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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.

J.U.math
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3 Answers3

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$\mathbb{H}$ 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, $\mathbb{H}$.

Andrew Dudzik
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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 $K \cong \mathbb{C}$.

Ethan Alwaise
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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

J.U.math
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