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Let $k$ be a field. I know that Clifford algebra $C(V,q)$ is central simple $k$-algebra where $\dim V$ is $2n$ and $q$ is non degenerate. But what does it mean $k$-algebra here? What is the $k$ action on $C(V,q)$? Is it $a\otimes w$ for $a \in k$ and $w\in C(V,q)$? If we want to prove $C(V,q)$ has central $k$ that is all element of $k$ commute with $ C(V,q)$, should I prove that $1\otimes e_i=e_i \otimes 1 $ for all basis element $e_i$ of $V$?

R_D
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Hui YI LU
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  • $k$-algebra means that 1) $C(V,q)$ is a vector space over $k$, and 2) its multiplication rule is bilinear over $k$. In other words, if $x,y,z\in C(V,q)$ and $a,b\in k$ are arbitrary, then $(ax+by)z=a(xz)+b(yz)$ and $x(ay+bz)=a(xy)+b(xz)$. For $C(V,q)$ to be central you need to show that if $z\in C(v,q)$ has the property that $zx=xz$ for all $x\in C(V,q)$, then $z$ has to be a $k$-multiple of $1_{C(V,q)}$. – Jyrki Lahtonen Oct 03 '16 at 13:20
  • IOW you don't need tensor products to get the $k$-action. An element of $C(V,q)$ is just a finite sum $a+\sum_i a_iei+\sum_{i,j}a_{i,j}e_i\otimes e_j+\cdots$. You multiply it by a scalar $b\in k$ by multiplying all the coefficients $a, a_i, a_{i,j},\ldots$. – Jyrki Lahtonen Oct 03 '16 at 13:23
  • I know that real and complex Clifford algebras resolve into central simple algebras for nondegenerate spaces of even dimension, but does it really hold for all fields? I just haven't thought about it enough to know for sure. Do you have a good reference? – rschwieb Oct 03 '16 at 13:31
  • https://en.wikipedia.org/wiki/Clifford_algebra I think wiki will be a good reference – Hui YI LU Oct 03 '16 at 13:40

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