# Thread:

1. ## P-adic number field

I've been investigating the p-adic field, and I was wondering how the p-adics form a field.

The part I'm stuck at is that from what I've learned, to qualify as a field, every element except the additive identity, should have a multiplicative inverse. But, from what I've read, in the p-adic field, any p-adic integer that ends in zero doesn't have a multiplicative inverse. So, I snooped around abit and found another definition for a field, which said that every non-zero element has a multiplicative inverse. I'm assuming that this means that the p-adic numbers that end in zero would be considered as zero elements.

What exactly are zero elements? (Or non-zero elements?) and how does it apply to the p-adic numbers?

2. Originally Posted by Bingk
I've been investigating the p-adic field, and I was wondering how the p-adics form a field.

The part I'm stuck at is that from what I've learned, to qualify as a field, every element except the additive identity, should have a multiplicative inverse. But, from what I've read, in the p-adic field, any p-adic integer that ends in zero doesn't have a multiplicative inverse. So, I snooped around abit and found another definition for a field, which said that every non-zero element has a multiplicative inverse. I'm assuming that this means that the p-adic numbers that end in zero would be considered as zero elements.

What exactly are zero elements? (Or non-zero elements?) and how does it apply to the p-adic numbers?
non-zero p-adic integers that end in zero don't have an inverse in $\mathbb{Z}_p,$ the ring of p-adic integers, but they do have an inverse in $\mathbb{Q}_p,$ the field of p-adic numbers. that is easy to see: a non-

zero p-adic integer which ends in zero is in the form $x=\sum_{i=k}^{\infty}x_i p^i,$ for some $k \geq 1$ with $x_k \neq 0.$ then, since $p$ is prime and $1 \leq x_k \leq p-1,$ there exists some $1 \leq y \leq p-1$ such

that $x_ky \equiv 1 \mod p.$ it's now easy to see that $x^{-1}=yp^{-k} + \cdots \in \mathbb{Q}_p.$

3. Ah! I think I get it, hehehe thanks!

Less formally, are you saying that basically you shift the digits so that it doesn't end in zero, get the inverse of that, and shift the digits to the appropriate place?
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