Every $4$-dimensional central simple algebra is a quaternion algebra. I am looking at the proof of the above theorem in which it starts like this:
Let $x$ be a non zero element of the central simple algebra such that $x \notin k$ and consider the subalgebra generated by $x$. It is easy to see that the minimal polynomial of $x$ is $P(z)= z^2 - t(x)z + n(x)$ and that $P(z) \in k[z]$ is irreducible.
I am not able to understand how this minimal polynomial was obtained. If anybody has some alternate proof, that also would be helpful.
The notes I am referring to can be found here and the theorem is 5.5 in these notes.
 A: By definition of $t(x)$ and $n(x)$, 
$(z-x)(z-\bar x)=z^2-z(x+\bar x) +x\bar x=z^2-t(x)z+n(x)$
$x$ is obviously a root, and if it weren't irreducible, it would factor into linear factors and thus $x\in k$.

As pointed out much later now by m.s., there does appear to be a problem in the cited text. $n$ and $t$ are both defined in terms of a basis which has not been assumed to exist. 
I originally interpreted this simply as a question about the minimality of a given polynomial, not as a challenge to the existence of the polynomial, and I was fairly certain the user's question was about why $x$ is a root.
Of course, after proving the theorem without this polynomial, one can conclude that this is precisely the polynomial, in terms of that basis's $n$ and $t$.
A: I understand your confusion, since in the notes you cited, the reduced norm and the reduced trace are defined by means of a quaternionic basis, so it is circular to use them the proof of a Theorem that asserts existence of such a basis.
Let $B$ be a four dimensional, central simple algebra over $k$. Then either $B  \cong M_2(k)$, which has a quaternionic basis, or $B$ is a division algebra. So we may assume that $B$ is a division algebra. Now the statement one needs is:

For any $x \in B$ with $x \notin k$, the sub-algebra $k[x]$ is a
  quadratic field extension of $k$.

To prove this, we first note that $k[x]$ is a commutative subalgebra of $B$ of finite dimension with no non-zero zero divisors (because $B$ has none). Therefore, $k[x]$ is a field. Since $B$ is not commutative, we have $k[x] \neq B$ and since $x \notin k$, we have $k[x] \neq k$. Hence $[k[x]: k] \in \{2,3\}$. Since $k[x]$ is a simple $k$-algebra, the double centralizer theorem  -- precisely, part b) of Theorem 4.14 in the notes you cited -- implies that $[k[x]: k][Z_{B}(k[x]):k] = 4$, hence $[k[x]: k] = [Z_{B}(k[x]):k] = 2$.
