Mixing
About $k$-algebra [closed]
$begingroup$
How to show that $ operatorname{Hom}(M,M)$ is $k$-algebra? Is it enough to show that it is $k$ -module? M is finitely generated.
linear-algebra abstract-algebra modules
edited Feb 2 at 10:03
Bernard
124k741117
asked Feb 2 at 7:18
maths studentmaths student
6321521
$endgroup$
closed as off-topic by user26857, Leucippus, Kemono Chen, Eevee Trainer, mrtaurho Feb 27 at 6:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user26857, Leucippus, Kemono Chen, Eevee Trainer, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How to show that $ operatorname{Hom}(M,M)$ is $k$-algebra? Is it enough to show that it is $k$ -module? M is finitely generated.
linear-algebra abstract-algebra modules
edited Feb 2 at 10:03
Bernard
124k741117
asked Feb 2 at 7:18
maths studentmaths student
6321521
$endgroup$
closed as off-topic by user26857, Leucippus, Kemono Chen, Eevee Trainer, mrtaurho Feb 27 at 6:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user26857, Leucippus, Kemono Chen, Eevee Trainer, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You need to show that it satisfies the entire definition of being a $k$-algebra which requires showing more than just the that it is a $k$-module.
$endgroup$
– Jonathan Dunay
Feb 2 at 7:27
$begingroup$
Can you explain it? @JonathanDunay
$endgroup$
– maths student
Feb 2 at 7:29
$begingroup$
Do you know what the definition of a $k$-algebra is?
$endgroup$
– Jonathan Dunay
Feb 2 at 7:33
$begingroup$
It means that we can write A as k-algebra if A = k[x1,x2,...xn]/ideal @JonathanDunay
$endgroup$
– maths student
Feb 2 at 7:39
add a comment |
$begingroup$
How to show that $ operatorname{Hom}(M,M)$ is $k$-algebra? Is it enough to show that it is $k$ -module? M is finitely generated.
linear-algebra abstract-algebra modules
edited Feb 2 at 10:03
Bernard
124k741117
asked Feb 2 at 7:18
maths studentmaths student
6321521
$endgroup$
How to show that $ operatorname{Hom}(M,M)$ is $k$-algebra? Is it enough to show that it is $k$ -module? M is finitely generated.
linear-algebra abstract-algebra modules
linear-algebra abstract-algebra modules
edited Feb 2 at 10:03
Bernard
124k741117
asked Feb 2 at 7:18
maths studentmaths student
6321521
edited Feb 2 at 10:03
Bernard
124k741117
asked Feb 2 at 7:18
maths studentmaths student
6321521
edited Feb 2 at 10:03
Bernard
124k741117
edited Feb 2 at 10:03
Bernard
124k741117
edited Feb 2 at 10:03
Bernard
124k741117
124k741117
asked Feb 2 at 7:18
maths studentmaths student
6321521
asked Feb 2 at 7:18
maths studentmaths student
6321521
asked Feb 2 at 7:18
maths studentmaths student
6321521
6321521
closed as off-topic by user26857, Leucippus, Kemono Chen, Eevee Trainer, mrtaurho Feb 27 at 6:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user26857, Leucippus, Kemono Chen, Eevee Trainer, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user26857, Leucippus, Kemono Chen, Eevee Trainer, mrtaurho Feb 27 at 6:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user26857, Leucippus, Kemono Chen, Eevee Trainer, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You need to show that it satisfies the entire definition of being a $k$-algebra which requires showing more than just the that it is a $k$-module.
$endgroup$
– Jonathan Dunay
Feb 2 at 7:27
$begingroup$
Can you explain it? @JonathanDunay
$endgroup$
– maths student
Feb 2 at 7:29
$begingroup$
Do you know what the definition of a $k$-algebra is?
$endgroup$
– Jonathan Dunay
Feb 2 at 7:33
$begingroup$
It means that we can write A as k-algebra if A = k[x1,x2,...xn]/ideal @JonathanDunay
$endgroup$
– maths student
Feb 2 at 7:39
add a comment |
$begingroup$
You need to show that it satisfies the entire definition of being a $k$-algebra which requires showing more than just the that it is a $k$-module.
$endgroup$
– Jonathan Dunay
Feb 2 at 7:27
$begingroup$
Can you explain it? @JonathanDunay
$endgroup$
– maths student
Feb 2 at 7:29
$begingroup$
Do you know what the definition of a $k$-algebra is?
$endgroup$
– Jonathan Dunay
Feb 2 at 7:33
$begingroup$
It means that we can write A as k-algebra if A = k[x1,x2,...xn]/ideal @JonathanDunay
$endgroup$
– maths student
Feb 2 at 7:39
$begingroup$
You need to show that it satisfies the entire definition of being a $k$-algebra which requires showing more than just the that it is a $k$-module.
$endgroup$
– Jonathan Dunay
Feb 2 at 7:27
$begingroup$
You need to show that it satisfies the entire definition of being a $k$-algebra which requires showing more than just the that it is a $k$-module.
$endgroup$
– Jonathan Dunay
Feb 2 at 7:27
$begingroup$
Can you explain it? @JonathanDunay
$endgroup$
– maths student
Feb 2 at 7:29
$begingroup$
Can you explain it? @JonathanDunay
$endgroup$
– maths student
Feb 2 at 7:29
$begingroup$
Do you know what the definition of a $k$-algebra is?
$endgroup$
– Jonathan Dunay
Feb 2 at 7:33
$begingroup$
Do you know what the definition of a $k$-algebra is?
$endgroup$
– Jonathan Dunay
Feb 2 at 7:33
$begingroup$
It means that we can write A as k-algebra if A = k[x1,x2,...xn]/ideal @JonathanDunay
$endgroup$
– maths student
Feb 2 at 7:39
$begingroup$
It means that we can write A as k-algebra if A = k[x1,x2,...xn]/ideal @JonathanDunay
$endgroup$
– maths student
Feb 2 at 7:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A $k$-Algebra $A$ is a $k$-module with an internal operation such that it is compatible with the structure of $k$-module and such that $A$ is a semigroup with respect to that operation.
In your case $hom(M,M)$ is obviously a $k$-module while the internal operation will be the composition of two functions:
$circ : hom(M,M)times hom(M,M)to hom(M,M)$
is defined as the operation that maps each $(f,g)$ to $g circ f$
In your case $hom(M,M)$ is also a $k$-algebra with unity because the identity map $id_M$ has the property of unity for the internal operation $circ$.
If you want see an abstract $k$-algebra as a quotient of the ring of polynomials with coefficient in $k$ you must impose the finite generation condition:
A $k$-algebra $A$ is finitely generated if there exists ${f_1,dots , f_n}$ in $A$ such that each $gin A$ can be written as a polynomial of $k[x_1,dots ,x_n]$ valued in $f_1,dots ,f_n$.
In other words $A$ is finitely generated by ${f_1, dots , f_n}$ if and only if it is the smallest $k$-algebra of $A$ that contains ${f_1, dots , f_n}$.
In this case there exists a natural morphism of $k$-algebra
$s :k[x_1, dots , x_n] to A$
that maps each $p(x_1,dots ,x_n)$ to $p(f_1, dots , f_n)$
The map is surjective because $A$ is finitely generated so by theorem of morphism we have that
$Asimeq k[x_1, dots , x_n] / ker(s)$
You can observe that if $A$ is also reduced, so there are not nilpotent elements with respect to $circ$, then
the ideal $ker(s)$ is also radical.
An important application of this theory is in Algebraic Geometry in which an algebraic set on $k^n$ is uniquely determined by your algebra of polynomials with coefficient on $k$.
In this case we have proved that every $k$-reduced algebra is uniquely associated (up to isomorphism) to an algebraic set on $k^n$.
edited Feb 2 at 10:06
Bernard
124k741117
answered Feb 2 at 7:55
Federico FalluccaFederico Fallucca
2,335210
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A $k$-Algebra $A$ is a $k$-module with an internal operation such that it is compatible with the structure of $k$-module and such that $A$ is a semigroup with respect to that operation.
In your case $hom(M,M)$ is obviously a $k$-module while the internal operation will be the composition of two functions:
$circ : hom(M,M)times hom(M,M)to hom(M,M)$
is defined as the operation that maps each $(f,g)$ to $g circ f$
In your case $hom(M,M)$ is also a $k$-algebra with unity because the identity map $id_M$ has the property of unity for the internal operation $circ$.
If you want see an abstract $k$-algebra as a quotient of the ring of polynomials with coefficient in $k$ you must impose the finite generation condition:
A $k$-algebra $A$ is finitely generated if there exists ${f_1,dots , f_n}$ in $A$ such that each $gin A$ can be written as a polynomial of $k[x_1,dots ,x_n]$ valued in $f_1,dots ,f_n$.
In other words $A$ is finitely generated by ${f_1, dots , f_n}$ if and only if it is the smallest $k$-algebra of $A$ that contains ${f_1, dots , f_n}$.
In this case there exists a natural morphism of $k$-algebra
$s :k[x_1, dots , x_n] to A$
that maps each $p(x_1,dots ,x_n)$ to $p(f_1, dots , f_n)$
The map is surjective because $A$ is finitely generated so by theorem of morphism we have that
$Asimeq k[x_1, dots , x_n] / ker(s)$
You can observe that if $A$ is also reduced, so there are not nilpotent elements with respect to $circ$, then
the ideal $ker(s)$ is also radical.
An important application of this theory is in Algebraic Geometry in which an algebraic set on $k^n$ is uniquely determined by your algebra of polynomials with coefficient on $k$.
In this case we have proved that every $k$-reduced algebra is uniquely associated (up to isomorphism) to an algebraic set on $k^n$.
edited Feb 2 at 10:06
Bernard
124k741117
answered Feb 2 at 7:55
Federico FalluccaFederico Fallucca
2,335210
$endgroup$
add a comment |
$begingroup$
A $k$-Algebra $A$ is a $k$-module with an internal operation such that it is compatible with the structure of $k$-module and such that $A$ is a semigroup with respect to that operation.
In your case $hom(M,M)$ is obviously a $k$-module while the internal operation will be the composition of two functions:
$circ : hom(M,M)times hom(M,M)to hom(M,M)$
is defined as the operation that maps each $(f,g)$ to $g circ f$
In your case $hom(M,M)$ is also a $k$-algebra with unity because the identity map $id_M$ has the property of unity for the internal operation $circ$.
If you want see an abstract $k$-algebra as a quotient of the ring of polynomials with coefficient in $k$ you must impose the finite generation condition:
A $k$-algebra $A$ is finitely generated if there exists ${f_1,dots , f_n}$ in $A$ such that each $gin A$ can be written as a polynomial of $k[x_1,dots ,x_n]$ valued in $f_1,dots ,f_n$.
In other words $A$ is finitely generated by ${f_1, dots , f_n}$ if and only if it is the smallest $k$-algebra of $A$ that contains ${f_1, dots , f_n}$.
In this case there exists a natural morphism of $k$-algebra
$s :k[x_1, dots , x_n] to A$
that maps each $p(x_1,dots ,x_n)$ to $p(f_1, dots , f_n)$
The map is surjective because $A$ is finitely generated so by theorem of morphism we have that
$Asimeq k[x_1, dots , x_n] / ker(s)$
You can observe that if $A$ is also reduced, so there are not nilpotent elements with respect to $circ$, then
the ideal $ker(s)$ is also radical.
An important application of this theory is in Algebraic Geometry in which an algebraic set on $k^n$ is uniquely determined by your algebra of polynomials with coefficient on $k$.
In this case we have proved that every $k$-reduced algebra is uniquely associated (up to isomorphism) to an algebraic set on $k^n$.
edited Feb 2 at 10:06
Bernard
124k741117
answered Feb 2 at 7:55
Federico FalluccaFederico Fallucca
2,335210
$endgroup$
add a comment |
$begingroup$
A $k$-Algebra $A$ is a $k$-module with an internal operation such that it is compatible with the structure of $k$-module and such that $A$ is a semigroup with respect to that operation.
In your case $hom(M,M)$ is obviously a $k$-module while the internal operation will be the composition of two functions:
$circ : hom(M,M)times hom(M,M)to hom(M,M)$
is defined as the operation that maps each $(f,g)$ to $g circ f$
In your case $hom(M,M)$ is also a $k$-algebra with unity because the identity map $id_M$ has the property of unity for the internal operation $circ$.
If you want see an abstract $k$-algebra as a quotient of the ring of polynomials with coefficient in $k$ you must impose the finite generation condition:
A $k$-algebra $A$ is finitely generated if there exists ${f_1,dots , f_n}$ in $A$ such that each $gin A$ can be written as a polynomial of $k[x_1,dots ,x_n]$ valued in $f_1,dots ,f_n$.
In other words $A$ is finitely generated by ${f_1, dots , f_n}$ if and only if it is the smallest $k$-algebra of $A$ that contains ${f_1, dots , f_n}$.
In this case there exists a natural morphism of $k$-algebra
$s :k[x_1, dots , x_n] to A$
that maps each $p(x_1,dots ,x_n)$ to $p(f_1, dots , f_n)$
The map is surjective because $A$ is finitely generated so by theorem of morphism we have that
$Asimeq k[x_1, dots , x_n] / ker(s)$
You can observe that if $A$ is also reduced, so there are not nilpotent elements with respect to $circ$, then
the ideal $ker(s)$ is also radical.
An important application of this theory is in Algebraic Geometry in which an algebraic set on $k^n$ is uniquely determined by your algebra of polynomials with coefficient on $k$.
In this case we have proved that every $k$-reduced algebra is uniquely associated (up to isomorphism) to an algebraic set on $k^n$.
edited Feb 2 at 10:06
Bernard
124k741117
answered Feb 2 at 7:55
Federico FalluccaFederico Fallucca
2,335210
$endgroup$
A $k$-Algebra $A$ is a $k$-module with an internal operation such that it is compatible with the structure of $k$-module and such that $A$ is a semigroup with respect to that operation.
In your case $hom(M,M)$ is obviously a $k$-module while the internal operation will be the composition of two functions:
$circ : hom(M,M)times hom(M,M)to hom(M,M)$
is defined as the operation that maps each $(f,g)$ to $g circ f$
In your case $hom(M,M)$ is also a $k$-algebra with unity because the identity map $id_M$ has the property of unity for the internal operation $circ$.
If you want see an abstract $k$-algebra as a quotient of the ring of polynomials with coefficient in $k$ you must impose the finite generation condition:
A $k$-algebra $A$ is finitely generated if there exists ${f_1,dots , f_n}$ in $A$ such that each $gin A$ can be written as a polynomial of $k[x_1,dots ,x_n]$ valued in $f_1,dots ,f_n$.
In other words $A$ is finitely generated by ${f_1, dots , f_n}$ if and only if it is the smallest $k$-algebra of $A$ that contains ${f_1, dots , f_n}$.
In this case there exists a natural morphism of $k$-algebra
$s :k[x_1, dots , x_n] to A$
that maps each $p(x_1,dots ,x_n)$ to $p(f_1, dots , f_n)$
The map is surjective because $A$ is finitely generated so by theorem of morphism we have that
$Asimeq k[x_1, dots , x_n] / ker(s)$
You can observe that if $A$ is also reduced, so there are not nilpotent elements with respect to $circ$, then
the ideal $ker(s)$ is also radical.
An important application of this theory is in Algebraic Geometry in which an algebraic set on $k^n$ is uniquely determined by your algebra of polynomials with coefficient on $k$.
In this case we have proved that every $k$-reduced algebra is uniquely associated (up to isomorphism) to an algebraic set on $k^n$.
edited Feb 2 at 10:06
Bernard
124k741117
answered Feb 2 at 7:55
Federico FalluccaFederico Fallucca
2,335210
edited Feb 2 at 10:06
Bernard
124k741117
edited Feb 2 at 10:06
Bernard
124k741117
edited Feb 2 at 10:06
Bernard
124k741117
124k741117
answered Feb 2 at 7:55
Federico FalluccaFederico Fallucca
2,335210
answered Feb 2 at 7:55
Federico FalluccaFederico Fallucca
2,335210
answered Feb 2 at 7:55
Federico FalluccaFederico Fallucca
2,335210
2,335210
add a comment |
add a comment |
$begingroup$
You need to show that it satisfies the entire definition of being a $k$-algebra which requires showing more than just the that it is a $k$-module.
$endgroup$
– Jonathan Dunay
Feb 2 at 7:27
$begingroup$
Can you explain it? @JonathanDunay
$endgroup$
– maths student
Feb 2 at 7:29
$begingroup$
Do you know what the definition of a $k$-algebra is?
$endgroup$
– Jonathan Dunay
Feb 2 at 7:33
$begingroup$
It means that we can write A as k-algebra if A = k[x1,x2,...xn]/ideal @JonathanDunay
$endgroup$
– maths student
Feb 2 at 7:39