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Let $k$ be a field which does not have characteristic 2. Let $M$ be the free $k$-vector space generated by two elements $\{ c, x \}$. Let $T(M)$ be the tensor algebra of $M$ and let $I$ be the ideal generated by the elements $$\{c^2 -1, x^2, cx + xc \}.$$

I am trying to show that $B = \{ 1, c, x, cx\}$ forms a basis of the quotient space $T(M) / I$. I can see why it spans the space but I cannot show that it is linearly independent. Any help would be appreciated.

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As you probably know, this is an example of a Clifford algebra attached to the quadratic form which gives two orthogonal basis elements $c,x$ respective "norms" $1$ and $0$, and $c,x$ anti-commute, etc.

The standard trick to show that the dimension is as large as possible begins with the case of the Clifford algebra of a one-dimensional vector space with bilinear form, and in that case we know that $k[x]/(x^2-a)$ is two-dimensional, in any case, because we have simpler control in the commutative case.

It is perhaps over-kill for the two-dimensional vectorspace case, but the induction step in proving that a Clifford algebra on an $n$-dimensional v.s. $V$ has dimension $2^n$ (rather than accidentally smaller) is to express $V$ as an orthogonal sum $V_1\oplus V_2$ where $V_1$ is one-dimensional. There is the parity grading into odd and even elements, inherited from the universal associative (tensor) algebra, and for monomials define a twisted multiplication $(x\otimes y)\cdot (x'\otimes y') = xx'\otimes yy' \cdot (-1)^{\deg x'\cdot \deg y}$ on $Cl(V_1)\otimes CL(V_2)$. One checks directly that this gives the Clifford multiplication, and we know the dimension of a tensor product (whether or not the multiplication is defined with $-1$'s in it).

(So this is not as serious a result as Poincare-Birkhoff-Witt for universal enveloping algebras, but it's not completely trivial.)

Edit: Actually, knowing the above, one can pretend to have the epiphany that this algebra is inside the $4\times 4$ matrix algebra over the ground-field (it has some degeneracy) so prove the four-dimensionality by mapping this (universal) object surjectively to a four-dimensional subalgebra of the matrix algebra. But this is a bit of a prank... :)

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Thanks for you answer. I wasn't aware it was a Clifford algebra, but I will look into this! – Paul Slevin Aug 30 '12 at 20:49
You're welcome. This smallish case is also some sort of "quaternion algebra", which indeed is the smallest non-commutative Clifford algebra. – paul garrett Aug 30 '12 at 20:58

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