# why Z5 in Qp different with Zp(5)

K = Qp(5, print_mode='digits') Z5 = K.integer_ring();Z5;Z5;Z5 ==Zp(5)

5-adic Ring with capped relative precision 20 5-adic Ring with capped relative precision 20 False

why Z5 in Qp different with Zp(5)

K = Qp(5, print_mode='digits') Z5 = K.integer_ring();Z5;Z5;Z5 ==Zp(5)

5-adic Ring with capped relative precision 20 5-adic Ring with capped relative precision 20 False

add a comment

1

As with your other question, please ask the question in the body of the posting and explain yourself.

The answer is that you created K=Q5 with a non-default option print_mode. So you have ended up with two different version of Z5, with different print modes. To illustrate this:http://ask.sagemath.org/question/3140/artin-decomposition-for-p-adic-numbers

```
sage: Z5(11).sqrt()
...433234102330200211
sage: Zp(5)(11).sqrt()
1 + 5 + 2*5^2 + 2*5^5 + 3*5^7 + 3*5^8 + 2*5^9 + 5^11 + 4*5^12 + 3*5^13 + 2*5^14 + 3*5^15 + 3*5^16 + 4*5^17 + O(5^20)
```

By contrast note that

```
sage: Qp(5).integer_ring() == Zp(5)
True
```

Asked: **
2013-12-14 20:12:15 -0600
**

Seen: **86 times**

Last updated: **Dec 16 '13**

how to a p-adic expansion of a rational function in sage?

Algorithm for finding a defining polynomial for an unramified extension?

programming of looping to print selected value of m

Finding p-adic valuations in high degree cyclotomic fields

Can't construct automorphisms of p-adic fields

Coding p-adic Newton's method to solve polynomials?

what PI and e in p-adic number?

$p$-adic extension of $n$th root of unity.

i want to find factorization ideal (3) in integral closure of Z_3 in Q_3(sqrt(2),sqrt(3))

Copyright Sage, 2010. Some rights reserved under creative commons license. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license.