6 URL updated

Update: now with a question 2 which is much more elementary (and should be well-known!).

Let $k$ be a commutative ring with $1$. Let $L$ be a $k$-module, and $g:L\to k$ be a quadratic form, i. e., a map for which the map $L\times L\to k,\ \left(x,y\right)\mapsto g\left(x+y\right)-g\left(x\right)-g\left(y\right)$ is $k$-bilinear and which satisfies

$g\left(\lambda x\right)=\lambda^2 g\left(x\right)$ for all $x\in L$ and $\lambda\in k$.

We define the Clifford algebra $\mathrm{Cl}\left(L,g\right)$ as the tensor algebra $\otimes L$ (all tensors are over $k$) divided by the two-sided ideal generated by $x\otimes x-g\left(x\right)$ for all $x\in L$.

Question 1: Is $\mathrm{Cl}\left(L,g\right)$ isomorphic to $\wedge L$ as a $k$-module? Is the associated graded object of $\mathrm{Cl}\left(L,g\right)$ isomorphic to $\wedge L$ as a $k$-algebra?

Remark: The answer is yes if the quadratic form $g$ comes from a bilinear (not necessarily symmetric!) form. As a consequence, the answer is yes if $L$ is a free $k$-module or, more generally, a direct sum of quotient $k$-modules of $k$ (in fact, it is easy to see that in this case, every quadratic form on $L$ comes from a bilinear form). (I think that in the case when $L$ is a finite-free $k$-module, the answer "yes" can also be proven by the diamond lemma, though I have not checked.) I am interested in the "perverse" cases when none of these holds, but I suffer from a lack of perversion: I can't name any such case. So here is the question which obviously needs to be addressed first:

Question 2: Find a commutative ring $k$ with $1$ and a $k$-module $L$ with a quadratic form (this is defined as above) which doesn't come from any bilinear (symmetric or not) form on $L$. (A quadratic form $g$ is said to come from a bilinear form $h$ if we have $g\left(v\right)=h\left(v,v\right)$ for every $v\in L$.)

Note: The Clifford algebra of a quadratic form is, in some sense, a "little sister" of the universal enveloping algebra of a Lie or pseudo-Lie algebra. ("Little sister" not in the historical sense, but in the sense of partly having the same properties, but them being easier to prove in the Clifford case than in the universal enveloping algebra case.) The above question asks for a kind of Poincaré-Birkhoff-Witt (PBW) theorem for the Clifford algebra. (Note that the PBW theorem itself requires some niceness conditions such as $L$ being finite-free or $k$ being a $\mathbb Q$-algebra, so it wouldn't surprise me if the Clifford case also doesn't work in full generality. But the opposite case wouldn't surprise me either, because PBW is substantially harder than PBW for Clifford algebras, even in characteristic $0$.)

5 split into 2 questions

Update: now with a question 2 which is much more elementary (and should be well-known!).

Question 1: Is $\mathrm{Cl}\left(L,g\right)$ isomorphic to $\wedge L$ as a $k$-module? Is the associated graded object of $\mathrm{Cl}\left(L,g\right)$ isomorphic to $\wedge L$ as a $k$-algebra?

Remark: The answer is yes if the quadratic form $g$ comes from a bilinear (not necessarily symmetric!) form. I also think As a consequence, the answer is yes if $L$ is a finite-free free $k$-module or, more generally, a direct sum of quotient $k$-modules of $k$ (in fact, it is easy to see that in this case, every quadratic form on $L$ comes from a bilinear form). (I think that in the case when $L$ is a finite-free $k$-module, the answer "yes" can also be proven by the diamond lemma, though I have not checked). checked.) I am interested in the "perverse" cases when none of these holds, but I suffer from a lack of perversion: I can't name any such case. So here is the question which obviously needs to be addressed first:

Question 2: Find a commutative ring $k$ with $1$ and a $k$-module $L$ with a quadratic form (this is defined as above) which doesn't come from any bilinear (symmetric or not) form on $L$. (A quadratic form $g$ is said to come from a bilinear form $h$ if we have $g\left(v\right)=h\left(v,v\right)$ for every $v\in L$.)

4 my old definition of a quadratic form was wrong even for fields (although it was true if the field is separable over its prime field), hence clearly inappropriate

Let $k$ be a commutative ring with $1$. Let $L$ be a $k$-module, and $g:L\to k$ be a quadratic form, i. e., a map satisfying

$g\left(x+y+z\right)-g\left(y+z\right)-g\left(z+x\right)-g\left(x+y\right)+g\left(x\right)+g\left(y\right)+g\left(z\right)=0$ for all which the map $x,y,z\in L$L\times L\to k,\ \left(x,y\right)\mapsto g\left(x+y\right)-g\left(x\right)-g\left(y\right)$is$k$-bilinear and which satisfies$g\left(\lambda x\right)=\lambda^2 g\left(x\right)$for all$x\in L$and$\lambda\in k$. We define the Clifford algebra$\mathrm{Cl}\left(L,g\right)$as the tensor algebra$\otimes L$(all tensors are over$k$) divided by the two-sided ideal generated by$x\otimes x-g\left(x\right)$for all$x\in L$. Question: Is$\mathrm{Cl}\left(L,g\right)$isomorphic to$\wedge L$as a$k$-module? Is the associated graded object of$\mathrm{Cl}\left(L,g\right)$isomorphic to$\wedge L$as a$k$-algebra? Remark: The answer is yes if the quadratic form$g$comes from a bilinear form. I also think the answer is yes if$L$is a finite-free$k$-module (by the diamond lemma, though I have not checked). I am interested in the "perverse" cases when none of these holds. Note: The Clifford algebra of a quadratic form is, in some sense, a "little sister" of the universal enveloping algebra of a Lie or pseudo-Lie algebra. ("Little sister" not in the historical sense, but in the sense of partly having the same properties, but them being easier to prove in the Clifford case than in the universal enveloping algebra case.) The above question asks for a kind of Poincaré-Birkhoff-Witt (PBW) theorem for the Clifford algebra. (Note that the PBW theorem itself requires some niceness conditions such as$L$being finite-free or$k$being a$\mathbb Q$-algebra, so it wouldn't surprise me if the Clifford case also doesn't work in full generality. But the opposite case wouldn't surprise me either, because PBW is substantially harder than PBW for Clifford algebras, even in characteristic$0\$.)

3 stupid typo is stupid
2 added 368 characters in body
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