Ideal of Clifford algebra additive group

Question

The definition of the Clifford algebra over $V$ with a quadratic form $Q$ via the tensor algebra $T(V)$ requires the ideal $$\mathcal{I}_Q=\{A\otimes(v\otimes v-Q(v))\otimes B|v\in V,A,B\in T(V)\}.$$ I see the property that make this an ideal but not why $X,Y\in\mathcal{I}_Q:X+Y\in\mathcal{I}_Q$. I just can't figure out how to cast $X+Y$ back into the form of an element of $\mathcal{I}_Q$. I'd be grateful for any help.

Answer 1

You did not correctly copy the source you are citing. It does not say

$$\mathcal{I}_Q=\{A\otimes(v\otimes v-Q(v))\otimes B|v\in V,A,B\in T(V)\}.$$

it says

$$\mathcal{I}_Q=\left\{\sum_k A_k\otimes(v\otimes v-Q(v))\otimes B_k\,\middle|\,v\in V,A_k,B_k\in T(V)\right\}$$

which makes a difference.

However, this is still not quite clear, because it does not adequately show why the sum of two such things is in $\mathcal{I}_Q$. In this formulation, it looks like only one $v$ is allowed in the summation, but actually it should be many $v_k\in V$. The author probably should have written

$$\mathcal{I}_Q=\left\{\sum_k A_k\otimes(v_k\otimes v_k-Q(v_k))\otimes B_k\,\middle|\,v_k\in V,A_k,B_k\in T(V)\right\}$$

which accurately describes the ideal generated by elements of the form $v\otimes v-Q(v)$. In this format, you can clearly see that the sum of two elements of that form is again of that form.

If you do not do this, then you get stuck precisely where you got stuck.

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If you doubt it, you should just shed the complexity above and think about what the form of an ideal of $R$ generated by a subset $S$ looks like:

$$\langle S\rangle=\left\{\sum_k r_ks_k r'_k\,\middle|\,s_k\in S,r_k,r'_k\in R\right\}$$

This is what is given in basically any ring theory book (which allows noncommutative rings) and appears in the wiki article here.

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I know you may find it dubious that an author like Lundholm would have not been precise on a basic definition, but then again this thing being defined has basically no importance after the page it appears on. The author simply moved on to more interesting things.

Answer 2

The Wikipedia article Clifford algebra states

and so a Clifford algebra would be the quotient of this tensor algebra by the two-sided ideal generated by elements of the form $\,v\otimes v-Q(v)\,$ for all elements $\,v\in V.$

What this means is that the definition as quoted is not exactly correct, even if the original source is quoted exactly. The ideal required is not just the set of elements noted, but includes a lot more elements according to the definition of a two-sided ideal generated by a set of elements. Of course, by definition of an ideal it would then have the property $\,X+Y\in \mathcal{I}_Q\,$ you mention.