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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
edited 23 hours ago
Sky
1,178212
asked 2 days ago
SSF
19510
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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
edited 23 hours ago
Sky
1,178212
asked 2 days ago
SSF
19510
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
edited 23 hours ago
Sky
1,178212
asked 2 days ago
SSF
19510
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 23 hours ago
Sky
1,178212
asked 2 days ago
SSF
19510
edited 23 hours ago
Sky
1,178212
asked 2 days ago
SSF
19510
edited 23 hours ago
Sky
1,178212
edited 23 hours ago
Sky
1,178212
edited 23 hours ago
Sky
1,178212
1,178212
asked 2 days ago
SSF
19510
asked 2 days ago
SSF
19510
asked 2 days ago
SSF
19510
19510
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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 $ninmathbb N_0$.
answered yesterday
Pedro Tamaroff♦
95.5k10149295
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up vote
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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.
edited 7 hours ago
user26857
39.1k123882
answered yesterday
Sky
1,178212
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
yesterday
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
yesterday
1
Rings and Categories of Modules.GTM13
– Sky
yesterday
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 $ninmathbb N_0$.
answered yesterday
Pedro Tamaroff♦
95.5k10149295
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 $ninmathbb N_0$.
answered yesterday
Pedro Tamaroff♦
95.5k10149295
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 $ninmathbb N_0$.
answered yesterday
Pedro Tamaroff♦
95.5k10149295
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 yesterday
Pedro Tamaroff♦
95.5k10149295
answered yesterday
Pedro Tamaroff♦
95.5k10149295
answered yesterday
Pedro Tamaroff♦
95.5k10149295
answered yesterday
Pedro Tamaroff♦
95.5k10149295
95.5k10149295
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.
edited 7 hours ago
user26857
39.1k123882
answered yesterday
Sky
1,178212
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
yesterday
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
yesterday
1
Rings and Categories of Modules.GTM13
– Sky
yesterday
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.
edited 7 hours ago
user26857
39.1k123882
answered yesterday
Sky
1,178212
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
yesterday
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
yesterday
1
Rings and Categories of Modules.GTM13
– Sky
yesterday
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.
edited 7 hours ago
user26857
39.1k123882
answered yesterday
Sky
1,178212
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 7 hours ago
user26857
39.1k123882
answered yesterday
Sky
1,178212
edited 7 hours ago
user26857
39.1k123882
edited 7 hours ago
user26857
39.1k123882
edited 7 hours ago
user26857
39.1k123882
39.1k123882
answered yesterday
Sky
1,178212
answered yesterday
Sky
1,178212
answered yesterday
Sky
1,178212
1,178212
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
yesterday
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
yesterday
1
Rings and Categories of Modules.GTM13
– Sky
yesterday
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
yesterday
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
yesterday
1
Rings and Categories of Modules.GTM13
– Sky
yesterday
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
yesterday
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
yesterday
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
yesterday
@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
yesterday
1
1
Rings and Categories of Modules.GTM13
– Sky
yesterday
Rings and Categories of Modules.GTM13
– Sky
yesterday
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