Showing that $M_n(R[x]) cong (M_n(R)[x]$
$begingroup$
I'm trying to show that $M_n(R[x]) cong (M_n(R)[x]$ so I consider the mapping that sends an element $A in M_n(R[x])$ to the polynomial whose coefficients are matrices in which the entries of those matrices are the coefficients of the polynomials of in $A$. So for example:
$$begin{bmatrix}
x^2 & x \
3 & 2x+1
end{bmatrix}
mapsto
begin{bmatrix}
1 & 0 \
0 & 0 \
end{bmatrix}
x^2 +
begin{bmatrix}
0 &1 \
0 & 2
end{bmatrix}
x +
begin{bmatrix}
0 & 0 \
3 & 1 \
end{bmatrix}
$$
Clearly $phi(A+B) = phi(A) + phi(B)$ however it doesn't seem to hold for multiplication. Is this the correct mapping I'm suppose to be considering or is there another that would work better, thanks in advance.
abstract-algebra ring-theory
asked Dec 10 '18 at 2:35
Justin StevensonJustin Stevenson
758517
$endgroup$
add a comment |
$begingroup$
I'm trying to show that $M_n(R[x]) cong (M_n(R)[x]$ so I consider the mapping that sends an element $A in M_n(R[x])$ to the polynomial whose coefficients are matrices in which the entries of those matrices are the coefficients of the polynomials of in $A$. So for example:
$$begin{bmatrix}
x^2 & x \
3 & 2x+1
end{bmatrix}
mapsto
begin{bmatrix}
1 & 0 \
0 & 0 \
end{bmatrix}
x^2 +
begin{bmatrix}
0 &1 \
0 & 2
end{bmatrix}
x +
begin{bmatrix}
0 & 0 \
3 & 1 \
end{bmatrix}
$$
Clearly $phi(A+B) = phi(A) + phi(B)$ however it doesn't seem to hold for multiplication. Is this the correct mapping I'm suppose to be considering or is there another that would work better, thanks in advance.
abstract-algebra ring-theory
asked Dec 10 '18 at 2:35
Justin StevensonJustin Stevenson
758517
$endgroup$
$begingroup$
a ring isomorphism would be a linear map as well between the vector spaces, so take a look at the standard bases in each of the rings. that should help you identify something.
$endgroup$
– dezdichado
Dec 10 '18 at 2:49
$begingroup$
@justin Why do you think it fails to be multiplicative? Do you have an example?
$endgroup$
– rschwieb
Dec 10 '18 at 14:38
add a comment |
$begingroup$
I'm trying to show that $M_n(R[x]) cong (M_n(R)[x]$ so I consider the mapping that sends an element $A in M_n(R[x])$ to the polynomial whose coefficients are matrices in which the entries of those matrices are the coefficients of the polynomials of in $A$. So for example:
$$begin{bmatrix}
x^2 & x \
3 & 2x+1
end{bmatrix}
mapsto
begin{bmatrix}
1 & 0 \
0 & 0 \
end{bmatrix}
x^2 +
begin{bmatrix}
0 &1 \
0 & 2
end{bmatrix}
x +
begin{bmatrix}
0 & 0 \
3 & 1 \
end{bmatrix}
$$
Clearly $phi(A+B) = phi(A) + phi(B)$ however it doesn't seem to hold for multiplication. Is this the correct mapping I'm suppose to be considering or is there another that would work better, thanks in advance.
abstract-algebra ring-theory
asked Dec 10 '18 at 2:35
Justin StevensonJustin Stevenson
758517
$endgroup$
I'm trying to show that $M_n(R[x]) cong (M_n(R)[x]$ so I consider the mapping that sends an element $A in M_n(R[x])$ to the polynomial whose coefficients are matrices in which the entries of those matrices are the coefficients of the polynomials of in $A$. So for example:
$$begin{bmatrix}
x^2 & x \
3 & 2x+1
end{bmatrix}
mapsto
begin{bmatrix}
1 & 0 \
0 & 0 \
end{bmatrix}
x^2 +
begin{bmatrix}
0 &1 \
0 & 2
end{bmatrix}
x +
begin{bmatrix}
0 & 0 \
3 & 1 \
end{bmatrix}
$$
Clearly $phi(A+B) = phi(A) + phi(B)$ however it doesn't seem to hold for multiplication. Is this the correct mapping I'm suppose to be considering or is there another that would work better, thanks in advance.
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Dec 10 '18 at 2:35
Justin StevensonJustin Stevenson
758517
asked Dec 10 '18 at 2:35
Justin StevensonJustin Stevenson
758517
asked Dec 10 '18 at 2:35
Justin StevensonJustin Stevenson
758517
asked Dec 10 '18 at 2:35
Justin StevensonJustin Stevenson
758517
asked Dec 10 '18 at 2:35
Justin StevensonJustin Stevenson
758517
758517
$begingroup$
a ring isomorphism would be a linear map as well between the vector spaces, so take a look at the standard bases in each of the rings. that should help you identify something.
$endgroup$
– dezdichado
Dec 10 '18 at 2:49
$begingroup$
@justin Why do you think it fails to be multiplicative? Do you have an example?
$endgroup$
– rschwieb
Dec 10 '18 at 14:38
add a comment |
$begingroup$
a ring isomorphism would be a linear map as well between the vector spaces, so take a look at the standard bases in each of the rings. that should help you identify something.
$endgroup$
– dezdichado
Dec 10 '18 at 2:49
$begingroup$
@justin Why do you think it fails to be multiplicative? Do you have an example?
$endgroup$
– rschwieb
Dec 10 '18 at 14:38
$begingroup$
a ring isomorphism would be a linear map as well between the vector spaces, so take a look at the standard bases in each of the rings. that should help you identify something.
$endgroup$
– dezdichado
Dec 10 '18 at 2:49
$begingroup$
a ring isomorphism would be a linear map as well between the vector spaces, so take a look at the standard bases in each of the rings. that should help you identify something.
$endgroup$
– dezdichado
Dec 10 '18 at 2:49
$begingroup$
@justin Why do you think it fails to be multiplicative? Do you have an example?
$endgroup$
– rschwieb
Dec 10 '18 at 14:38
$begingroup$
@justin Why do you think it fails to be multiplicative? Do you have an example?
$endgroup$
– rschwieb
Dec 10 '18 at 14:38
add a comment |
1 Answer
1
active
oldest
votes
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Note that $M_n(R[x])$ is a free $R[x]$ module, and has basis the elementary matrices $E_{ij}$ which are all $0$ except for the $ij$th entry, which is 1. Thus a basis for $M_n(R[x])$ as a free $R$ module is $x^kE_{ij}$. Similarly, a basis for $M_n(R)[x]$ is $E_{ij}x^k$, where I'm using the order of multiplication to distinguish these two rings.
You're trying to show that these $R$-algebras are isomorphic as (presumably) $R$-algebras, so the map should be $R$-linear. Thus such a map is defined by what it does on an $R$-generating set, or in this case since the $R$-algebras are free modules, an $R$-basis. Your map $phi$ is the map defined by sending $x^kE_{ij}$ to $E_{ij}x^k$. Since it is $R$-linear, it certainly preserves addition, so we just need to check that this map does indeed preserve multiplication. For this, it suffices to check on the basis.
$$phinewcommandof[1]{left({#1}right)}of{of{x^kE_{ij}}of{x^{k'}E_{i'j'}}}
=phiof{x^{k+k'}E_{ij}E_{i'j'}}=phiof{delta_{ji'}x^{k+k'}E_{ij'}}=delta_{ji'}E_{ij'}x^{k+k'},$$
and
$$phiof{x^kE_{ij}}phiof{x^{k'}E_{i'j'}}=of{E_{ij}x^k}of{E_{i'j'}x^{k'}}=E_{ij}E_{i'j'}x^{k+k'}=delta_{ji'}E_{ij'}x^{k+k'}.$$
Thus $phi$ preserves the multiplication as well.
If you think I've made an error, do let me know.
answered Dec 10 '18 at 14:25
jgonjgon
13.7k22041
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add a comment |
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1 Answer
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$begingroup$
Note that $M_n(R[x])$ is a free $R[x]$ module, and has basis the elementary matrices $E_{ij}$ which are all $0$ except for the $ij$th entry, which is 1. Thus a basis for $M_n(R[x])$ as a free $R$ module is $x^kE_{ij}$. Similarly, a basis for $M_n(R)[x]$ is $E_{ij}x^k$, where I'm using the order of multiplication to distinguish these two rings.
You're trying to show that these $R$-algebras are isomorphic as (presumably) $R$-algebras, so the map should be $R$-linear. Thus such a map is defined by what it does on an $R$-generating set, or in this case since the $R$-algebras are free modules, an $R$-basis. Your map $phi$ is the map defined by sending $x^kE_{ij}$ to $E_{ij}x^k$. Since it is $R$-linear, it certainly preserves addition, so we just need to check that this map does indeed preserve multiplication. For this, it suffices to check on the basis.
$$phinewcommandof[1]{left({#1}right)}of{of{x^kE_{ij}}of{x^{k'}E_{i'j'}}}
=phiof{x^{k+k'}E_{ij}E_{i'j'}}=phiof{delta_{ji'}x^{k+k'}E_{ij'}}=delta_{ji'}E_{ij'}x^{k+k'},$$
and
$$phiof{x^kE_{ij}}phiof{x^{k'}E_{i'j'}}=of{E_{ij}x^k}of{E_{i'j'}x^{k'}}=E_{ij}E_{i'j'}x^{k+k'}=delta_{ji'}E_{ij'}x^{k+k'}.$$
Thus $phi$ preserves the multiplication as well.
If you think I've made an error, do let me know.
answered Dec 10 '18 at 14:25
jgonjgon
13.7k22041
$endgroup$
add a comment |
$begingroup$
Note that $M_n(R[x])$ is a free $R[x]$ module, and has basis the elementary matrices $E_{ij}$ which are all $0$ except for the $ij$th entry, which is 1. Thus a basis for $M_n(R[x])$ as a free $R$ module is $x^kE_{ij}$. Similarly, a basis for $M_n(R)[x]$ is $E_{ij}x^k$, where I'm using the order of multiplication to distinguish these two rings.
You're trying to show that these $R$-algebras are isomorphic as (presumably) $R$-algebras, so the map should be $R$-linear. Thus such a map is defined by what it does on an $R$-generating set, or in this case since the $R$-algebras are free modules, an $R$-basis. Your map $phi$ is the map defined by sending $x^kE_{ij}$ to $E_{ij}x^k$. Since it is $R$-linear, it certainly preserves addition, so we just need to check that this map does indeed preserve multiplication. For this, it suffices to check on the basis.
$$phinewcommandof[1]{left({#1}right)}of{of{x^kE_{ij}}of{x^{k'}E_{i'j'}}}
=phiof{x^{k+k'}E_{ij}E_{i'j'}}=phiof{delta_{ji'}x^{k+k'}E_{ij'}}=delta_{ji'}E_{ij'}x^{k+k'},$$
and
$$phiof{x^kE_{ij}}phiof{x^{k'}E_{i'j'}}=of{E_{ij}x^k}of{E_{i'j'}x^{k'}}=E_{ij}E_{i'j'}x^{k+k'}=delta_{ji'}E_{ij'}x^{k+k'}.$$
Thus $phi$ preserves the multiplication as well.
If you think I've made an error, do let me know.
answered Dec 10 '18 at 14:25
jgonjgon
13.7k22041
$endgroup$
add a comment |
$begingroup$
Note that $M_n(R[x])$ is a free $R[x]$ module, and has basis the elementary matrices $E_{ij}$ which are all $0$ except for the $ij$th entry, which is 1. Thus a basis for $M_n(R[x])$ as a free $R$ module is $x^kE_{ij}$. Similarly, a basis for $M_n(R)[x]$ is $E_{ij}x^k$, where I'm using the order of multiplication to distinguish these two rings.
You're trying to show that these $R$-algebras are isomorphic as (presumably) $R$-algebras, so the map should be $R$-linear. Thus such a map is defined by what it does on an $R$-generating set, or in this case since the $R$-algebras are free modules, an $R$-basis. Your map $phi$ is the map defined by sending $x^kE_{ij}$ to $E_{ij}x^k$. Since it is $R$-linear, it certainly preserves addition, so we just need to check that this map does indeed preserve multiplication. For this, it suffices to check on the basis.
$$phinewcommandof[1]{left({#1}right)}of{of{x^kE_{ij}}of{x^{k'}E_{i'j'}}}
=phiof{x^{k+k'}E_{ij}E_{i'j'}}=phiof{delta_{ji'}x^{k+k'}E_{ij'}}=delta_{ji'}E_{ij'}x^{k+k'},$$
and
$$phiof{x^kE_{ij}}phiof{x^{k'}E_{i'j'}}=of{E_{ij}x^k}of{E_{i'j'}x^{k'}}=E_{ij}E_{i'j'}x^{k+k'}=delta_{ji'}E_{ij'}x^{k+k'}.$$
Thus $phi$ preserves the multiplication as well.
If you think I've made an error, do let me know.
answered Dec 10 '18 at 14:25
jgonjgon
13.7k22041
$endgroup$
Note that $M_n(R[x])$ is a free $R[x]$ module, and has basis the elementary matrices $E_{ij}$ which are all $0$ except for the $ij$th entry, which is 1. Thus a basis for $M_n(R[x])$ as a free $R$ module is $x^kE_{ij}$. Similarly, a basis for $M_n(R)[x]$ is $E_{ij}x^k$, where I'm using the order of multiplication to distinguish these two rings.
You're trying to show that these $R$-algebras are isomorphic as (presumably) $R$-algebras, so the map should be $R$-linear. Thus such a map is defined by what it does on an $R$-generating set, or in this case since the $R$-algebras are free modules, an $R$-basis. Your map $phi$ is the map defined by sending $x^kE_{ij}$ to $E_{ij}x^k$. Since it is $R$-linear, it certainly preserves addition, so we just need to check that this map does indeed preserve multiplication. For this, it suffices to check on the basis.
$$phinewcommandof[1]{left({#1}right)}of{of{x^kE_{ij}}of{x^{k'}E_{i'j'}}}
=phiof{x^{k+k'}E_{ij}E_{i'j'}}=phiof{delta_{ji'}x^{k+k'}E_{ij'}}=delta_{ji'}E_{ij'}x^{k+k'},$$
and
$$phiof{x^kE_{ij}}phiof{x^{k'}E_{i'j'}}=of{E_{ij}x^k}of{E_{i'j'}x^{k'}}=E_{ij}E_{i'j'}x^{k+k'}=delta_{ji'}E_{ij'}x^{k+k'}.$$
Thus $phi$ preserves the multiplication as well.
If you think I've made an error, do let me know.
answered Dec 10 '18 at 14:25
jgonjgon
13.7k22041
answered Dec 10 '18 at 14:25
jgonjgon
13.7k22041
answered Dec 10 '18 at 14:25
jgonjgon
13.7k22041
answered Dec 10 '18 at 14:25
jgonjgon
13.7k22041
13.7k22041
add a comment |
add a comment |
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$begingroup$
a ring isomorphism would be a linear map as well between the vector spaces, so take a look at the standard bases in each of the rings. that should help you identify something.
$endgroup$
– dezdichado
Dec 10 '18 at 2:49
$begingroup$
@justin Why do you think it fails to be multiplicative? Do you have an example?
$endgroup$
– rschwieb
Dec 10 '18 at 14:38