Some Tor and Ext calculations

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

Let

be a field. I'd like to calculate $\text{Tor}_{k[x]}(k,k)$ and $\text{Ext}_{k[x]}(k,k)$

A good projective (free) resolution of

as a

-module is given by
$0 \to \cdots \to k[x] \stackrel{\eta}{\to} k[x] \stackrel{\epsilon} {\to}\to k \to 0$

where $\epsilon$ is the augmentation $\epsilon(x)=0$ and $\eta$ is multiplication by

.

So to calculate Tor we get the sequence
$0 \to k[x] \otimes_{k[x]} k \to k[x]\otimes_{k[x]} k \to 0$ which is just
$0 \to k \to k \to 0$

My question is what is the map $k \to k$ ? It comes from $\eta$ , but what happens when we use the isomoprhism $k[x] \otimes_{k[x]} k \simeq k$ ? (I think the map must be the zero map?)

Similar for $\text{Ext}$ . We end up with $\text{Hom}_{k[x]}(k[x],k)$ . Since a homomorphism $k[x] \to k$ is determined by where we send

and it can go anywhere in

, I think that $\text{Hom}_{k[x]}(k[x],k) \simeq k$ , but then I'm not sure what the map $k \to k$ should be.

Shadow · **Posted:** Mon, 19 Mar 2012 00:19:03 UTC

The tensor works by modding out by the action of x, so your multiplication by x is now multiplication by 0.

outermeasure · **Posted:** Mon, 19 Mar 2012 02:22:37 UTC

qwirk wrote:

Let

be a field. I'd like to calculate $\text{Tor}_{k[x]}(k,k)$ and $\text{Ext}_{k[x]}(k,k)$

A good projective (free) resolution of

as a

-module is given by
$0 \to \cdots \to k[x] \stackrel{\eta}{\to} k[x] \stackrel{\epsilon} {\to}\to k \to 0$

where $\epsilon$ is the augmentation $\epsilon(x)=0$ and $\eta$ is multiplication by

.

So to calculate Tor we get the sequence
$0 \to k[x] \otimes_{k[x]} k \to k[x]\otimes_{k[x]} k \to 0$ which is just
$0 \to k \to k \to 0$

My question is what is the map $k \to k$ ? It comes from $\eta$ , but what happens when we use the isomoprhism $k[x] \otimes_{k[x]} k \simeq k$ ? (I think the map must be the zero map?)

Similar for $\text{Ext}$ . We end up with $\text{Hom}_{k[x]}(k[x],k)$ . Since a homomorphism $k[x] \to k$ is determined by where we send

and it can go anywhere in

, I think that $\text{Hom}_{k[x]}(k[x],k) \simeq k$ , but then I'm not sure what the map $k \to k$ should be.

Hmm... you haven't seen dual map and the induced map on tensor products when you were doing linear algebra?

Shadow has done the Tor, so I'll do the Ext.

Recall the map $\mathop{\mathrm{Hom}}_R(P^j,B)\to\mathop{\mathrm{Hom}}_R(P^{j+1},B)$ is defined as $(P^j\xrightarrow{f}B)\mapsto(P^{j+1}\to P^j\xrightarrow{f}B)$ .

So the map $\mathop{\mathrm{Hom}}_{k[x]}(\underbrace{k[x]}_{P^0},k)\xrightarrow{\eta^*}\mathop{\mathrm{Hom}}_{k[x]}(\underbrace{k[x]}_{P^1},k)$ is defined as taking the map "evaluate at $a\in k$ " and mapping it to "multiply by

then evaluate at $a\in k$ ". Of course, the multiplication by

is absorbed in the identification $\mathop{\mathrm{Hom}}_{k[x]}(P^j,k)\cong k$ (j=0,1), so $k\to k$ is the identity map.

Indeed, you wouldn't be asking the question had you written xk[x]

instead of k[x]

for your

(although you will then have to be careful when you do the Tor functor calculation not to fall into the obvious trap).

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

Thanks a lot to both of you!

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Some Tor and Ext calculations

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