How can i prove that the finite extension field of real number is itself or the field which is isomorphic to complex number? 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.
$\mathbb{H}$ is not a field, because it is not commutative.  Fraleigh is correct that $\mathbb{R}$ and $\mathbb{C}$ are the only finite field extensions of the reals. (Hint: an odd degree polynomial over $\mathbb{R}$ has a real root.)

If you consider division algebras instead of fields, however, there is exactly one more possibility: the one you found, $\mathbb{H}$.
In deed, i did solve that simply.
complex number field is algebraic closure of real number field. So If $M$ is a finite extension field of real number field including complex number field. Then $M$ is also a finite extension field of complex number field. It is a contradiction that complex number field algebraic clusure(it does not have any its algebraic extension field)
I would like to consider another opinion