Infinity plus Infinity
Let $a \in \mathbb{C}$. Ahlfors says we let $a + \infty = \infty$ and $a \cdot \infty = \infty$. But we cannot define $\infty + \infty$ without violating the laws of arithimetic (i.e. field axioms).
I don't see why this is. Don't we have $\infty + \infty = \infty$ by applying the distributive law to $2\cdot \infty$? What am I misunderstanding?
The misunderstanding is assuming that the distributive law applies without exception to the extended arithmetic you're defining here. It doesn't, and your reasoning shows why.
Without defining $\infty+\infty$ you can make the field axioms hold to the extent that the expressions in the axioms are defined at all. The point of not defining $\infty+\infty$ is exactly to avoid expressions for which different applications of the field axioms would give multiple inconsistent values.
Ah, but are you sure you have the distributive law? :)
The proof of non-existence is that the arithmetic operations are defined by continuous extension. So, we need to check whether or not the following limit exists: $$ \lim_{(x,y) \mapsto (\infty, \infty)} x + y $$ If it existed, we could compute it by taking the limit a particular path. The first chooses $x=y$, and the second $x=-y$: $$ \lim_{(x,y) \mapsto (\infty, \infty)} x + y = \lim_{x \mapsto \infty} x + x = \infty $$ $$ \lim_{(x,y) \mapsto (\infty, \infty)} x + y = \lim_{x \mapsto \infty} x - x = 0 $$
Assume you had such a $\infty$ consistant with field axioms. Then $ \infty = \infty + \infty \Longrightarrow \infty - \infty = 0$. This follows from the fact that in a field, a unique additive inverse exists for every element, so that also exists for $ \infty$. You wanted to define $\infty = \infty + \infty$, and if we add the additive inverse on both sides, we get $ 0 = \infty + (-\infty) = (\infty + \infty) + (-\infty)$. By associativity, we can swap the brackets on the right to couple $\infty$ with its additive inverse and get $\infty + 0 = \infty$
But then $ 1 = 1 + \infty + (-\infty) = \infty + (-\infty) = 0$, a contradicition.