Homological degrees of polynomials Given a continuous map $f:S^n \to S^n$, the homological degree of $f$ is defined to be the integer $\deg f_*$ such that $f_*(\alpha)=(\deg f_*)\alpha$ for any $\alpha \in H_n(S^n) \approx \mathbb{Z}$.
Exercise 2.2.8 in Hatcher's text asks the reader to consider a complex polynomial $p:\mathbb{C} \to \mathbb{C}$. If this polynomial is nonconstant, then setting $p(\infty)=\infty$ gives an extension of $p$ to the one point compactification $p:\hat{\mathbb{C}} \to \hat{\mathbb{C}}$. But $\hat{\mathbb C} \approx S^2$, so one can consider the homological degree of $p$. Turns out, this degree is equal to the degree of $p$ as a polynomial. $$\deg p_* = \deg p$$
If you do this whole thing over $\mathbb{R}$ instead, with one point compacitication $\hat{\mathbb R} \approx S^1$, you don't get an analagous result. If $p(x)=a_nx^n + \dotsb +a_1x+a_0$ with $a_i \in \mathbb{R}$ (and $a_n \not= 0$), then
$$\deg p_* = \begin{cases} 0 & \text{if $n$ is even} \\ \operatorname{sign} a_n & \text{if $n$ is odd} \end{cases}$$
where $\operatorname{sign}$ is the $\pm 1$ sign of a nonzero real number.
This led me to wonder what happens in other scenarios. For instance, what happens over the quaternions $\mathbb{H}$ with $\hat{\mathbb H} \approx S^4$? I'm not even sure what "quaternionic polynomial" would mean since the quaternions aren't commutative, so expressions like $i \cdot q \cdot i$ and $(i^2)q=-q$ aren't equivalent.
What about other complete fields, like $p$-adics $\mathbb{Q}_p$? These spaces are much worse topologically, in particular totally disconnected. But one can still consider the projective lines $\mathbb{P}^1(\mathbb{Q}_p)$ (which is the same as the one point compactification in the $\mathbb{R}$ and $\mathbb{C}$ case at least). I imagine that number theorists and algebraic geometers have invented some appropriate analog of singular homology to study maps of these things.
 A: For arbitrary fields, we have the étale cohomology, which is more well-behaved for algebraically closed fields.
Let $k$ be an algebraically closed field and $n$ a positive integer that is not zero in $k$. Then we consider the group $H^2(\Bbb P^1_k, \mu_n) \cong \Bbb Z/n\Bbb Z$, where $\mu_n$ is the group of $n$th roots of unity, which is non-canonically isomorphic to $\Bbb Z/n\Bbb Z$.
Now let $f: \Bbb P^1_k \to \Bbb P^1_k$ be the map induced by a polynomial $p = \sum a_n x^n$. From the computation of $H^2(\Bbb P^1_k, \mu_n)$, we have the following diagram:
$\require{AMScd}$
$$\begin{CD}
\operatorname{Pic}(\Bbb P^1_k) @>>> \operatorname{Pic}(\Bbb P^1_k) @>>> H^2(\operatorname{Pic}(\Bbb P^1_k), \mu_n) @>>> 0 \\
@VgVV @VgVV @Vf^\ast VV @VVV \\
\operatorname{Pic}(\Bbb P^1_k) @>>> \operatorname{Pic}(\Bbb P^1_k) @>>> H^2(\operatorname{Pic}(\Bbb P^1_k), \mu_n) @>>> 0
\end{CD}$$
where $\operatorname{Pic}(\Bbb P^1_k) \cong \Bbb Z$ denotes the Picard group.
where $g$ sends the point $[z]$ to $p^{-1}([z])$, so $g$ is just multiplication by $\deg p$.
Therefore, $f^\ast$ is also multiplication by $\deg p$.

Now if $k$ is not algebraically closed, we can use the Hochschild--Serre spectral sequence:
$$H^r(\operatorname{Gal}(\overline{k}/k), H^s(\Bbb P^1_{\overline k}, \mu_n)) \implies H^{r+s}(\Bbb P^1_k, \mu_n)$$
Note that $\mu_n$ is no longer isomorphic to $\Bbb Z/n\Bbb Z$, because we now consider them as $\operatorname{Gal}(\overline{k}/k)$-modules.
Now $H^0(\Bbb P^1_{\overline k}, \mu_n) = \mu_n$ and $H^2(\Bbb P^1_{\overline k}, \mu_n) = \Bbb Z/n\Bbb Z$ (these are the only two non-trivial groups among $H^s(\Bbb P^1_{\overline k}, \mu_n)$), and:
$$H^r(\operatorname{Gal}(\overline{k}/k),\mu_n) =
\begin{cases}
\mu_n \cap k & r = 0 \\
k^\times / k^{\times n} & r = 1 \\
0 & r \ge 2
\end{cases}$$
Therefore,
$$H^r(\Bbb P^1_k,\mu_n) =
\begin{cases}
\mu_n \cap k & r = 0 \\
k^\times / k^{\times n} & r = 1 \\
H^{r-2}(\operatorname{Gal}(\overline{k}/k), \Bbb Z/n\Bbb Z) & r \ge 2
\end{cases}$$
So $H^2(\Bbb P^1_k,\mu_n)$ is still $\Bbb Z/n\Bbb Z$, and we still recover $f^\ast$ being multiplication by $\deg p$.
