About lattice of finitely generated projective module The Next CEO of Stack OverflowAbout Inner product spaceAbout symmetric inner product spaceCan we contruct a basis in a finitely generated moduleBasis of a subset of finitely generated torsion free moduleProve module $M$ is finitely generated if $N$ and $M/N$ are finitely generatedProof about finitely generated torsion-free R-module M is free, where R is a PID(Zelmanowitz) Regular Module but not ProjectiveFinitely generated module over an orderprojective module which is a submodule of a finitely generated free moduleWhat can be said of the lattices of submodules of a noetherian/finitely generated module?Rings over which torsion free module is projectiveFinitely generated free module is projective.
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About lattice of finitely generated projective module
The Next CEO of Stack OverflowAbout Inner product spaceAbout symmetric inner product spaceCan we contruct a basis in a finitely generated moduleBasis of a subset of finitely generated torsion free moduleProve module $M$ is finitely generated if $N$ and $M/N$ are finitely generatedProof about finitely generated torsion-free R-module M is free, where R is a PID(Zelmanowitz) Regular Module but not ProjectiveFinitely generated module over an orderprojective module which is a submodule of a finitely generated free moduleWhat can be said of the lattices of submodules of a noetherian/finitely generated module?Rings over which torsion free module is projectiveFinitely generated free module is projective.
$begingroup$
Let $A$ be euclidian ring and $K$ be its field of a fraction.Let $(V,B)$ be a nonzero IPS (inner product space) over $K$. A finitely generated A-submodule
$L ⊆ V $is said to be an $A$ -lattice in $V$ if $L$ contains a $K$-basis of $V$ . As we have
already observed, $L$ must be $A$-free since it is torsion-free, and it is easy to verify that
an $A$-basis for $L$ will automatically be a $K$-basis for $V$.
Attempt: I know that $V$ is finitely generated projective module since $(V, B)$ is inner product space and $B$ is regular bilinear form. Now question is what is the meaning of $L$ contains a $K$-basis of $V$ and how to verify $A$-basis for L will automatically be a $K$-basis for $V$ ; it certainly means we can extend from basis of $L$ to basis of $V$?
abstract-algebra commutative-algebra modules projective-module
asked yesterday
maths studentmaths student
6321521
$endgroup$
add a comment |
$begingroup$
Let $A$ be euclidian ring and $K$ be its field of a fraction.Let $(V,B)$ be a nonzero IPS (inner product space) over $K$. A finitely generated A-submodule
$L ⊆ V $is said to be an $A$ -lattice in $V$ if $L$ contains a $K$-basis of $V$ . As we have
already observed, $L$ must be $A$-free since it is torsion-free, and it is easy to verify that
an $A$-basis for $L$ will automatically be a $K$-basis for $V$.
Attempt: I know that $V$ is finitely generated projective module since $(V, B)$ is inner product space and $B$ is regular bilinear form. Now question is what is the meaning of $L$ contains a $K$-basis of $V$ and how to verify $A$-basis for L will automatically be a $K$-basis for $V$ ; it certainly means we can extend from basis of $L$ to basis of $V$?
abstract-algebra commutative-algebra modules projective-module
asked yesterday
maths studentmaths student
6321521
$endgroup$
add a comment |
$begingroup$
Let $A$ be euclidian ring and $K$ be its field of a fraction.Let $(V,B)$ be a nonzero IPS (inner product space) over $K$. A finitely generated A-submodule
$L ⊆ V $is said to be an $A$ -lattice in $V$ if $L$ contains a $K$-basis of $V$ . As we have
already observed, $L$ must be $A$-free since it is torsion-free, and it is easy to verify that
an $A$-basis for $L$ will automatically be a $K$-basis for $V$.
Attempt: I know that $V$ is finitely generated projective module since $(V, B)$ is inner product space and $B$ is regular bilinear form. Now question is what is the meaning of $L$ contains a $K$-basis of $V$ and how to verify $A$-basis for L will automatically be a $K$-basis for $V$ ; it certainly means we can extend from basis of $L$ to basis of $V$?
abstract-algebra commutative-algebra modules projective-module
asked yesterday
maths studentmaths student
6321521
$endgroup$
Let $A$ be euclidian ring and $K$ be its field of a fraction.Let $(V,B)$ be a nonzero IPS (inner product space) over $K$. A finitely generated A-submodule
$L ⊆ V $is said to be an $A$ -lattice in $V$ if $L$ contains a $K$-basis of $V$ . As we have
already observed, $L$ must be $A$-free since it is torsion-free, and it is easy to verify that
an $A$-basis for $L$ will automatically be a $K$-basis for $V$.
Attempt: I know that $V$ is finitely generated projective module since $(V, B)$ is inner product space and $B$ is regular bilinear form. Now question is what is the meaning of $L$ contains a $K$-basis of $V$ and how to verify $A$-basis for L will automatically be a $K$-basis for $V$ ; it certainly means we can extend from basis of $L$ to basis of $V$?
abstract-algebra commutative-algebra modules projective-module
abstract-algebra commutative-algebra modules projective-module
asked yesterday
maths studentmaths student
6321521
asked yesterday
maths studentmaths student
6321521
asked yesterday
maths studentmaths student
6321521
asked yesterday
maths studentmaths student
6321521
asked yesterday
maths studentmaths student
6321521
6321521
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The meaning of $L$ containing a $K$-basis for $V$ is that there are $ell_1,ldots,ell_nin Lsubseteq V$ forming a basis for $V$ as a $K$-vector space (where $n=dim_K V$).
Now suppose that we have $l_1,ldots,l_rin L$ are a basis for $L$ as a free $A$-module. (Note the different style of $l$). We want to prove that $r=n$ and that the elements $l_i$ are linearly independent over $K$, so that they form a $K$-basis.
If $$sum_i=1^r c_il_i=0,$$
with $c_iin K$, we can multiply by a common denominator $dne 0in A$, so that $c_idin A$. Then we get
$$sum_i=1^r (c_id)l_i=0,$$
but the $l_i$ were an $A$-basis for $L$, so $c_id=0$ for all $i$, and since $dne 0$, $c_i=0$ for all $i$.
As for why $n=r$, observe that $V=KL=Kotimes_A L=Kotimes_A A^r = K^r$, so $r=dim K^r=dim V =n$.
answered yesterday
jgonjgon
16.1k32143
$endgroup$
$begingroup$
@mathsstudent I don't usually participate in chat, if you have another question, feel free to ask it, and drop me a comment linking to it and I'll check it out.
$endgroup$
– jgon
yesterday
$begingroup$
@mathsstudent Yes
$endgroup$
– jgon
yesterday
$begingroup$
I ask doubt below sir.
$endgroup$
– maths student
yesterday
$begingroup$
math.stackexchange.com/questions/3164762/… I have ask it?
$endgroup$
– maths student
yesterday
$begingroup$
I have got corollary 3.9 can you please explain to me how to prove note below corollary 3.9.
$endgroup$
– maths student
yesterday
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
The meaning of $L$ containing a $K$-basis for $V$ is that there are $ell_1,ldots,ell_nin Lsubseteq V$ forming a basis for $V$ as a $K$-vector space (where $n=dim_K V$).
Now suppose that we have $l_1,ldots,l_rin L$ are a basis for $L$ as a free $A$-module. (Note the different style of $l$). We want to prove that $r=n$ and that the elements $l_i$ are linearly independent over $K$, so that they form a $K$-basis.
If $$sum_i=1^r c_il_i=0,$$
with $c_iin K$, we can multiply by a common denominator $dne 0in A$, so that $c_idin A$. Then we get
$$sum_i=1^r (c_id)l_i=0,$$
but the $l_i$ were an $A$-basis for $L$, so $c_id=0$ for all $i$, and since $dne 0$, $c_i=0$ for all $i$.
As for why $n=r$, observe that $V=KL=Kotimes_A L=Kotimes_A A^r = K^r$, so $r=dim K^r=dim V =n$.
answered yesterday
jgonjgon
16.1k32143
$endgroup$
$begingroup$
@mathsstudent I don't usually participate in chat, if you have another question, feel free to ask it, and drop me a comment linking to it and I'll check it out.
$endgroup$
– jgon
yesterday
$begingroup$
@mathsstudent Yes
$endgroup$
– jgon
yesterday
$begingroup$
I ask doubt below sir.
$endgroup$
– maths student
yesterday
$begingroup$
math.stackexchange.com/questions/3164762/… I have ask it?
$endgroup$
– maths student
yesterday
$begingroup$
I have got corollary 3.9 can you please explain to me how to prove note below corollary 3.9.
$endgroup$
– maths student
yesterday
add a comment |
$begingroup$
The meaning of $L$ containing a $K$-basis for $V$ is that there are $ell_1,ldots,ell_nin Lsubseteq V$ forming a basis for $V$ as a $K$-vector space (where $n=dim_K V$).
Now suppose that we have $l_1,ldots,l_rin L$ are a basis for $L$ as a free $A$-module. (Note the different style of $l$). We want to prove that $r=n$ and that the elements $l_i$ are linearly independent over $K$, so that they form a $K$-basis.
If $$sum_i=1^r c_il_i=0,$$
with $c_iin K$, we can multiply by a common denominator $dne 0in A$, so that $c_idin A$. Then we get
$$sum_i=1^r (c_id)l_i=0,$$
but the $l_i$ were an $A$-basis for $L$, so $c_id=0$ for all $i$, and since $dne 0$, $c_i=0$ for all $i$.
As for why $n=r$, observe that $V=KL=Kotimes_A L=Kotimes_A A^r = K^r$, so $r=dim K^r=dim V =n$.
answered yesterday
jgonjgon
16.1k32143
$endgroup$
$begingroup$
@mathsstudent I don't usually participate in chat, if you have another question, feel free to ask it, and drop me a comment linking to it and I'll check it out.
$endgroup$
– jgon
yesterday
$begingroup$
@mathsstudent Yes
$endgroup$
– jgon
yesterday
$begingroup$
I ask doubt below sir.
$endgroup$
– maths student
yesterday
$begingroup$
math.stackexchange.com/questions/3164762/… I have ask it?
$endgroup$
– maths student
yesterday
$begingroup$
I have got corollary 3.9 can you please explain to me how to prove note below corollary 3.9.
$endgroup$
– maths student
yesterday
add a comment |
$begingroup$
The meaning of $L$ containing a $K$-basis for $V$ is that there are $ell_1,ldots,ell_nin Lsubseteq V$ forming a basis for $V$ as a $K$-vector space (where $n=dim_K V$).
Now suppose that we have $l_1,ldots,l_rin L$ are a basis for $L$ as a free $A$-module. (Note the different style of $l$). We want to prove that $r=n$ and that the elements $l_i$ are linearly independent over $K$, so that they form a $K$-basis.
If $$sum_i=1^r c_il_i=0,$$
with $c_iin K$, we can multiply by a common denominator $dne 0in A$, so that $c_idin A$. Then we get
$$sum_i=1^r (c_id)l_i=0,$$
but the $l_i$ were an $A$-basis for $L$, so $c_id=0$ for all $i$, and since $dne 0$, $c_i=0$ for all $i$.
As for why $n=r$, observe that $V=KL=Kotimes_A L=Kotimes_A A^r = K^r$, so $r=dim K^r=dim V =n$.
answered yesterday
jgonjgon
16.1k32143
$endgroup$
The meaning of $L$ containing a $K$-basis for $V$ is that there are $ell_1,ldots,ell_nin Lsubseteq V$ forming a basis for $V$ as a $K$-vector space (where $n=dim_K V$).
Now suppose that we have $l_1,ldots,l_rin L$ are a basis for $L$ as a free $A$-module. (Note the different style of $l$). We want to prove that $r=n$ and that the elements $l_i$ are linearly independent over $K$, so that they form a $K$-basis.
If $$sum_i=1^r c_il_i=0,$$
with $c_iin K$, we can multiply by a common denominator $dne 0in A$, so that $c_idin A$. Then we get
$$sum_i=1^r (c_id)l_i=0,$$
but the $l_i$ were an $A$-basis for $L$, so $c_id=0$ for all $i$, and since $dne 0$, $c_i=0$ for all $i$.
As for why $n=r$, observe that $V=KL=Kotimes_A L=Kotimes_A A^r = K^r$, so $r=dim K^r=dim V =n$.
answered yesterday
jgonjgon
16.1k32143
answered yesterday
jgonjgon
16.1k32143
answered yesterday
jgonjgon
16.1k32143
answered yesterday
jgonjgon
16.1k32143
16.1k32143
$begingroup$
@mathsstudent I don't usually participate in chat, if you have another question, feel free to ask it, and drop me a comment linking to it and I'll check it out.
$endgroup$
– jgon
yesterday
$begingroup$
@mathsstudent Yes
$endgroup$
– jgon
yesterday
$begingroup$
I ask doubt below sir.
$endgroup$
– maths student
yesterday
$begingroup$
math.stackexchange.com/questions/3164762/… I have ask it?
$endgroup$
– maths student
yesterday
$begingroup$
I have got corollary 3.9 can you please explain to me how to prove note below corollary 3.9.
$endgroup$
– maths student
yesterday
add a comment |
$begingroup$
@mathsstudent I don't usually participate in chat, if you have another question, feel free to ask it, and drop me a comment linking to it and I'll check it out.
$endgroup$
– jgon
yesterday
$begingroup$
@mathsstudent Yes
$endgroup$
– jgon
yesterday
$begingroup$
I ask doubt below sir.
$endgroup$
– maths student
yesterday
$begingroup$
math.stackexchange.com/questions/3164762/… I have ask it?
$endgroup$
– maths student
yesterday
$begingroup$
I have got corollary 3.9 can you please explain to me how to prove note below corollary 3.9.
$endgroup$
– maths student
yesterday
$begingroup$
@mathsstudent I don't usually participate in chat, if you have another question, feel free to ask it, and drop me a comment linking to it and I'll check it out.
$endgroup$
– jgon
yesterday
$begingroup$
@mathsstudent I don't usually participate in chat, if you have another question, feel free to ask it, and drop me a comment linking to it and I'll check it out.
$endgroup$
– jgon
yesterday
$begingroup$
@mathsstudent Yes
$endgroup$
– jgon
yesterday
$begingroup$
@mathsstudent Yes
$endgroup$
– jgon
yesterday
$begingroup$
I ask doubt below sir.
$endgroup$
– maths student
yesterday
$begingroup$
I ask doubt below sir.
$endgroup$
– maths student
yesterday
$begingroup$
math.stackexchange.com/questions/3164762/… I have ask it?
$endgroup$
– maths student
yesterday
$begingroup$
math.stackexchange.com/questions/3164762/… I have ask it?
$endgroup$
– maths student
yesterday
$begingroup$
I have got corollary 3.9 can you please explain to me how to prove note below corollary 3.9.
$endgroup$
– maths student
yesterday
$begingroup$
I have got corollary 3.9 can you please explain to me how to prove note below corollary 3.9.
$endgroup$
– maths student
yesterday
add a comment |
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