TITLE: How can i prove that the finite extension field of real number is itself or the field which is isomorphic to complex number?
QUESTION [6 upvotes]: How can i prove that the finite extension field of real number is itself or the field which is isomorphic to complex number ? In deed, this example is included in Fraleght . Abstract Algebra text.
I did try the followings:
 $\mathbb{R}$ is real number. Then $\mathbb{C}$ is explassd as the smallest extension field including $ \mathbb{R} \cup ${$i$}
How about considering this set . Let set $\mathbb{H}$ is the smallest field including 
$\mathbb{R} \cup${$i,j,k$} where $i, j, k$ are called Hamilton number or quaternion
their square are equal to $-1$. Firstly, I do know that this set is a ring. But i check that this set is a field. 
Of course, $\mathbb{H}$ may be not a field. Becasue, if that is true, then The Fraleght text book is wrong. However, I would like to know the specific reasons and Example's solution .
Please help me to get this.

REPLY [2 votes]: If $K$ is a finite extension of $\mathbb{R}[x]$, then
$$K \cong \frac{\mathbb{R}[x]}{(f(x))}$$
for some irreducible polynomial $f \in \mathbb{R}[x]$. Thus $f$ has degree at most $2$. If $f$ has degree $1$, then $K \cong \mathbb{R}$. If $f$ has degree $2$, you can use the quadratic formula to show that $K \cong \mathbb{C}$.