$M$ is an irreducible $R$ module $iff$ $M$ is a cyclic module and every nonzero element is a generator
$begingroup$
$M$ is an irreducible $R$ module $iff$ $M$ is a cyclic module and every nonzero element is a generator.
($rightarrow$) If $M$ is an irreducible $R$-module then it's obvious that $M$ is a cylclic module where every nonzero element is a generator, otherwise $mR <_R M$ would be a submodule of M.
($leftarrow$) So, this would be obvious if we knew that every $M$ submodule is somehow related to the cyclic submodules of $M$, like a direct product or something. If $M$ was free then this would be easy, but as it is this suprises me, I feel like there should be some weird counterexample where there is some module $M$ with some submodule not related to its cyclic submodules. Why is this not possible?
abstract-algebra
asked Dec 9 '18 at 20:34
Math is hardMath is hard
817211
$endgroup$
add a comment |
$begingroup$
$M$ is an irreducible $R$ module $iff$ $M$ is a cyclic module and every nonzero element is a generator.
($rightarrow$) If $M$ is an irreducible $R$-module then it's obvious that $M$ is a cylclic module where every nonzero element is a generator, otherwise $mR <_R M$ would be a submodule of M.
($leftarrow$) So, this would be obvious if we knew that every $M$ submodule is somehow related to the cyclic submodules of $M$, like a direct product or something. If $M$ was free then this would be easy, but as it is this suprises me, I feel like there should be some weird counterexample where there is some module $M$ with some submodule not related to its cyclic submodules. Why is this not possible?
abstract-algebra
asked Dec 9 '18 at 20:34
Math is hardMath is hard
817211
$endgroup$
1
$begingroup$
Let $N$ be any nonzero submodule of $M$. Pick any nonzero element $x in N$. $x$ generates $M$, thus $M subseteq N$. End of the proof.
$endgroup$
– Crostul
Dec 9 '18 at 20:45
add a comment |
$begingroup$
$M$ is an irreducible $R$ module $iff$ $M$ is a cyclic module and every nonzero element is a generator.
($rightarrow$) If $M$ is an irreducible $R$-module then it's obvious that $M$ is a cylclic module where every nonzero element is a generator, otherwise $mR <_R M$ would be a submodule of M.
($leftarrow$) So, this would be obvious if we knew that every $M$ submodule is somehow related to the cyclic submodules of $M$, like a direct product or something. If $M$ was free then this would be easy, but as it is this suprises me, I feel like there should be some weird counterexample where there is some module $M$ with some submodule not related to its cyclic submodules. Why is this not possible?
abstract-algebra
asked Dec 9 '18 at 20:34
Math is hardMath is hard
817211
$endgroup$
$M$ is an irreducible $R$ module $iff$ $M$ is a cyclic module and every nonzero element is a generator.
($rightarrow$) If $M$ is an irreducible $R$-module then it's obvious that $M$ is a cylclic module where every nonzero element is a generator, otherwise $mR <_R M$ would be a submodule of M.
($leftarrow$) So, this would be obvious if we knew that every $M$ submodule is somehow related to the cyclic submodules of $M$, like a direct product or something. If $M$ was free then this would be easy, but as it is this suprises me, I feel like there should be some weird counterexample where there is some module $M$ with some submodule not related to its cyclic submodules. Why is this not possible?
abstract-algebra
abstract-algebra
asked Dec 9 '18 at 20:34
Math is hardMath is hard
817211
asked Dec 9 '18 at 20:34
Math is hardMath is hard
817211
asked Dec 9 '18 at 20:34
Math is hardMath is hard
817211
asked Dec 9 '18 at 20:34
Math is hardMath is hard
817211
asked Dec 9 '18 at 20:34
Math is hardMath is hard
817211
817211
1
$begingroup$
Let $N$ be any nonzero submodule of $M$. Pick any nonzero element $x in N$. $x$ generates $M$, thus $M subseteq N$. End of the proof.
$endgroup$
– Crostul
Dec 9 '18 at 20:45
add a comment |
1
$begingroup$
Let $N$ be any nonzero submodule of $M$. Pick any nonzero element $x in N$. $x$ generates $M$, thus $M subseteq N$. End of the proof.
$endgroup$
– Crostul
Dec 9 '18 at 20:45
1
1
$begingroup$
Let $N$ be any nonzero submodule of $M$. Pick any nonzero element $x in N$. $x$ generates $M$, thus $M subseteq N$. End of the proof.
$endgroup$
– Crostul
Dec 9 '18 at 20:45
$begingroup$
Let $N$ be any nonzero submodule of $M$. Pick any nonzero element $x in N$. $x$ generates $M$, thus $M subseteq N$. End of the proof.
$endgroup$
– Crostul
Dec 9 '18 at 20:45
add a comment |
1 Answer
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oldest
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$begingroup$
Note that the statement is not really correct: a module $M$ is irreducible if and only if it is nonzero and every nonzero element of $M$ is a generator.
You're looking too far, in my opinion.
Suppose every nonzero element of $M$ (a nonzero module) is a generator. Let $L$ be a nonzero submodule of $M$; then take $xin L$, $xne0$; then $M=xRsubseteq L$, forcing $L=M$.
For the converse: let $xin M$, $xne0$; then $xR$ is a nonzero submodule of $M$, so $xR=M$ and $x$ is a generator.
answered Dec 9 '18 at 20:45
egregegreg
181k1485203
$endgroup$
add a comment |
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$begingroup$
Note that the statement is not really correct: a module $M$ is irreducible if and only if it is nonzero and every nonzero element of $M$ is a generator.
You're looking too far, in my opinion.
Suppose every nonzero element of $M$ (a nonzero module) is a generator. Let $L$ be a nonzero submodule of $M$; then take $xin L$, $xne0$; then $M=xRsubseteq L$, forcing $L=M$.
For the converse: let $xin M$, $xne0$; then $xR$ is a nonzero submodule of $M$, so $xR=M$ and $x$ is a generator.
answered Dec 9 '18 at 20:45
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
Note that the statement is not really correct: a module $M$ is irreducible if and only if it is nonzero and every nonzero element of $M$ is a generator.
You're looking too far, in my opinion.
Suppose every nonzero element of $M$ (a nonzero module) is a generator. Let $L$ be a nonzero submodule of $M$; then take $xin L$, $xne0$; then $M=xRsubseteq L$, forcing $L=M$.
For the converse: let $xin M$, $xne0$; then $xR$ is a nonzero submodule of $M$, so $xR=M$ and $x$ is a generator.
answered Dec 9 '18 at 20:45
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
Note that the statement is not really correct: a module $M$ is irreducible if and only if it is nonzero and every nonzero element of $M$ is a generator.
You're looking too far, in my opinion.
Suppose every nonzero element of $M$ (a nonzero module) is a generator. Let $L$ be a nonzero submodule of $M$; then take $xin L$, $xne0$; then $M=xRsubseteq L$, forcing $L=M$.
For the converse: let $xin M$, $xne0$; then $xR$ is a nonzero submodule of $M$, so $xR=M$ and $x$ is a generator.
answered Dec 9 '18 at 20:45
egregegreg
181k1485203
$endgroup$
Note that the statement is not really correct: a module $M$ is irreducible if and only if it is nonzero and every nonzero element of $M$ is a generator.
You're looking too far, in my opinion.
Suppose every nonzero element of $M$ (a nonzero module) is a generator. Let $L$ be a nonzero submodule of $M$; then take $xin L$, $xne0$; then $M=xRsubseteq L$, forcing $L=M$.
For the converse: let $xin M$, $xne0$; then $xR$ is a nonzero submodule of $M$, so $xR=M$ and $x$ is a generator.
answered Dec 9 '18 at 20:45
egregegreg
181k1485203
answered Dec 9 '18 at 20:45
egregegreg
181k1485203
answered Dec 9 '18 at 20:45
egregegreg
181k1485203
answered Dec 9 '18 at 20:45
egregegreg
181k1485203
181k1485203
add a comment |
add a comment |
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$begingroup$
Let $N$ be any nonzero submodule of $M$. Pick any nonzero element $x in N$. $x$ generates $M$, thus $M subseteq N$. End of the proof.
$endgroup$
– Crostul
Dec 9 '18 at 20:45