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A quick search here shows me this question: No extension to complex numbers? where they explain there's no 3D or 4D analogue of complex numbers. Then there's a mention of quaternions but the multiplication is non-commutative. This has left me a bit "confused."

Anyway, I just learnt a theorem that if $F$ is field and $p(x) \in F[x]$ is an irreducible polynomial of degree $n$ then $K=F[x]/(p(x))$ is an $n$-dimensional vector space over $F$.

So what if $F=\mathbb{R}$ and $p(x)=x^4+1$, then $\mathbb{C} \subset K$ and $[K:\mathbb{R}]=4$, right?

Specifically, if $\theta=x + (p(x))$ then $\{1, \theta, \theta^2, \theta^3\}$ are the basis for $K$ and $\mathbb{C} \cong \{a+b\theta^2 : a,b \in \mathbb{R}\}$.

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Are you familiar with the Fundamental Theorem of Algebra? It follows from it that any irreducible polynomial over $\mathbb R$ has degree at most 2... –  N. S. Feb 1 '13 at 3:40
Now that I think of it I can see how that follows. I hope we'll prove the Fundamental Theorem of Algebra in both Complex Analysis & Modern Algebra. –  genepeer Feb 1 '13 at 21:36

3 Answers 3

up vote 6 down vote accepted

Massive hint: How sure are you that $p(x) = x^4 + 1$ is irreducible over $\Bbb{R}$? Do you know that

$$x^4 + 1 = -(-x^2+\sqrt{2} x-1) (x^2+\sqrt{2} x+1)?$$

The resulting quotient ring is isomorphic to $\Bbb{C} \times \Bbb{C}$ (using the CRT) which is not even an integral domain.

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I forgot to check that, oops. –  genepeer Feb 1 '13 at 3:28
I read somewhere that Leibniz didn’t know that factorization either, so you’re in good company. –  Lubin Feb 1 '13 at 5:27
I'm more disappointed by how I was so close to finding the factorization but failed to see it. Using $\theta = (1+i)/\sqrt{2}$, I knew that $\theta = (1+\theta^2)/\sqrt{2}$. Since $\theta$ and $\theta^2$ had to be linearly independent (they're in a basis), I concluded that $\theta \notin \mathbb{C}$. Even worse, I justified this by assuming $p(x)$ can have more than 4 roots in $K$, although it has only 4 roots in $\mathbb{C}$. I've learnt my lesson haha. –  genepeer Feb 1 '13 at 21:52

Hint $\rm\ \ x^4\! + 1 = (x^2\!+\!1)^2 - 2x^2\, =\, (x^2\!+\!1)^2 - (\sqrt{2}\, x)^2 = \: \ldots\: $ (factor the difference of squares), therefore the quotient ring is neither a domain nor a field.

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Weird, I had set $\theta = (1+i)/\sqrt{2}$ and found that $\{1, \theta, \theta^2, \theta^3\}$ were not linearly independent, $\theta = (1+\theta^2)/\sqrt{2}$. But instead of backtracking and using this to show $p(x)$ was reducible, I just assumed that $\theta$ must be outside $\mathbb{C}$. Shame. –  genepeer Feb 1 '13 at 3:43

That $\Bbb R[x]/(x^4+1)$ does not produce a field has been cleared up, but I'm surprised nobody has mentioned the connection to the Frobenius theorem.

It says that the only finite dimensional associative division algebras over $\Bbb R$ are $\Bbb R$, $\Bbb C$ and $\Bbb H$ the (quaternions). That's why the quaternions showed up in the other post, since they share this relationship with $\Bbb R$.

Since it looks like you implicitly wanted the analogue for the complex numbers to be a field extension of $\Bbb R$, and since the quaternions are noncommutative, the only possibility left is $\Bbb C$.

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