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Let $F$ be a field and $g(x)$ in $F[x]$. Prove if $1+g(x)^2$ has an irreducible factor of odd degree then there exists $a$ in $F$ such that $a^2 =-1$.

I didn't get too far on this problem. It doesn't seem like it's a Galois Theory or field extension problem since we don't know much about our field.

I first started looking at roots of $1+g(x)^2$ since then $g(r)^2 =-1$ for root $r$ as we want. I didn't really get anywhere with that.

We know $F[x]$ is a Euclidean Domain, and it seems like the Euclidean Algorithm might be useful. We know $1+g(x)^2 =f(x)p(x)$ for irreducible $f(x)$ and so $1=f(x)p(x)+(-g(x))g(x)$.

Any ideas?

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Let $\alpha$ be a zero of $f$ in some extension of $F$. Let $a = g(\alpha)$.

Now, $a \in F[\alpha]$, and $a^2 = -1$ since $1 + a^2 = f(\alpha)\cdot p(\alpha) = 0$.

We already have $a \in F$, because $[F(\alpha) : F] = \deg f \equiv 1 \pmod{2}$, so the minimal polynomial of $a$ has odd degree. Since it is a divisor of $X^2+1$, the degree is $1$.

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Hint: Let $f(x)$ be the irreducible factor. Consider the extension field $K=F[x]/(f(x))$. Using the last line of your question prove that $-1$ is a square in the field $K$. Because $[K:F]$ is odd, show that the norm $N^K_F(\sqrt{-1})$ is a square root of $-1$ in $F$.

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Looks like Daniel has a better idea for that last step :-) – Jyrki Lahtonen Jul 14 '13 at 21:52

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