct? If not correct, then are there any possible conditions on $M$ or types of $R$-modules $M$ for which $Na=(0:_Nb)$ holds for all $Nleq_RM$?
ring-theory modules
asked Jan 27 at 8:40
mariammariam
294
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add a comment |
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Suppose that for a fixed $ain R$ there exists a $bin R$ such that $Ma = (0:_Mb)$, where $(0:_Mb)=lbrace min M | mb=0_Mrbrace$. Is it possible that for any $R$-submodule, $Nleq_RM$, of $M$ $Na=(0:_Nb)=lbrace nin N | nb=0_Nrbrace$? If not, is there any counter example?
In my proof I am able to show that if $Nleq_RM$, then $Nasubseteq Ma=(0:_Mb)$ and $Nasubseteq N$ so that $Nab=0_N$. Thus, $Nasubseteq (0:_Nb)$. To prove the reverse inclusion, let $xin (0:_Nb)$ so that $xb=0_N$ for some $xin N$. Then $x=na$ for some $nin N$.
This is where I have issues:
(1) Is it right to choose $x=na$? Aren't there any other elements $yin N$ for which $yb=0_N$ but $yneq ta, tin N$?
(1) Is the equality $Na=(0:_Nb)$ correct? If not correct, then are there any possible conditions on $M$ or types of $R$-modules $M$ for which $Na=(0:_Nb)$ holds for all $Nleq_RM$?
ring-theory modules
asked Jan 27 at 8:40
mariammariam
294
$endgroup$
add a comment |
$begingroup$
Suppose that for a fixed $ain R$ there exists a $bin R$ such that $Ma = (0:_Mb)$, where $(0:_Mb)=lbrace min M | mb=0_Mrbrace$. Is it possible that for any $R$-submodule, $Nleq_RM$, of $M$ $Na=(0:_Nb)=lbrace nin N | nb=0_Nrbrace$? If not, is there any counter example?
In my proof I am able to show that if $Nleq_RM$, then $Nasubseteq Ma=(0:_Mb)$ and $Nasubseteq N$ so that $Nab=0_N$. Thus, $Nasubseteq (0:_Nb)$. To prove the reverse inclusion, let $xin (0:_Nb)$ so that $xb=0_N$ for some $xin N$. Then $x=na$ for some $nin N$.
This is where I have issues:
(1) Is it right to choose $x=na$? Aren't there any other elements $yin N$ for which $yb=0_N$ but $yneq ta, tin N$?
(1) Is the equality $Na=(0:_Nb)$ correct? If not correct, then are there any possible conditions on $M$ or types of $R$-modules $M$ for which $Na=(0:_Nb)$ holds for all $Nleq_RM$?
ring-theory modules
asked Jan 27 at 8:40
mariammariam
294
$endgroup$
Suppose that for a fixed $ain R$ there exists a $bin R$ such that $Ma = (0:_Mb)$, where $(0:_Mb)=lbrace min M | mb=0_Mrbrace$. Is it possible that for any $R$-submodule, $Nleq_RM$, of $M$ $Na=(0:_Nb)=lbrace nin N | nb=0_Nrbrace$? If not, is there any counter example?
In my proof I am able to show that if $Nleq_RM$, then $Nasubseteq Ma=(0:_Mb)$ and $Nasubseteq N$ so that $Nab=0_N$. Thus, $Nasubseteq (0:_Nb)$. To prove the reverse inclusion, let $xin (0:_Nb)$ so that $xb=0_N$ for some $xin N$. Then $x=na$ for some $nin N$.
This is where I have issues:
(1) Is it right to choose $x=na$? Aren't there any other elements $yin N$ for which $yb=0_N$ but $yneq ta, tin N$?
(1) Is the equality $Na=(0:_Nb)$ correct? If not correct, then are there any possible conditions on $M$ or types of $R$-modules $M$ for which $Na=(0:_Nb)$ holds for all $Nleq_RM$?
ring-theory modules
ring-theory modules
asked Jan 27 at 8:40
mariammariam
294
asked Jan 27 at 8:40
mariammariam
294
edited Feb 1 at 22:18
mariam
asked Jan 27 at 8:40
mariammariam
294
asked Jan 27 at 8:40
mariammariam
294
asked Jan 27 at 8:40
mariammariam
294
294
add a comment |
add a comment |
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