Computing projective resolutions over quotients of polynomial rings
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I'd like to find projective resolutions for $k$ considered as an $R$-module,
where $k$ is a field and $R$ as follows:
(i) $R=frac{k[x]}{(x^n)}$;
(ii) $R=k[x,y]$
(iii) $R=frac{k[x,y]}{(x^n,y^m)}$;
(iv) $ R=frac{k[x,y]}{(x^2,y^2,xy)}$.
I try to find free resolutions, because I don't know any obvious projective non-free modules over these rings:
For (i), I've found:
$$ ldots rightarrow R xrightarrow{cdot x} R xrightarrow{cdot x^{n-1}} R xrightarrow{cdot x} R rightarrow k rightarrow 0$$
For (ii), I've found:
$$ 0 xrightarrow{cdot x} R xrightarrow{1mapsto(y,-x)} Roplus R xrightarrow{(1,0)mapsto x, (0,1) mapsto y} R rightarrow k rightarrow 0$$
For (iii), I think it should end similarly to (ii), but I'm not sure what comes before:
$$ ? xrightarrow{} R xrightarrow{1mapsto(y,-x)} Roplus R xrightarrow{(1,0)mapsto x, (0,1) mapsto y} R rightarrow k rightarrow 0$$
Are these correct?
Thanks in advance.
ring-theory commutative-algebra modules homology-cohomology homological-algebra
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up vote
3
down vote
favorite
I'd like to find projective resolutions for $k$ considered as an $R$-module,
where $k$ is a field and $R$ as follows:
(i) $R=frac{k[x]}{(x^n)}$;
(ii) $R=k[x,y]$
(iii) $R=frac{k[x,y]}{(x^n,y^m)}$;
(iv) $ R=frac{k[x,y]}{(x^2,y^2,xy)}$.
I try to find free resolutions, because I don't know any obvious projective non-free modules over these rings:
For (i), I've found:
$$ ldots rightarrow R xrightarrow{cdot x} R xrightarrow{cdot x^{n-1}} R xrightarrow{cdot x} R rightarrow k rightarrow 0$$
For (ii), I've found:
$$ 0 xrightarrow{cdot x} R xrightarrow{1mapsto(y,-x)} Roplus R xrightarrow{(1,0)mapsto x, (0,1) mapsto y} R rightarrow k rightarrow 0$$
For (iii), I think it should end similarly to (ii), but I'm not sure what comes before:
$$ ? xrightarrow{} R xrightarrow{1mapsto(y,-x)} Roplus R xrightarrow{(1,0)mapsto x, (0,1) mapsto y} R rightarrow k rightarrow 0$$
Are these correct?
Thanks in advance.
ring-theory commutative-algebra modules homology-cohomology homological-algebra
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'd like to find projective resolutions for $k$ considered as an $R$-module,
where $k$ is a field and $R$ as follows:
(i) $R=frac{k[x]}{(x^n)}$;
(ii) $R=k[x,y]$
(iii) $R=frac{k[x,y]}{(x^n,y^m)}$;
(iv) $ R=frac{k[x,y]}{(x^2,y^2,xy)}$.
I try to find free resolutions, because I don't know any obvious projective non-free modules over these rings:
For (i), I've found:
$$ ldots rightarrow R xrightarrow{cdot x} R xrightarrow{cdot x^{n-1}} R xrightarrow{cdot x} R rightarrow k rightarrow 0$$
For (ii), I've found:
$$ 0 xrightarrow{cdot x} R xrightarrow{1mapsto(y,-x)} Roplus R xrightarrow{(1,0)mapsto x, (0,1) mapsto y} R rightarrow k rightarrow 0$$
For (iii), I think it should end similarly to (ii), but I'm not sure what comes before:
$$ ? xrightarrow{} R xrightarrow{1mapsto(y,-x)} Roplus R xrightarrow{(1,0)mapsto x, (0,1) mapsto y} R rightarrow k rightarrow 0$$
Are these correct?
Thanks in advance.
ring-theory commutative-algebra modules homology-cohomology homological-algebra
I'd like to find projective resolutions for $k$ considered as an $R$-module,
where $k$ is a field and $R$ as follows:
(i) $R=frac{k[x]}{(x^n)}$;
(ii) $R=k[x,y]$
(iii) $R=frac{k[x,y]}{(x^n,y^m)}$;
(iv) $ R=frac{k[x,y]}{(x^2,y^2,xy)}$.
I try to find free resolutions, because I don't know any obvious projective non-free modules over these rings:
For (i), I've found:
$$ ldots rightarrow R xrightarrow{cdot x} R xrightarrow{cdot x^{n-1}} R xrightarrow{cdot x} R rightarrow k rightarrow 0$$
For (ii), I've found:
$$ 0 xrightarrow{cdot x} R xrightarrow{1mapsto(y,-x)} Roplus R xrightarrow{(1,0)mapsto x, (0,1) mapsto y} R rightarrow k rightarrow 0$$
For (iii), I think it should end similarly to (ii), but I'm not sure what comes before:
$$ ? xrightarrow{} R xrightarrow{1mapsto(y,-x)} Roplus R xrightarrow{(1,0)mapsto x, (0,1) mapsto y} R rightarrow k rightarrow 0$$
Are these correct?
Thanks in advance.
ring-theory commutative-algebra modules homology-cohomology homological-algebra
ring-theory commutative-algebra modules homology-cohomology homological-algebra
edited Nov 20 at 0:17
Sky
1,228212
1,228212
asked Nov 18 at 22:22
SSF
338110
338110
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2 Answers
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For (i), your resolution works. Let me call it $P(n) to mathbb k$. Your resolution for (ii) works too, it is called the Koszul resolution. Now note that if the corresponding algebra is $A(n)$, then $A(n)otimes A(m)$ is the algebra in (iii). Since tensoring over $mathbb k$ is exact, it follows that $P(n)otimes P(m)$ is a resolution of the trival module in this case. This is a complex that has in degree $d$ the sum $P(n)_iotimes P(n)_j$ where $i+j=d$.
Finally, let me do (iv) explicitly. This algebra $A$ has dimension 3, spanned by 1,$x$ and $y$, and we can begin the resolution by $A^2to Ato mathbb k$ where $f:A^2to A$ covers $K=ker(Ato mathbb k) = (x,y)$ by sending $(p,q)$ to $xp+yq$. The kernel of this is easy to work out: since in this algebra $xp = xp(0,0)$ and $y q= y q(0,0)$, this map has kernel the pairs $(p,q)$ that both vanish at zero. Thus the kernel of $f$ is $Koplus K$, which we can cover by $foplus f$. Repeating, we arrive at a minimal free resolution
$$ cdots to A^8to A^4to A^2to A.$$
This is consistent with the fact that we can present $A$ by two free generators $x,y$ and monomial relations $x^2,xy,yx,y^2$, which generate $2^{n+1}$ Anick $n$-chains for each $ninmathbb N_0$.
add a comment |
up vote
1
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I will give part solution to your question, this is what I can do.
For (i), and (ii) the resolution you gave is right. More general, you can get the free resolution of $k[x_1,...,x_n]$ by tensoring $n$-copies $k[x_1,...,x_n]xrightarrow {x_i}k[x_1,...,x_n]$ then we can get an acyclic complex which is called Koszul complex, and it is a resolution of $k$.
For (iv) $k[x,y]/{(x,y)^2}$ is a finite dimensional local algebra and radical square zero, so any kernel of a projective cover is semisimple. There is only one simple module $k$. If we get the short exact sequence $0rightarrow kcoprod kcong (x,y)/{(x,y)^2}xrightarrow {inc}k[x,y]/{(x,y)^2}rightarrow krightarrow 0$. So the projective dimension of $k$ is infinite.
For (iii), giving a concrete projective resolution is difficult I think. I am expecting someone can give one. Denote $A=k[x,y]/{(x^n,y^m)}$ is finite dimensional local algebra. Next I will give a proof that $A$ is self-injective and projective dimension of $k$ is $infty$. Remark that $k[x,y]/{(x,y)^2}$ is not self-injective, this can be check directly by Baer's criterion.
In general, suppose $B$ is finite dimensional local commutative k-algebra, where k is a field. If $mathrm{soc}(B)$ is simple, then $B$ is self-injective.
Proof: $mathrm{soc}(B)rightarrow B$ is essential extension. Select a injective envelope of $mathrm{soc}(B)$, and denote it by $I(mathrm{soc}(B))$, so there is a natural induced map $alpha :Brightarrow I(mathrm{soc}(B))$. This is a monomorphism since essential extension. Remark $D=mathrm{Hom}_k(-,k):B-mathrm{mod}rightarrow B-mathrm{mod}$ is equivalence between finite generated modules. So $D(I(mathrm{soc}(B))$ and $B$ are both projective covers of the unique simple module of $B$. Hence they have the same length. So $alpha $ is isomorphism.
The global dimension of a self-injective algebra is $0$ or $infty$.
So we know projective dimension of $k$ as $A$ is infinite since there is only one simple module $A-mathrm{mod}$ and $A$ is finite dimensional algebra.
Hi, thanks very much for the reply. I'm not familiar with lots of these terms (local algebra, projective cover, soc, self-injective algebra). May I ask what good textbook/source is there to learn these terms?
– SSF
Nov 19 at 1:52
1
@SSF local means it has only one maximal ideal.Radical,socle ,projective cover you can find in Wikipedia ,homological algebra ,Rings modules and categories.Of course,Wikipedia is very convenient.
– Sky
Nov 19 at 2:12
1
Rings and Categories of Modules.GTM13
– Sky
Nov 19 at 2:21
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
For (i), your resolution works. Let me call it $P(n) to mathbb k$. Your resolution for (ii) works too, it is called the Koszul resolution. Now note that if the corresponding algebra is $A(n)$, then $A(n)otimes A(m)$ is the algebra in (iii). Since tensoring over $mathbb k$ is exact, it follows that $P(n)otimes P(m)$ is a resolution of the trival module in this case. This is a complex that has in degree $d$ the sum $P(n)_iotimes P(n)_j$ where $i+j=d$.
Finally, let me do (iv) explicitly. This algebra $A$ has dimension 3, spanned by 1,$x$ and $y$, and we can begin the resolution by $A^2to Ato mathbb k$ where $f:A^2to A$ covers $K=ker(Ato mathbb k) = (x,y)$ by sending $(p,q)$ to $xp+yq$. The kernel of this is easy to work out: since in this algebra $xp = xp(0,0)$ and $y q= y q(0,0)$, this map has kernel the pairs $(p,q)$ that both vanish at zero. Thus the kernel of $f$ is $Koplus K$, which we can cover by $foplus f$. Repeating, we arrive at a minimal free resolution
$$ cdots to A^8to A^4to A^2to A.$$
This is consistent with the fact that we can present $A$ by two free generators $x,y$ and monomial relations $x^2,xy,yx,y^2$, which generate $2^{n+1}$ Anick $n$-chains for each $ninmathbb N_0$.
add a comment |
up vote
0
down vote
accepted
For (i), your resolution works. Let me call it $P(n) to mathbb k$. Your resolution for (ii) works too, it is called the Koszul resolution. Now note that if the corresponding algebra is $A(n)$, then $A(n)otimes A(m)$ is the algebra in (iii). Since tensoring over $mathbb k$ is exact, it follows that $P(n)otimes P(m)$ is a resolution of the trival module in this case. This is a complex that has in degree $d$ the sum $P(n)_iotimes P(n)_j$ where $i+j=d$.
Finally, let me do (iv) explicitly. This algebra $A$ has dimension 3, spanned by 1,$x$ and $y$, and we can begin the resolution by $A^2to Ato mathbb k$ where $f:A^2to A$ covers $K=ker(Ato mathbb k) = (x,y)$ by sending $(p,q)$ to $xp+yq$. The kernel of this is easy to work out: since in this algebra $xp = xp(0,0)$ and $y q= y q(0,0)$, this map has kernel the pairs $(p,q)$ that both vanish at zero. Thus the kernel of $f$ is $Koplus K$, which we can cover by $foplus f$. Repeating, we arrive at a minimal free resolution
$$ cdots to A^8to A^4to A^2to A.$$
This is consistent with the fact that we can present $A$ by two free generators $x,y$ and monomial relations $x^2,xy,yx,y^2$, which generate $2^{n+1}$ Anick $n$-chains for each $ninmathbb N_0$.
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
For (i), your resolution works. Let me call it $P(n) to mathbb k$. Your resolution for (ii) works too, it is called the Koszul resolution. Now note that if the corresponding algebra is $A(n)$, then $A(n)otimes A(m)$ is the algebra in (iii). Since tensoring over $mathbb k$ is exact, it follows that $P(n)otimes P(m)$ is a resolution of the trival module in this case. This is a complex that has in degree $d$ the sum $P(n)_iotimes P(n)_j$ where $i+j=d$.
Finally, let me do (iv) explicitly. This algebra $A$ has dimension 3, spanned by 1,$x$ and $y$, and we can begin the resolution by $A^2to Ato mathbb k$ where $f:A^2to A$ covers $K=ker(Ato mathbb k) = (x,y)$ by sending $(p,q)$ to $xp+yq$. The kernel of this is easy to work out: since in this algebra $xp = xp(0,0)$ and $y q= y q(0,0)$, this map has kernel the pairs $(p,q)$ that both vanish at zero. Thus the kernel of $f$ is $Koplus K$, which we can cover by $foplus f$. Repeating, we arrive at a minimal free resolution
$$ cdots to A^8to A^4to A^2to A.$$
This is consistent with the fact that we can present $A$ by two free generators $x,y$ and monomial relations $x^2,xy,yx,y^2$, which generate $2^{n+1}$ Anick $n$-chains for each $ninmathbb N_0$.
For (i), your resolution works. Let me call it $P(n) to mathbb k$. Your resolution for (ii) works too, it is called the Koszul resolution. Now note that if the corresponding algebra is $A(n)$, then $A(n)otimes A(m)$ is the algebra in (iii). Since tensoring over $mathbb k$ is exact, it follows that $P(n)otimes P(m)$ is a resolution of the trival module in this case. This is a complex that has in degree $d$ the sum $P(n)_iotimes P(n)_j$ where $i+j=d$.
Finally, let me do (iv) explicitly. This algebra $A$ has dimension 3, spanned by 1,$x$ and $y$, and we can begin the resolution by $A^2to Ato mathbb k$ where $f:A^2to A$ covers $K=ker(Ato mathbb k) = (x,y)$ by sending $(p,q)$ to $xp+yq$. The kernel of this is easy to work out: since in this algebra $xp = xp(0,0)$ and $y q= y q(0,0)$, this map has kernel the pairs $(p,q)$ that both vanish at zero. Thus the kernel of $f$ is $Koplus K$, which we can cover by $foplus f$. Repeating, we arrive at a minimal free resolution
$$ cdots to A^8to A^4to A^2to A.$$
This is consistent with the fact that we can present $A$ by two free generators $x,y$ and monomial relations $x^2,xy,yx,y^2$, which generate $2^{n+1}$ Anick $n$-chains for each $ninmathbb N_0$.
answered Nov 19 at 14:39
Pedro Tamaroff♦
95.8k10150295
95.8k10150295
add a comment |
add a comment |
up vote
1
down vote
I will give part solution to your question, this is what I can do.
For (i), and (ii) the resolution you gave is right. More general, you can get the free resolution of $k[x_1,...,x_n]$ by tensoring $n$-copies $k[x_1,...,x_n]xrightarrow {x_i}k[x_1,...,x_n]$ then we can get an acyclic complex which is called Koszul complex, and it is a resolution of $k$.
For (iv) $k[x,y]/{(x,y)^2}$ is a finite dimensional local algebra and radical square zero, so any kernel of a projective cover is semisimple. There is only one simple module $k$. If we get the short exact sequence $0rightarrow kcoprod kcong (x,y)/{(x,y)^2}xrightarrow {inc}k[x,y]/{(x,y)^2}rightarrow krightarrow 0$. So the projective dimension of $k$ is infinite.
For (iii), giving a concrete projective resolution is difficult I think. I am expecting someone can give one. Denote $A=k[x,y]/{(x^n,y^m)}$ is finite dimensional local algebra. Next I will give a proof that $A$ is self-injective and projective dimension of $k$ is $infty$. Remark that $k[x,y]/{(x,y)^2}$ is not self-injective, this can be check directly by Baer's criterion.
In general, suppose $B$ is finite dimensional local commutative k-algebra, where k is a field. If $mathrm{soc}(B)$ is simple, then $B$ is self-injective.
Proof: $mathrm{soc}(B)rightarrow B$ is essential extension. Select a injective envelope of $mathrm{soc}(B)$, and denote it by $I(mathrm{soc}(B))$, so there is a natural induced map $alpha :Brightarrow I(mathrm{soc}(B))$. This is a monomorphism since essential extension. Remark $D=mathrm{Hom}_k(-,k):B-mathrm{mod}rightarrow B-mathrm{mod}$ is equivalence between finite generated modules. So $D(I(mathrm{soc}(B))$ and $B$ are both projective covers of the unique simple module of $B$. Hence they have the same length. So $alpha $ is isomorphism.
The global dimension of a self-injective algebra is $0$ or $infty$.
So we know projective dimension of $k$ as $A$ is infinite since there is only one simple module $A-mathrm{mod}$ and $A$ is finite dimensional algebra.
Hi, thanks very much for the reply. I'm not familiar with lots of these terms (local algebra, projective cover, soc, self-injective algebra). May I ask what good textbook/source is there to learn these terms?
– SSF
Nov 19 at 1:52
1
@SSF local means it has only one maximal ideal.Radical,socle ,projective cover you can find in Wikipedia ,homological algebra ,Rings modules and categories.Of course,Wikipedia is very convenient.
– Sky
Nov 19 at 2:12
1
Rings and Categories of Modules.GTM13
– Sky
Nov 19 at 2:21
add a comment |
up vote
1
down vote
I will give part solution to your question, this is what I can do.
For (i), and (ii) the resolution you gave is right. More general, you can get the free resolution of $k[x_1,...,x_n]$ by tensoring $n$-copies $k[x_1,...,x_n]xrightarrow {x_i}k[x_1,...,x_n]$ then we can get an acyclic complex which is called Koszul complex, and it is a resolution of $k$.
For (iv) $k[x,y]/{(x,y)^2}$ is a finite dimensional local algebra and radical square zero, so any kernel of a projective cover is semisimple. There is only one simple module $k$. If we get the short exact sequence $0rightarrow kcoprod kcong (x,y)/{(x,y)^2}xrightarrow {inc}k[x,y]/{(x,y)^2}rightarrow krightarrow 0$. So the projective dimension of $k$ is infinite.
For (iii), giving a concrete projective resolution is difficult I think. I am expecting someone can give one. Denote $A=k[x,y]/{(x^n,y^m)}$ is finite dimensional local algebra. Next I will give a proof that $A$ is self-injective and projective dimension of $k$ is $infty$. Remark that $k[x,y]/{(x,y)^2}$ is not self-injective, this can be check directly by Baer's criterion.
In general, suppose $B$ is finite dimensional local commutative k-algebra, where k is a field. If $mathrm{soc}(B)$ is simple, then $B$ is self-injective.
Proof: $mathrm{soc}(B)rightarrow B$ is essential extension. Select a injective envelope of $mathrm{soc}(B)$, and denote it by $I(mathrm{soc}(B))$, so there is a natural induced map $alpha :Brightarrow I(mathrm{soc}(B))$. This is a monomorphism since essential extension. Remark $D=mathrm{Hom}_k(-,k):B-mathrm{mod}rightarrow B-mathrm{mod}$ is equivalence between finite generated modules. So $D(I(mathrm{soc}(B))$ and $B$ are both projective covers of the unique simple module of $B$. Hence they have the same length. So $alpha $ is isomorphism.
The global dimension of a self-injective algebra is $0$ or $infty$.
So we know projective dimension of $k$ as $A$ is infinite since there is only one simple module $A-mathrm{mod}$ and $A$ is finite dimensional algebra.
Hi, thanks very much for the reply. I'm not familiar with lots of these terms (local algebra, projective cover, soc, self-injective algebra). May I ask what good textbook/source is there to learn these terms?
– SSF
Nov 19 at 1:52
1
@SSF local means it has only one maximal ideal.Radical,socle ,projective cover you can find in Wikipedia ,homological algebra ,Rings modules and categories.Of course,Wikipedia is very convenient.
– Sky
Nov 19 at 2:12
1
Rings and Categories of Modules.GTM13
– Sky
Nov 19 at 2:21
add a comment |
up vote
1
down vote
up vote
1
down vote
I will give part solution to your question, this is what I can do.
For (i), and (ii) the resolution you gave is right. More general, you can get the free resolution of $k[x_1,...,x_n]$ by tensoring $n$-copies $k[x_1,...,x_n]xrightarrow {x_i}k[x_1,...,x_n]$ then we can get an acyclic complex which is called Koszul complex, and it is a resolution of $k$.
For (iv) $k[x,y]/{(x,y)^2}$ is a finite dimensional local algebra and radical square zero, so any kernel of a projective cover is semisimple. There is only one simple module $k$. If we get the short exact sequence $0rightarrow kcoprod kcong (x,y)/{(x,y)^2}xrightarrow {inc}k[x,y]/{(x,y)^2}rightarrow krightarrow 0$. So the projective dimension of $k$ is infinite.
For (iii), giving a concrete projective resolution is difficult I think. I am expecting someone can give one. Denote $A=k[x,y]/{(x^n,y^m)}$ is finite dimensional local algebra. Next I will give a proof that $A$ is self-injective and projective dimension of $k$ is $infty$. Remark that $k[x,y]/{(x,y)^2}$ is not self-injective, this can be check directly by Baer's criterion.
In general, suppose $B$ is finite dimensional local commutative k-algebra, where k is a field. If $mathrm{soc}(B)$ is simple, then $B$ is self-injective.
Proof: $mathrm{soc}(B)rightarrow B$ is essential extension. Select a injective envelope of $mathrm{soc}(B)$, and denote it by $I(mathrm{soc}(B))$, so there is a natural induced map $alpha :Brightarrow I(mathrm{soc}(B))$. This is a monomorphism since essential extension. Remark $D=mathrm{Hom}_k(-,k):B-mathrm{mod}rightarrow B-mathrm{mod}$ is equivalence between finite generated modules. So $D(I(mathrm{soc}(B))$ and $B$ are both projective covers of the unique simple module of $B$. Hence they have the same length. So $alpha $ is isomorphism.
The global dimension of a self-injective algebra is $0$ or $infty$.
So we know projective dimension of $k$ as $A$ is infinite since there is only one simple module $A-mathrm{mod}$ and $A$ is finite dimensional algebra.
I will give part solution to your question, this is what I can do.
For (i), and (ii) the resolution you gave is right. More general, you can get the free resolution of $k[x_1,...,x_n]$ by tensoring $n$-copies $k[x_1,...,x_n]xrightarrow {x_i}k[x_1,...,x_n]$ then we can get an acyclic complex which is called Koszul complex, and it is a resolution of $k$.
For (iv) $k[x,y]/{(x,y)^2}$ is a finite dimensional local algebra and radical square zero, so any kernel of a projective cover is semisimple. There is only one simple module $k$. If we get the short exact sequence $0rightarrow kcoprod kcong (x,y)/{(x,y)^2}xrightarrow {inc}k[x,y]/{(x,y)^2}rightarrow krightarrow 0$. So the projective dimension of $k$ is infinite.
For (iii), giving a concrete projective resolution is difficult I think. I am expecting someone can give one. Denote $A=k[x,y]/{(x^n,y^m)}$ is finite dimensional local algebra. Next I will give a proof that $A$ is self-injective and projective dimension of $k$ is $infty$. Remark that $k[x,y]/{(x,y)^2}$ is not self-injective, this can be check directly by Baer's criterion.
In general, suppose $B$ is finite dimensional local commutative k-algebra, where k is a field. If $mathrm{soc}(B)$ is simple, then $B$ is self-injective.
Proof: $mathrm{soc}(B)rightarrow B$ is essential extension. Select a injective envelope of $mathrm{soc}(B)$, and denote it by $I(mathrm{soc}(B))$, so there is a natural induced map $alpha :Brightarrow I(mathrm{soc}(B))$. This is a monomorphism since essential extension. Remark $D=mathrm{Hom}_k(-,k):B-mathrm{mod}rightarrow B-mathrm{mod}$ is equivalence between finite generated modules. So $D(I(mathrm{soc}(B))$ and $B$ are both projective covers of the unique simple module of $B$. Hence they have the same length. So $alpha $ is isomorphism.
The global dimension of a self-injective algebra is $0$ or $infty$.
So we know projective dimension of $k$ as $A$ is infinite since there is only one simple module $A-mathrm{mod}$ and $A$ is finite dimensional algebra.
edited Nov 20 at 15:34
user26857
39.2k123882
39.2k123882
answered Nov 19 at 1:18
Sky
1,228212
1,228212
Hi, thanks very much for the reply. I'm not familiar with lots of these terms (local algebra, projective cover, soc, self-injective algebra). May I ask what good textbook/source is there to learn these terms?
– SSF
Nov 19 at 1:52
1
@SSF local means it has only one maximal ideal.Radical,socle ,projective cover you can find in Wikipedia ,homological algebra ,Rings modules and categories.Of course,Wikipedia is very convenient.
– Sky
Nov 19 at 2:12
1
Rings and Categories of Modules.GTM13
– Sky
Nov 19 at 2:21
add a comment |
Hi, thanks very much for the reply. I'm not familiar with lots of these terms (local algebra, projective cover, soc, self-injective algebra). May I ask what good textbook/source is there to learn these terms?
– SSF
Nov 19 at 1:52
1
@SSF local means it has only one maximal ideal.Radical,socle ,projective cover you can find in Wikipedia ,homological algebra ,Rings modules and categories.Of course,Wikipedia is very convenient.
– Sky
Nov 19 at 2:12
1
Rings and Categories of Modules.GTM13
– Sky
Nov 19 at 2:21
Hi, thanks very much for the reply. I'm not familiar with lots of these terms (local algebra, projective cover, soc, self-injective algebra). May I ask what good textbook/source is there to learn these terms?
– SSF
Nov 19 at 1:52
Hi, thanks very much for the reply. I'm not familiar with lots of these terms (local algebra, projective cover, soc, self-injective algebra). May I ask what good textbook/source is there to learn these terms?
– SSF
Nov 19 at 1:52
1
1
@SSF local means it has only one maximal ideal.Radical,socle ,projective cover you can find in Wikipedia ,homological algebra ,Rings modules and categories.Of course,Wikipedia is very convenient.
– Sky
Nov 19 at 2:12
@SSF local means it has only one maximal ideal.Radical,socle ,projective cover you can find in Wikipedia ,homological algebra ,Rings modules and categories.Of course,Wikipedia is very convenient.
– Sky
Nov 19 at 2:12
1
1
Rings and Categories of Modules.GTM13
– Sky
Nov 19 at 2:21
Rings and Categories of Modules.GTM13
– Sky
Nov 19 at 2:21
add a comment |
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