Ideal generated by $v\otimes v - \Phi(v)1$ I'm trying to understand the proof of the existence of a Clifford algebra. Let $V$ be a vector space over $\mathbb{K}$, $\varphi: V\times V \to \mathbb{K}$ a bilinear form and $\Phi(v) = \varphi(v,v)$ its associated quadratic form. The prove existence of a Clifford algebra, one takes the quotient of the tensor algebra $T(V)$ with the ideal generated by elements of the form $v\otimes v - \Phi(v)1$, $v \in V$ and $1 \in \mathbb{K}$.
I don't have background in algebra, so I'd like to know what exactly is "the ideal generated by elements of the form $v \otimes v - \Phi(v)$". Also, is this the same as the ideal generated by $u\otimes v + v\otimes u - \varphi(u,v)1$ (which is also found in some references)?
 A: In general, if $A$ is an algebra, and $S$ is a subset, then the ideal generated by $S$, is the smallest ideal $I_{S} \subseteq A$ that contains $S$.
In this specific case, the set $S$ is
\begin{equation*}
 S = \{ v \otimes v - \Phi(v)1 \in T(V) \mid v \in V \}.
\end{equation*}
One way to get your hands on $I_{S}$ in this case is to simply set
\begin{equation*}
 I_{S} = \{ a \in T(V) \mid \exists (b_{i},b_{i}' \in T(V), s_{i} \in S): a = \sum_{i} b_{i}s_{i}b_{i}' \}.
\end{equation*}
To answer your question about the relation to the other ideal, we'd have to consider
\begin{equation*}
 S' = \{ u \otimes v + v \otimes u - 2\varphi(u,v)1 \mid u,v, \in V \}
\end{equation*}
and show that $I_{S} = I_{S'}$.
(You definitely need the factor 2, just consider the case $v=u$.)
So, let $V$ and $\varphi$ be as you stated, and let's define $\mathrm{Cl}(V)$ to be algebra $T(V)/I_{S}$.
Now we'd like to prove that this satisfies the universal property.
To that end, let $A$ be an associative, unital algebra, and let $f: V \rightarrow A$ be an arbitrary map.
Then, the map $f: V \rightarrow A$ extends to an algebra homomorphism $F: T(V) \rightarrow A$.
Now, suppose that $f$ was in fact a Clifford map, i.e.,
\begin{equation*}
 f(v)^{2} = \Phi(v) 1_{A},
\end{equation*}
for all $v \in V$.
One can use this equation that the map $F: T(V) \rightarrow A$ vanishes on the ideal $I_{S}$, and thus descends to an algebra homomorphism $T(V)/I_{S} = \mathrm{Cl}(V) \rightarrow A$.
If you want to know more about Clifford algebras, I strongly recommend the first chapter of Lawson & Michelsohn's Spin Geometry.
A: @Peter's answer is great for your specific case, but some insight on the more general algebraic construct you're dealing with should help, I think, because it helped me a lot back in the day.
I think the concept you are lacking is the idea of "an algebraic quotient by an equivalence class", and how this is linked to ideals of rings.
I'll take it that you're familiar with the very basics of set theory/abstract algebra/category theory: ie, that you at least know what an equivalence relation, a partial order, a binary operator, closure, associativity, identity, invertibility, commutativity, a monoid, a group, a ring, a field, a vector space, and a $\mathbb{K}$-algebra are. [If you're not acquainted with each of these concepts, I've written a concise introduction to this subject that might interest you, and that's freely available online; but such introductions abound on the web in any case.]
An ideal is a particular subset of a ring (or a $\mathbb{K}$-algebra, which is very similar to a ring). Ideals are special because they are subsets of a ring that are closed (stable) under both addition and multiplication. This gives them some important properties. Ideals of a ring $R$ are generally expressed as a set $eR = \{ e * r \; | \; r \in R \}$ (note that in non-commutative rings, you have to distinguish left-ideals from right-ideals). Thus, an ideal can generally be understood as "the set of multiples of $e$ in $R$". For example $4\mathbb{Z} = \{\dots, -8, -4, 0, 4, 8, \dots\}$, and $(x^2 + 1)\mathbb{R}[X]$ is the set of all polynomials which you can factor by $(x^2 + 1)$ (all polynomials that have $i$ and $-i$ amongst their list of roots).
One very important property of ideals is that they separate a ring into a "subspace" (the ideal itself; which is not exactly a ring, because it lacks the multiplicative identity in general) and a list of "affine subspaces" (the ideal plus some element of the ring) which are parallel to the "subspace".  For example,
$4\mathbb{Z} = \{\dots, -8, -4, 0, 4, 8, \dots\}$
$4\mathbb{Z} + 1 = \{\dots, -7, -3, 1, 5, 9, \dots\}$
$4\mathbb{Z} + 2 = \{\dots, -6, -2, 2, 6, 10, \dots\}$
$4\mathbb{Z} + 3 = \{\dots, -5, -1, 3, 7, 11, \dots\}$
$4\mathbb{Z} + 4 = 4\mathbb{Z}$
With this, we can define an equivalence relation over elements of the base ring (here $\mathbb{Z}$), based on any ideal: we consider any element to be equivalent if they are in the same affine space for that ideal. This is expressed algebraically as $e \sim f \Leftrightarrow e - f \in I$, where I is the ideal. The "algebraic quotient of a ring by an ideal" is then the homomorphism mapping an element of the input ring to its equivalence class. The output set, generally written $R/e$ or $R/eR$, has the same number of elements as there were "subspaces". $\mathbb{Z}/4\mathbb{Z}$ is obtained by mapping $\mathbb{Z}$ via the "modulo $4$" operator (the remainder of the Euclidean division by four); it is a space which functions like a clock with $4$ hours. $\mathbb{R}[X]/(x^2 + 1)$ is isomorphic to $\mathbb{C}$. The kernel of this homomorphism is ALWAYS the ideal itself.
Algebraic quotients come up all over mathematics; everywhere you see an equivalence relation. They're not limited to rings. Parallelism is an equivalence relation, and for example, taking the set of lines $\mathbb{R}^2$, and quotienting it by parallelism, you'll get a "looping half-circle" isomorphic to $S^1$. You can define the $u - v \in W$ equivalence relation for any subspace $W$ of any vector space $V$. Most spaces you'll see have their "most elegant construction" done through algebraic quotients. Generally, you use a free functor to "get an abstract block of marble" from some other structure (for example, the free functor from $Vect_\mathbb{K}$ to $Alg_\mathbb{K}$, which turns a vector space into its tensor algebra, which is the "least specific/most general way" of building an algebra from a vector space) and then you use algebraic quotients to "inject desired mathematical properties" into your quotient space (for example, the ideal generated by the elements of the form $a \otimes b + b \otimes a = 0$ injects anticommutavity into your tensor algebra over $V$; turning into the exterior algebra over $V$).
In the case of $\mathbb{R}[X]/(x^2 + 1)$, there's "only one element of the form $(x^2 + 1)$", the polynomial $I = x^2 + 1$; but this also means every polynomial which is a multiple of $I$ is part of the kernel; so in $\mathbb{R}[X]/(x^2 + 1)$, any $Q = (x^2 + 1)P = 0$. This makes the property $x^2 + 1 = 0$ be injected into $\mathbb{R}[X]$. Consequently, there exists at least one element $i$ such that $i^2 + 1 = 0$.
Now, in your case, you have an ideal generated by multiple elements, but the principle is the same. For every vector $v$ in $V$, you're going to inject the property that the "abstract, propertyless (safe for bilinearity)" multiplication of the tensor algebra must respect $v \otimes v = \Phi(v)$ (ie, the product of two vectors is a globally defined quadratic, euclidean (pseudo)norm). It is equivalent, @Peter shows, to defining the (pseudo)norm via the use of a (generalized) inner product (ie, $u \otimes v + v \otimes u = \phi(u,v)$).
