03/07/2023, 11:19 PM
I've definitely lost the plot so far. I'm pretty shit at p-adic stuff. But \(V(0) = -\infty\) seems like a non issue. This is absolutely standard. The same way every valuation satisfies this--this is how we make p-adic norms. Even adding in the fact that \(^N 0 = (N \pmod{2})\). The reason this happens is because you have assumed that \(0^1 = 0\) and \(0^0 = 1\). And we have a strange oscillation. This is still a singular behaviour though, which is better written with \(V(0) = - \infty\)...
I tend to lean towards Ember's analysis here. As there's room for error about \(V(0)\)... and \(V(0) = -\infty\) seems perfectly natural. Especially because it appears to be acting as a valuation. The same way the degree of a constant polynomial \(\deg( C) = 0\) and the degree of the zero polynomial is \(\deg(0) = -\infty\)... It may seem wrong at first glance, but it describes the algebra perfectly...
I'm out of my depth here though, so correct me if I'm being an idiot! But I see it as being much more natural to just set \(V(0) = -\infty\)....
Regards, James
I tend to lean towards Ember's analysis here. As there's room for error about \(V(0)\)... and \(V(0) = -\infty\) seems perfectly natural. Especially because it appears to be acting as a valuation. The same way the degree of a constant polynomial \(\deg( C) = 0\) and the degree of the zero polynomial is \(\deg(0) = -\infty\)... It may seem wrong at first glance, but it describes the algebra perfectly...
I'm out of my depth here though, so correct me if I'm being an idiot! But I see it as being much more natural to just set \(V(0) = -\infty\)....
Regards, James