troubles showing existence of Clifford-algebra

Question

We had the following definition in class:

Let $V$ be a vector space, $K$ a field and $Q$ be a quadratic form. We call $(C(V,Q),j)=C$ a Clifford-algebra if:

1. $C$ is an assoziative algebra with 1,
2. $j:V\to C$ is linear with $j(x)^2=Q(x)\cdot 1$ for all $x\in V$,
3. if $A$ is another algebra with 1 and $u: V\to A$ linear with $u(x)^2=Q(x)\cdot 1$, then there exists a unique algebra homomorphism $U:C\to A$ with $u=U\circ j$

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We didn't prove the existence - which was left as an exercise for the reader. So I want to try it but having some troubles. My attempt:

Let $TV=K\oplus V\oplus(V\otimes V)\oplus...$ be the tensor algebra and $I$ the two sided ideal generated by the elements of the form $x\otimes x-Q(x)\cdot 1$ with $x\in V$.

I claim that $C=TV/I$ is a Clifford-algebra. So I have to check 1. to 3. Clearly 1. holds. For 2. I want to consider $$ j:V\xrightarrow{i}TV\xrightarrow{\pi}C $$with the canonical projection $\pi$ and canonical map $i$. Then cleary $i$ is linear. But how do you get $j(x)^2=Q(x)\cdot 1$?

To 3.: Let now $A$ be any algebra, $u: V\to A$ linear and $u(x)^2=Q(x)\cdot 1$. Then $$U(x_1\otimes...\otimes x_k)=u(x_1)\cdots u(x_k) $$ is a homomorphism $U: TV\to A$. Now I claim $I\subset \ker U$ but don't know why it holds. I've tried to calculate $U(x\otimes x-Q(x)\cdot 1)$ but didn't get along. Any idea?

If $I\subset \ker U$ then we get a homomorphism $\tilde U: C\to A$ as desired.

So let $\tilde W:C\to A$ be another homorphism. I want to show $\tilde W=\tilde U$. Since elements of $V$ generate $TV$ and therefor also $C$ (multiplicative) and we have $u=\tilde W\circ j=\tilde U\circ j$, can you follow $\tilde U=\tilde W$? Why?

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To sum up my questions:

How do you get $j(x)^2=Q(x)\cdot 1$ above?

How do you get $I\subset \ker U$ above?

How do you get the conclusion $\tilde U=\tilde W$?

Sorry for writing that much but I hope for any ideas!

Answer 1

I think you need to spend a little more time thinking about these. The first two answers follow directly from the definitions.

How do you get $j(x)^2=Q(x)⋅1$ above?

By definition of $C$, $j(x)^2=(\pi i(x))^2=(x+ I)\otimes(x+ I)=(x\otimes x)+ I= Q(x)⋅1 + I$. So the things are all equal in the quotient.

How do you get $I⊂\ker U$ above?

By definition of $U$, $U(x\otimes x - Q(x))=u(x)\cdot u(x)-Q(x)1_A=Q(x)1_A-Q(x)1_A=0$. All the generators are in the kernel, so $I$ is in the kernel.

can you follow $\bar U=\bar W$?

Yes: if the disagree on an element of $C$, they must disagree on some element of $V$. But that is ruled out by $\bar Uj=\bar Wj$.