Basis of Quaternion Algebras I was reading some notes in which the definition of quaternion algebra is given as :
For $a,b \in k^{*}$ ,we define $k$-algebra
by generators and relations as follows: it has two genrators $i$ and $j$ and is subject to the relations $i^2 = a,  j^2 = b, ij=-ij$.Any k-algebra isomophic to this is called a quaternion algebra.
Now,to prove that $ {{1,i,j,ij}}$ is the basis for the quaternion algebra,it is easy to check that this is the spanning set.But to show the $k$-linear independence,they say that it is enough to check that the $k$-algebra with ${1,i,j,ij}$ as basis is associative. 
I am not getting this that how will the associativity of the algebra will imply the linear independence.(it might be simple but I am stuck at this).

It is unclear to me as well, especially because $1, i, j, ij$ need not be linearly independent when the characteristic of $k$ is $2$, so linear independence cannot be a purely algebraic consequence. In the characteristic $2$ case, it depends on the precise definition of the algebra: if all you require is that $i$ and $j$ generate it, then one could simply have $i, j \in k$ and the algebra is equal to $k$. With the definition below however, they are linearly independent:
A formal definition of that algebra would be
$$\frac{k\langle i, j \rangle}{(i^2-a, j^2-b, ij + ji)},$$
a quotient of the free associative algebra on two generators $i, j$. It is then not hard to see that $1, i, j, ij$ form a basis, by looking at the degree of elements of $k \langle x, y \rangle$. The argument also works in characterstic $2$, but the algebra you get is always isomorphic to the commutative algebra
$$\frac{k[ i, j ]}{(i^2, j^2)}$$
which is not a quaternion algebra (it is not central nor simple).
