Mixing
On a proof of free modules over a commutative ring having all basis of the same cardinal
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
As the title states, I'm going over a proof of the following statement
Let $R$ be a commutative ring and $M$ a free $R$-module. If $mathcalB_1, mathcalB_2$ are two bases of $M$, then $|mathcalB_1| = |mathcalB_2|$.
The proof goes as follows: since $mathcalB_1$ and $mathcalB_2$ are two bases for $M$, we have that
$$
R^(mathcalB_1) simeq M simeq R^(mathcalB_2)
$$
and so there exist $A_1, A_2 in R^(mathcalB_1 times mathcalB_2)$ such that $A_1A_2 = 1$ and $A_2A_1 = 1$. Now, let $m subseteq R$ be a maximal ideal of $R$, and let $mathbbk := R/m$ which is a field since $R$ is commutative.
If we now consider $pi : R to mathbbk$ the projection to the quotient, this induces a ring morphism $tau : R^(mathcalB_1 times mathcalB_2) to mathbbk^(mathcalB_1 times mathcalB_2)$ and we have that:
$$tau(A_1)tau(A_2) = 1 = tau(A_2)tau(A_2)$$
and thus $mathbbk^(mathcalB_1 times mathcalB_2)$ has an invertible element which implies $mathbbk^(mathcalB_1) simeq mathbbk^(mathcalB_2)$, which in turn proves that $|mathcalB_1| = |mathcalB_2|$ because the latter is a $mathbbk$-vector space isomorphism.
I mostly understand this argument except for a key point, which is going from an invertible element of $ operatornameHom_R(R^(mathcalB_1),R^(mathcalB_2))$ to an invertible element of $R^(mathcalB_1 times mathcalB_2)$ and again from an invertible element of $mathbbk^(mathcalB_1 times mathcalB_2)$ to an invertible element of $ operatornameHom_mathbbk(mathbbk^(mathcalB_1),mathbbk^(mathcalB_2))$. How is this justified?
abstract-algebra commutative-algebra modules proof-explanation
edited Jul 20 at 9:52
Bernard
110k635103
asked Jul 20 at 9:37
Guido A.
3,754624
add a comment |Â
up vote
1
down vote
favorite
As the title states, I'm going over a proof of the following statement
Let $R$ be a commutative ring and $M$ a free $R$-module. If $mathcalB_1, mathcalB_2$ are two bases of $M$, then $|mathcalB_1| = |mathcalB_2|$.
The proof goes as follows: since $mathcalB_1$ and $mathcalB_2$ are two bases for $M$, we have that
$$
R^(mathcalB_1) simeq M simeq R^(mathcalB_2)
$$
and so there exist $A_1, A_2 in R^(mathcalB_1 times mathcalB_2)$ such that $A_1A_2 = 1$ and $A_2A_1 = 1$. Now, let $m subseteq R$ be a maximal ideal of $R$, and let $mathbbk := R/m$ which is a field since $R$ is commutative.
If we now consider $pi : R to mathbbk$ the projection to the quotient, this induces a ring morphism $tau : R^(mathcalB_1 times mathcalB_2) to mathbbk^(mathcalB_1 times mathcalB_2)$ and we have that:
$$tau(A_1)tau(A_2) = 1 = tau(A_2)tau(A_2)$$
and thus $mathbbk^(mathcalB_1 times mathcalB_2)$ has an invertible element which implies $mathbbk^(mathcalB_1) simeq mathbbk^(mathcalB_2)$, which in turn proves that $|mathcalB_1| = |mathcalB_2|$ because the latter is a $mathbbk$-vector space isomorphism.
I mostly understand this argument except for a key point, which is going from an invertible element of $ operatornameHom_R(R^(mathcalB_1),R^(mathcalB_2))$ to an invertible element of $R^(mathcalB_1 times mathcalB_2)$ and again from an invertible element of $mathbbk^(mathcalB_1 times mathcalB_2)$ to an invertible element of $ operatornameHom_mathbbk(mathbbk^(mathcalB_1),mathbbk^(mathcalB_2))$. How is this justified?
abstract-algebra commutative-algebra modules proof-explanation
edited Jul 20 at 9:52
Bernard
110k635103
asked Jul 20 at 9:37
Guido A.
3,754624
1
Suppose you have a linear map between finite-dimensional vector spaces and bases of both spaces. A linear map is then completely determined by its action on a basis and each image of a basis vector is completely determined by its coefficients w.r.t. that basis. The same holds for maps between free $R$-modules and gives the natural isomorphisms you are looking for.
– Mathematician 42
Jul 20 at 9:52
I see it now, thanks!
– Guido A.
Jul 20 at 10:00
@Mathematician42 One last question: iirc, the direct sum of infinitely many rings will not be a ring (with unity), so the equalities $A_1A_2 = A_2A_2 = 1$ are in $R^mathcalB_1 times mathcalB_1$, right?
– Guido A.
Jul 20 at 10:13
Yes that's correct
– Mathematician 42
Jul 20 at 10:29
Great. Thanks again for taking the time to answer.
– Guido A.
Jul 20 at 10:36
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
As the title states, I'm going over a proof of the following statement
Let $R$ be a commutative ring and $M$ a free $R$-module. If $mathcalB_1, mathcalB_2$ are two bases of $M$, then $|mathcalB_1| = |mathcalB_2|$.
The proof goes as follows: since $mathcalB_1$ and $mathcalB_2$ are two bases for $M$, we have that
$$
R^(mathcalB_1) simeq M simeq R^(mathcalB_2)
$$
and so there exist $A_1, A_2 in R^(mathcalB_1 times mathcalB_2)$ such that $A_1A_2 = 1$ and $A_2A_1 = 1$. Now, let $m subseteq R$ be a maximal ideal of $R$, and let $mathbbk := R/m$ which is a field since $R$ is commutative.
If we now consider $pi : R to mathbbk$ the projection to the quotient, this induces a ring morphism $tau : R^(mathcalB_1 times mathcalB_2) to mathbbk^(mathcalB_1 times mathcalB_2)$ and we have that:
$$tau(A_1)tau(A_2) = 1 = tau(A_2)tau(A_2)$$
and thus $mathbbk^(mathcalB_1 times mathcalB_2)$ has an invertible element which implies $mathbbk^(mathcalB_1) simeq mathbbk^(mathcalB_2)$, which in turn proves that $|mathcalB_1| = |mathcalB_2|$ because the latter is a $mathbbk$-vector space isomorphism.
I mostly understand this argument except for a key point, which is going from an invertible element of $ operatornameHom_R(R^(mathcalB_1),R^(mathcalB_2))$ to an invertible element of $R^(mathcalB_1 times mathcalB_2)$ and again from an invertible element of $mathbbk^(mathcalB_1 times mathcalB_2)$ to an invertible element of $ operatornameHom_mathbbk(mathbbk^(mathcalB_1),mathbbk^(mathcalB_2))$. How is this justified?
abstract-algebra commutative-algebra modules proof-explanation
edited Jul 20 at 9:52
Bernard
110k635103
asked Jul 20 at 9:37
Guido A.
3,754624
As the title states, I'm going over a proof of the following statement
Let $R$ be a commutative ring and $M$ a free $R$-module. If $mathcalB_1, mathcalB_2$ are two bases of $M$, then $|mathcalB_1| = |mathcalB_2|$.
The proof goes as follows: since $mathcalB_1$ and $mathcalB_2$ are two bases for $M$, we have that
$$
R^(mathcalB_1) simeq M simeq R^(mathcalB_2)
$$
and so there exist $A_1, A_2 in R^(mathcalB_1 times mathcalB_2)$ such that $A_1A_2 = 1$ and $A_2A_1 = 1$. Now, let $m subseteq R$ be a maximal ideal of $R$, and let $mathbbk := R/m$ which is a field since $R$ is commutative.
If we now consider $pi : R to mathbbk$ the projection to the quotient, this induces a ring morphism $tau : R^(mathcalB_1 times mathcalB_2) to mathbbk^(mathcalB_1 times mathcalB_2)$ and we have that:
$$tau(A_1)tau(A_2) = 1 = tau(A_2)tau(A_2)$$
and thus $mathbbk^(mathcalB_1 times mathcalB_2)$ has an invertible element which implies $mathbbk^(mathcalB_1) simeq mathbbk^(mathcalB_2)$, which in turn proves that $|mathcalB_1| = |mathcalB_2|$ because the latter is a $mathbbk$-vector space isomorphism.
I mostly understand this argument except for a key point, which is going from an invertible element of $ operatornameHom_R(R^(mathcalB_1),R^(mathcalB_2))$ to an invertible element of $R^(mathcalB_1 times mathcalB_2)$ and again from an invertible element of $mathbbk^(mathcalB_1 times mathcalB_2)$ to an invertible element of $ operatornameHom_mathbbk(mathbbk^(mathcalB_1),mathbbk^(mathcalB_2))$. How is this justified?
abstract-algebra commutative-algebra modules proof-explanation
edited Jul 20 at 9:52
Bernard
110k635103
asked Jul 20 at 9:37
Guido A.
3,754624
edited Jul 20 at 9:52
Bernard
110k635103
edited Jul 20 at 9:52
Bernard
110k635103
edited Jul 20 at 9:52
Bernard
110k635103
110k635103
asked Jul 20 at 9:37
Guido A.
3,754624
asked Jul 20 at 9:37
Guido A.
3,754624
asked Jul 20 at 9:37
Guido A.
3,754624
3,754624
1
Suppose you have a linear map between finite-dimensional vector spaces and bases of both spaces. A linear map is then completely determined by its action on a basis and each image of a basis vector is completely determined by its coefficients w.r.t. that basis. The same holds for maps between free $R$-modules and gives the natural isomorphisms you are looking for.
– Mathematician 42
Jul 20 at 9:52
I see it now, thanks!
– Guido A.
Jul 20 at 10:00
@Mathematician42 One last question: iirc, the direct sum of infinitely many rings will not be a ring (with unity), so the equalities $A_1A_2 = A_2A_2 = 1$ are in $R^mathcalB_1 times mathcalB_1$, right?
– Guido A.
Jul 20 at 10:13
Yes that's correct
– Mathematician 42
Jul 20 at 10:29
Great. Thanks again for taking the time to answer.
– Guido A.
Jul 20 at 10:36
add a comment |Â
1
Suppose you have a linear map between finite-dimensional vector spaces and bases of both spaces. A linear map is then completely determined by its action on a basis and each image of a basis vector is completely determined by its coefficients w.r.t. that basis. The same holds for maps between free $R$-modules and gives the natural isomorphisms you are looking for.
– Mathematician 42
Jul 20 at 9:52
I see it now, thanks!
– Guido A.
Jul 20 at 10:00
@Mathematician42 One last question: iirc, the direct sum of infinitely many rings will not be a ring (with unity), so the equalities $A_1A_2 = A_2A_2 = 1$ are in $R^mathcalB_1 times mathcalB_1$, right?
– Guido A.
Jul 20 at 10:13
Yes that's correct
– Mathematician 42
Jul 20 at 10:29
Great. Thanks again for taking the time to answer.
– Guido A.
Jul 20 at 10:36
1
1
Suppose you have a linear map between finite-dimensional vector spaces and bases of both spaces. A linear map is then completely determined by its action on a basis and each image of a basis vector is completely determined by its coefficients w.r.t. that basis. The same holds for maps between free $R$-modules and gives the natural isomorphisms you are looking for.
– Mathematician 42
Jul 20 at 9:52
Suppose you have a linear map between finite-dimensional vector spaces and bases of both spaces. A linear map is then completely determined by its action on a basis and each image of a basis vector is completely determined by its coefficients w.r.t. that basis. The same holds for maps between free $R$-modules and gives the natural isomorphisms you are looking for.
– Mathematician 42
Jul 20 at 9:52
I see it now, thanks!
– Guido A.
Jul 20 at 10:00
I see it now, thanks!
– Guido A.
Jul 20 at 10:00
@Mathematician42 One last question: iirc, the direct sum of infinitely many rings will not be a ring (with unity), so the equalities $A_1A_2 = A_2A_2 = 1$ are in $R^mathcalB_1 times mathcalB_1$, right?
– Guido A.
Jul 20 at 10:13
@Mathematician42 One last question: iirc, the direct sum of infinitely many rings will not be a ring (with unity), so the equalities $A_1A_2 = A_2A_2 = 1$ are in $R^mathcalB_1 times mathcalB_1$, right?
– Guido A.
Jul 20 at 10:13
Yes that's correct
– Mathematician 42
Jul 20 at 10:29
Yes that's correct
– Mathematician 42
Jul 20 at 10:29
Great. Thanks again for taking the time to answer.
– Guido A.
Jul 20 at 10:36
Great. Thanks again for taking the time to answer.
– Guido A.
Jul 20 at 10:36
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857466%2fon-a-proof-of-free-modules-over-a-commutative-ring-having-all-basis-of-the-same%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Suppose you have a linear map between finite-dimensional vector spaces and bases of both spaces. A linear map is then completely determined by its action on a basis and each image of a basis vector is completely determined by its coefficients w.r.t. that basis. The same holds for maps between free $R$-modules and gives the natural isomorphisms you are looking for.
– Mathematician 42
Jul 20 at 9:52
I see it now, thanks!
– Guido A.
Jul 20 at 10:00
@Mathematician42 One last question: iirc, the direct sum of infinitely many rings will not be a ring (with unity), so the equalities $A_1A_2 = A_2A_2 = 1$ are in $R^mathcalB_1 times mathcalB_1$, right?
– Guido A.
Jul 20 at 10:13
Yes that's correct
– Mathematician 42
Jul 20 at 10:29
Great. Thanks again for taking the time to answer.
– Guido A.
Jul 20 at 10:36