Let $(V,Q)$ be a $\mathbb{C}$ or $\mathbb{R}$ linear vector space with a symmetric bilinear form $Q$. Then we can construct a Clifford algebra by taking the tensor algebra $T(V)$, and taking the quotient by the ideal $I$ generated by the subsets:
$$I=\{v\otimes v+Q(v,v):v\in V\}$$
Let $\gamma=\pi\circ i$, where $i:V\rightarrow T(V)$ is the inclusion map, and $\pi:T(V)\rightarrow T(V)/I$ is the quotient map. If one could show that $I\cap V=\{0\}$ it should then follow that the map $\gamma:V\rightarrow T(V)/I$ is injective. Let $v\in I\cap V$, then:
$$v=\sum_{i}a_i\otimes (v_i\otimes v_i+Q(v_i,v_i))\otimes b_i$$
Since $v\in V$ we can assume that for each $i$ the degree of $a_i$ and $b_i$ sum to $1$, i.e. either $a_i$ is a vector, and $b_i$ is a scalar or vice versa. Since $v\in V$, we need:
$$\sum_ia_i\otimes (v_i\otimes v_i)\otimes b_i=0$$ My question is then, can we use this to show that: $$\sum_{i}a_i\otimes b_i Q(v_i,v_i)=0$$ I don't see a way forward to prove this, and I suspect these conditions are a little to0 general to see why this should be immediately clear. I believe it should be true, since there are other ways to show that $\gamma$ is an injection.
One thought I had was to consider the map:
$$Q':V^{\otimes 4}\rightarrow V^{\otimes 2}$$
given on simple tensors by:
$$Q'v_1\otimes v_2\otimes v_3\otimes v_4\rightarrow v_1\otimes v_4 \dot Q(v_2,v_3)$$
but then I quickly remembered that one of $a_i$ or $b_i$ is a scalar, so I would really need a map from $V^{\otimes 3}\rightarrow V$, and since $a_i$ or $b_i$ could be a vector on either side of $v_i\otimes v_i$, I don't think there is a consistent way of obtaining a map like that. Any help would be appreciated.
$ \newcommand\Ext{\mathop{\textstyle\bigwedge}} \newcommand\Cl{\mathrm{Cl}} \newcommand\gtensor{\mathbin{\hat\otimes}} \newcommand\gen[1]{\langle#1\rangle} $I'm having a surprisingly hard time finding this result (though one should really expect I'd find it if I keep looking). Pertti Lounesto references Bourbaki 1959 and T.Y. Lam 1973 in Clifford Algebras and Spinors, but I've not had success finding a copy of either. Ian Porteous goes a much different route in Clifford Algebras and the Classical Groups by constructing real/complex Clifford algebras as matrix algebras or direct sums of matrix algebras.
Larry Grove gives the following "proof" in Classic Groups and Geometric Algebra (which I paraphrase here):
This doesn't make any sense to me, and indeed if this argument worked then it would work just as well if $V$ were a module over an arbitary ring; but Jacques Helmstetter and Artibano Micali give Example 3.1.3 on page 106 of Quadratic Mappings and Clifford Algebra as a counterexample.
So below I give my proof in terms of the ideal $I$, and then also a corresponding proof in terms of universal properties and the fundamental isomorphism $\Cl(V,Q) \cong \Cl(V_1,Q)\gtensor\Cl(V_2,Q)$ where $V = V_1\oplus V_2$ is an orthogonal decomposition.
In Terms of $I$
We write $\Cl(V, Q) = T(V)/I$ and $Q(w) = Q(w,w)$.
The result is almost trivial when $Q$ is non-degenerate. Suppose there is nonzero $v$ such that $\pi(v) = 0$. For any $w \in V$ we have $$ 2Q(v, w) = \pi(v + w)^2 - \pi(v)^2 - \pi(w)^2 = \pi(v)\pi(w) + \pi(w)\pi(v) = 0. $$ Thus $v$ must be in the radical $V^\perp$, and so we have
Lemma. If $Q$ is non-degenerate then $\pi$ restricted to $V$ is injective. Equivalently, $V\cap I = \{0\}$.
Note: the following argument is likely broken, see the comments. However, I believe the argument in terms of universal properties further down is sound.
Now consider an orthogonal decomposition $V = V_1 \oplus V^\perp$; it follows that $V_1\cap I_1 = \{0\}$ where $I_1$ is the ideal of $T(V_1) \subseteq T(V)$ generated by elements of the form $v\otimes v - Q(v)$ for $v \in V_1$.
Every $w \in V$ can be uniquely written $w = u + r$ for $u \in V_1$ and $r \in V^\perp$. Thus if $v \in V\cap I$ we may write $$ v = \sum_ia_i\otimes X_i\otimes b_i $$ where $$ X_i = w_i\otimes w_i - Q(w_i) = u_i\otimes u_i - Q(u_i) + r_i\otimes r_i + u_i\otimes r_i + r_i\otimes u_i $$ with $u_i \in V_1$ and $r_i \in V^\perp$. We may assume that each term of the sum in $v$ has grade 1; WLOG we may assume $(a_i, b_i) = (v'_i, 1)$ for $v'_i \in V_1$ or $v'_i \in V^\perp$.
Let $X'_i = u_i\otimes u_i - Q(u_i)$. Since $v'_i\otimes X_i$ has grade 1 it follows that $w\otimes X_i$ has grade 1 for all $w \in V$, in particular $w \in V_1$. But then $w\otimes X'_i$ is in $V_1\cap I_1$ and so must be zero: $$ 0 = w\otimes X'_i = -Q(u_i)w. $$ Whether or not $V_1$ is trivial we get $Q(u_i) = 0$. So now $$ X_i = u_i\otimes u_i + r_i\otimes r_i + u_i\otimes r_i + r_i\otimes u_i $$ and every term of $v$ is of grade at least 2, hence $v = 0$.
In Terms of Universal Properties
Essentially what I will do now is show that the subalgebra generated by $V^\perp$ is isomorphic to its exterior algebra while also preserving the map $\pi : V^\perp \to \Cl(V,Q)$. Then since the natural map $V \to \Ext V^\perp$ is injective we get that $\pi$ is injective on $V^\perp$. First we prove the all-important universal property.
Theorem 1. $\Cl(V,Q)$ satisfies the following universal propery: for any other associative algebra $A$ and linear $f : V \to A$ such that $f(w)^2 = Q(w)$ there is a unique algebra homomorphism $F : \Cl(V,Q) \to A$ such that $F(\pi(w)) = f(w)$ for all $w \in V$.
Proof. By the universal propery of $T(V)$ there is a unique algebra homomorphism $G : T(V) \to A$ such that $G(w) = f(w)$ for all $w \in V$. We write $\iota \in I$ as $$ \iota = \sum_ia_i\otimes(v_i\otimes v_i - Q(v_i))\otimes b_i $$ whence $$ G(\iota) = \sum_iG(a_i)(f(v_i)^2 - Q(v_i))G(b_i) = 0. $$ Thus $G$ descends to the desired homomorphism $F : \Cl(V,Q) \to A$. $\quad\Box$
Definition. The $\mathbb Z/2\mathbb Z$-graded tensor product $A\gtensor B$ of two $\mathbb Z/2\mathbb Z$-graded algebras $A, B$ is a $\mathbb Z/2\mathbb Z$-graded algebra defined with underlying set $A\otimes B$ and such that $$ (a\otimes b)(a'\otimes b') = (-1)^{\partial a'\partial b}(aa')\otimes(bb') $$ where $b, a'$ are homogeneous with degrees $\partial b, \partial a'$.
Theorem 2. Let $V = V_1\oplus V_2$ be an orthogonal decomposition. Identify $T(V_1)$ and $T(V_2)$ as subalgebras of $T(V)$ and let $\pi_i$ denote the natural projection of $\Cl(V_i, Q)$. Then there is an algebra isomorphism $$ \psi : C = \Cl(V_1,Q)\gtensor\Cl(V_2, Q) \cong \Cl(V,Q) $$ such that $\psi(\pi_1(w)) = \pi(w)$ for all $w \in V_1$ and $\psi(\pi_2(w)) = \pi(w)$ for $w \in V_2$.
Corollary. $\pi$ is injective on $V$ for any $Q$.
Proof. We choose $V_2 = V^\perp$ whence $\Cl(V_2, Q) = \Ext V^\perp$. If $\Ext V^\perp = T(V^\perp)/I_\perp$ with $I_\perp$ the appropriate ideal, then if $v \in V^\perp\cap I_\perp$ we see that $v = 0$ since every element of $I_\perp$ has degree at least 2. Hence $\pi_2$ is injective on $V^\perp$ and so $\pi = \psi\circ\pi_2$ is injective on $V^\perp$. Similarly $\pi_1$ is injective on $V_1$ by the Lemma and so $\pi$ is injective on $V_1$. Hence $\pi$ is injective on $V = V_1\oplus V_2$. $\quad\Box$
Proof of Theorem 2. For each $i = 1,2$, by the universal property of $\Cl(V_i, Q)$ we get a homomorphism $\psi_i : \Cl(V_i,Q) \to \Cl(V,Q)$ such that $\psi_i(\pi_i(w)) = \pi(w)$ for $w \in V_i$. Define $\psi = \psi_1\otimes\psi_2$. By the universal property of $\Cl(V,Q)$ we get $\phi : \Cl(V,Q) \to C$ such that $$ \phi(\pi(w_i)) = \pi_i(w_i)\quad\text{when }w_i \in V_i. $$ If $X_1 \in T(V_1)$ and $X_2 \in T(V_2)$ it now easily follows that $$\begin{aligned} \phi\Bigl(\psi(\pi_1(X_1)\otimes\pi_2(X_2))\Bigr) &= \phi\Bigl(\psi_1\bigl(\pi_1(X_1))\otimes\psi_2(\pi_2(X_2)\bigr)\Bigr) \\ &= \phi(\pi(X_1))\otimes\phi(\pi(X_2)) \\ &= \pi_1(X_1)\otimes\pi_2(X_2). \end{aligned}$$ Every element $C$ is a sum of elements of the form $\pi_1(X_1)\otimes\pi_2(X_2)$, so by linearity $\psi = \phi^{-1}$ is an isomorphism with the requisite properties. $\quad\Box$