Mixing
Let $R$ be an integral domain with a field of fractions $F$. Show that if $M$ is a $R$-module, then …
$begingroup$
Let $R$ be an integral domain with a field of fractions $F$. Show that if $M$ is a $R$-module, then:
$1)$ $M_F=Fotimes_R M$ is a divisible $R$-module
$2)$ $xrightarrow 1otimes x$ is an essential monomorphism of $M$ into $M_F$ if $M$ is torsion-free.
Attempts: I have read over several of the other posts to piece the above together, as well as other resources, but I am still having trouble for both parts of this particular question. I was focused on the following: Let $R$ be an integral domain and let $S=Rsetminus 0$. Then $Tor(R)=0$ such that $S^{-1}R$ is the field of fractions of $R$.
Any help is greatly appreciated.
abstract-algebra commutative-algebra modules injective-module
asked Jan 27 at 8:10
user565684user565684
707
$endgroup$
add a comment |
$begingroup$
Let $R$ be an integral domain with a field of fractions $F$. Show that if $M$ is a $R$-module, then:
$1)$ $M_F=Fotimes_R M$ is a divisible $R$-module
$2)$ $xrightarrow 1otimes x$ is an essential monomorphism of $M$ into $M_F$ if $M$ is torsion-free.
Attempts: I have read over several of the other posts to piece the above together, as well as other resources, but I am still having trouble for both parts of this particular question. I was focused on the following: Let $R$ be an integral domain and let $S=Rsetminus 0$. Then $Tor(R)=0$ such that $S^{-1}R$ is the field of fractions of $R$.
Any help is greatly appreciated.
abstract-algebra commutative-algebra modules injective-module
asked Jan 27 at 8:10
user565684user565684
707
$endgroup$
$begingroup$
What is your definition of an essential monomorphism? I've assumed for now that it means the image is an essential submodule of $M_F$.
$endgroup$
– Servaes
Jan 27 at 9:00
$begingroup$
Yes, that is correct. Thank you.
$endgroup$
– user565684
Jan 27 at 9:05
add a comment |
$begingroup$
Let $R$ be an integral domain with a field of fractions $F$. Show that if $M$ is a $R$-module, then:
$1)$ $M_F=Fotimes_R M$ is a divisible $R$-module
$2)$ $xrightarrow 1otimes x$ is an essential monomorphism of $M$ into $M_F$ if $M$ is torsion-free.
Attempts: I have read over several of the other posts to piece the above together, as well as other resources, but I am still having trouble for both parts of this particular question. I was focused on the following: Let $R$ be an integral domain and let $S=Rsetminus 0$. Then $Tor(R)=0$ such that $S^{-1}R$ is the field of fractions of $R$.
Any help is greatly appreciated.
abstract-algebra commutative-algebra modules injective-module
asked Jan 27 at 8:10
user565684user565684
707
$endgroup$
Let $R$ be an integral domain with a field of fractions $F$. Show that if $M$ is a $R$-module, then:
$1)$ $M_F=Fotimes_R M$ is a divisible $R$-module
$2)$ $xrightarrow 1otimes x$ is an essential monomorphism of $M$ into $M_F$ if $M$ is torsion-free.
Attempts: I have read over several of the other posts to piece the above together, as well as other resources, but I am still having trouble for both parts of this particular question. I was focused on the following: Let $R$ be an integral domain and let $S=Rsetminus 0$. Then $Tor(R)=0$ such that $S^{-1}R$ is the field of fractions of $R$.
Any help is greatly appreciated.
abstract-algebra commutative-algebra modules injective-module
abstract-algebra commutative-algebra modules injective-module
asked Jan 27 at 8:10
user565684user565684
707
asked Jan 27 at 8:10
user565684user565684
707
asked Jan 27 at 8:10
user565684user565684
707
asked Jan 27 at 8:10
user565684user565684
707
asked Jan 27 at 8:10
user565684user565684
707
707
$begingroup$
What is your definition of an essential monomorphism? I've assumed for now that it means the image is an essential submodule of $M_F$.
$endgroup$
– Servaes
Jan 27 at 9:00
$begingroup$
Yes, that is correct. Thank you.
$endgroup$
– user565684
Jan 27 at 9:05
add a comment |
$begingroup$
What is your definition of an essential monomorphism? I've assumed for now that it means the image is an essential submodule of $M_F$.
$endgroup$
– Servaes
Jan 27 at 9:00
$begingroup$
Yes, that is correct. Thank you.
$endgroup$
– user565684
Jan 27 at 9:05
$begingroup$
What is your definition of an essential monomorphism? I've assumed for now that it means the image is an essential submodule of $M_F$.
$endgroup$
– Servaes
Jan 27 at 9:00
$begingroup$
What is your definition of an essential monomorphism? I've assumed for now that it means the image is an essential submodule of $M_F$.
$endgroup$
– Servaes
Jan 27 at 9:00
$begingroup$
Yes, that is correct. Thank you.
$endgroup$
– user565684
Jan 27 at 9:05
$begingroup$
Yes, that is correct. Thank you.
$endgroup$
– user565684
Jan 27 at 9:05
add a comment |
1 Answer
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active
oldest
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$begingroup$
By definition an $R$-module $A$ is divisible if for every $xin A$ and every non-zero divisor $rin R$ there exists $yin A$ such that $x=ry$. As $F$ is the field of fractions of $R$ and $M_F=Fotimes_RM$, every $xin M_F$ is of the form
$$x=sum_{i=1}^nf_iotimes m_i=sum_{i=1}^nfrac{r_i}{s_i}otimes m_i,$$
where $m_iin M$ and $f_iin F$ and $r_i,s_iin R$ with $s_ineq0$, because $R$ is an integral domain. It follows that for every non-zero $rin R$ we have
$$y:=rcdotsum_{i=1}^nfrac{r_i}{rs_i}otimes m_i,$$
satisfies $x=ry$, so $M_F$ is divisible.
It is clear that the map
$$M longrightarrow M_F: x longmapsto 1otimes x,$$
is a monomorphism, and its image (which I will also denote $M$) is an essential submodule of $M_F$ because if $Nsubset M_F$ is a submodule of $M_F$ such that $Mcap N=0$, then as before, for every $nin N$ we can write
$$n=sum_{i=1}^kfrac{r_i}{s_i}otimes m_i,$$
for some $kinBbb{N}$ and $r_i,s_iin R$ with $s_ineq0$ and $m_iin M$. Clearing denominators and setting $t_i:=frac{r_i}{s_i}prod_{i=j}^ks_jin R$ we find that
$$(s_1cdots s_k)n=sum_{i=1}^kt_iotimes m_i=sum_{i=1}^k1otimes t_im_1=1otimessum_{i=1}^kt_im_i,$$
and so $(s_1cdots s_k)n=0$. Because $M$ is torsion-free it follows that $n=0$, and hence that $N=0$. This means $M$ is an essential submodule of $M_F$.
answered Jan 27 at 8:59
ServaesServaes
28.5k34099
$endgroup$
$begingroup$
Thank you for explaining so well.
$endgroup$
– user565684
Jan 27 at 16:34
add a comment |
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1 Answer
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1 Answer
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$begingroup$
By definition an $R$-module $A$ is divisible if for every $xin A$ and every non-zero divisor $rin R$ there exists $yin A$ such that $x=ry$. As $F$ is the field of fractions of $R$ and $M_F=Fotimes_RM$, every $xin M_F$ is of the form
$$x=sum_{i=1}^nf_iotimes m_i=sum_{i=1}^nfrac{r_i}{s_i}otimes m_i,$$
where $m_iin M$ and $f_iin F$ and $r_i,s_iin R$ with $s_ineq0$, because $R$ is an integral domain. It follows that for every non-zero $rin R$ we have
$$y:=rcdotsum_{i=1}^nfrac{r_i}{rs_i}otimes m_i,$$
satisfies $x=ry$, so $M_F$ is divisible.
It is clear that the map
$$M longrightarrow M_F: x longmapsto 1otimes x,$$
is a monomorphism, and its image (which I will also denote $M$) is an essential submodule of $M_F$ because if $Nsubset M_F$ is a submodule of $M_F$ such that $Mcap N=0$, then as before, for every $nin N$ we can write
$$n=sum_{i=1}^kfrac{r_i}{s_i}otimes m_i,$$
for some $kinBbb{N}$ and $r_i,s_iin R$ with $s_ineq0$ and $m_iin M$. Clearing denominators and setting $t_i:=frac{r_i}{s_i}prod_{i=j}^ks_jin R$ we find that
$$(s_1cdots s_k)n=sum_{i=1}^kt_iotimes m_i=sum_{i=1}^k1otimes t_im_1=1otimessum_{i=1}^kt_im_i,$$
and so $(s_1cdots s_k)n=0$. Because $M$ is torsion-free it follows that $n=0$, and hence that $N=0$. This means $M$ is an essential submodule of $M_F$.
answered Jan 27 at 8:59
ServaesServaes
28.5k34099
$endgroup$
$begingroup$
Thank you for explaining so well.
$endgroup$
– user565684
Jan 27 at 16:34
add a comment |
$begingroup$
By definition an $R$-module $A$ is divisible if for every $xin A$ and every non-zero divisor $rin R$ there exists $yin A$ such that $x=ry$. As $F$ is the field of fractions of $R$ and $M_F=Fotimes_RM$, every $xin M_F$ is of the form
$$x=sum_{i=1}^nf_iotimes m_i=sum_{i=1}^nfrac{r_i}{s_i}otimes m_i,$$
where $m_iin M$ and $f_iin F$ and $r_i,s_iin R$ with $s_ineq0$, because $R$ is an integral domain. It follows that for every non-zero $rin R$ we have
$$y:=rcdotsum_{i=1}^nfrac{r_i}{rs_i}otimes m_i,$$
satisfies $x=ry$, so $M_F$ is divisible.
It is clear that the map
$$M longrightarrow M_F: x longmapsto 1otimes x,$$
is a monomorphism, and its image (which I will also denote $M$) is an essential submodule of $M_F$ because if $Nsubset M_F$ is a submodule of $M_F$ such that $Mcap N=0$, then as before, for every $nin N$ we can write
$$n=sum_{i=1}^kfrac{r_i}{s_i}otimes m_i,$$
for some $kinBbb{N}$ and $r_i,s_iin R$ with $s_ineq0$ and $m_iin M$. Clearing denominators and setting $t_i:=frac{r_i}{s_i}prod_{i=j}^ks_jin R$ we find that
$$(s_1cdots s_k)n=sum_{i=1}^kt_iotimes m_i=sum_{i=1}^k1otimes t_im_1=1otimessum_{i=1}^kt_im_i,$$
and so $(s_1cdots s_k)n=0$. Because $M$ is torsion-free it follows that $n=0$, and hence that $N=0$. This means $M$ is an essential submodule of $M_F$.
answered Jan 27 at 8:59
ServaesServaes
28.5k34099
$endgroup$
$begingroup$
Thank you for explaining so well.
$endgroup$
– user565684
Jan 27 at 16:34
add a comment |
$begingroup$
By definition an $R$-module $A$ is divisible if for every $xin A$ and every non-zero divisor $rin R$ there exists $yin A$ such that $x=ry$. As $F$ is the field of fractions of $R$ and $M_F=Fotimes_RM$, every $xin M_F$ is of the form
$$x=sum_{i=1}^nf_iotimes m_i=sum_{i=1}^nfrac{r_i}{s_i}otimes m_i,$$
where $m_iin M$ and $f_iin F$ and $r_i,s_iin R$ with $s_ineq0$, because $R$ is an integral domain. It follows that for every non-zero $rin R$ we have
$$y:=rcdotsum_{i=1}^nfrac{r_i}{rs_i}otimes m_i,$$
satisfies $x=ry$, so $M_F$ is divisible.
It is clear that the map
$$M longrightarrow M_F: x longmapsto 1otimes x,$$
is a monomorphism, and its image (which I will also denote $M$) is an essential submodule of $M_F$ because if $Nsubset M_F$ is a submodule of $M_F$ such that $Mcap N=0$, then as before, for every $nin N$ we can write
$$n=sum_{i=1}^kfrac{r_i}{s_i}otimes m_i,$$
for some $kinBbb{N}$ and $r_i,s_iin R$ with $s_ineq0$ and $m_iin M$. Clearing denominators and setting $t_i:=frac{r_i}{s_i}prod_{i=j}^ks_jin R$ we find that
$$(s_1cdots s_k)n=sum_{i=1}^kt_iotimes m_i=sum_{i=1}^k1otimes t_im_1=1otimessum_{i=1}^kt_im_i,$$
and so $(s_1cdots s_k)n=0$. Because $M$ is torsion-free it follows that $n=0$, and hence that $N=0$. This means $M$ is an essential submodule of $M_F$.
answered Jan 27 at 8:59
ServaesServaes
28.5k34099
$endgroup$
By definition an $R$-module $A$ is divisible if for every $xin A$ and every non-zero divisor $rin R$ there exists $yin A$ such that $x=ry$. As $F$ is the field of fractions of $R$ and $M_F=Fotimes_RM$, every $xin M_F$ is of the form
$$x=sum_{i=1}^nf_iotimes m_i=sum_{i=1}^nfrac{r_i}{s_i}otimes m_i,$$
where $m_iin M$ and $f_iin F$ and $r_i,s_iin R$ with $s_ineq0$, because $R$ is an integral domain. It follows that for every non-zero $rin R$ we have
$$y:=rcdotsum_{i=1}^nfrac{r_i}{rs_i}otimes m_i,$$
satisfies $x=ry$, so $M_F$ is divisible.
It is clear that the map
$$M longrightarrow M_F: x longmapsto 1otimes x,$$
is a monomorphism, and its image (which I will also denote $M$) is an essential submodule of $M_F$ because if $Nsubset M_F$ is a submodule of $M_F$ such that $Mcap N=0$, then as before, for every $nin N$ we can write
$$n=sum_{i=1}^kfrac{r_i}{s_i}otimes m_i,$$
for some $kinBbb{N}$ and $r_i,s_iin R$ with $s_ineq0$ and $m_iin M$. Clearing denominators and setting $t_i:=frac{r_i}{s_i}prod_{i=j}^ks_jin R$ we find that
$$(s_1cdots s_k)n=sum_{i=1}^kt_iotimes m_i=sum_{i=1}^k1otimes t_im_1=1otimessum_{i=1}^kt_im_i,$$
and so $(s_1cdots s_k)n=0$. Because $M$ is torsion-free it follows that $n=0$, and hence that $N=0$. This means $M$ is an essential submodule of $M_F$.
answered Jan 27 at 8:59
ServaesServaes
28.5k34099
edited Jan 27 at 16:41
answered Jan 27 at 8:59
ServaesServaes
28.5k34099
answered Jan 27 at 8:59
ServaesServaes
28.5k34099
answered Jan 27 at 8:59
ServaesServaes
28.5k34099
28.5k34099
$begingroup$
Thank you for explaining so well.
$endgroup$
– user565684
Jan 27 at 16:34
add a comment |
$begingroup$
Thank you for explaining so well.
$endgroup$
– user565684
Jan 27 at 16:34
$begingroup$
Thank you for explaining so well.
$endgroup$
– user565684
Jan 27 at 16:34
$begingroup$
Thank you for explaining so well.
$endgroup$
– user565684
Jan 27 at 16:34
add a comment |
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$begingroup$
What is your definition of an essential monomorphism? I've assumed for now that it means the image is an essential submodule of $M_F$.
$endgroup$
– Servaes
Jan 27 at 9:00
$begingroup$
Yes, that is correct. Thank you.
$endgroup$
– user565684
Jan 27 at 9:05