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Assume that (K,v) is a valued field with

Γ_v=(Z^2 ;+, <_lex).
Let p_1(x,y)=x, p_2(x,y)=y maps from
Z^2 —> Z.

a. Prove that w_i(a)=p_i(v(a)), i=1,2 are both valuations on K with Γ_w=Z.

b. Prove that one of the valuations w_i induces the same topology as v on K while the other induces a different valuation.

c. Deduce that when two valuations induce the same toplogy on a field K they need not to be equivalent.

*def: we say that v1 :K–>Γ_v1 ∪  {∞} and

 v2 :K–>Γ_v2  ∪ {∞} are equivalent if there is an isomorphism of groups
a: (Γ_v1,+,<)–>(Γ_v2,+,<)  such that for all x in K*:  v2(x) = a(v1(x)).

I know how to show the axioms of a valuation bue here somehow I did totally  manage to do this!

1.v is a valuation so we know that v(a)=∞ iff a=0_K. And we need to show that w_i(a)=∞ iff a=0_K. But then
w_i(0)=p_i(v(0))=p_i(∞) and now it is seems undefined!.

Fow w_i(xy)=w_i(x)+w_i(y) and
 w_i(x+y) \geq min {w_i(x), w_i(y)},

For example,

2.Let x, y in K,  then:


Here I wrote v(x)=(a1,b1) and v(y)=(a2,b2)  (where a1, a2, b1, b2 in Z) since v: K—> Z^2.

3.I think in this part we need to use the order <_lex which is given by:

(a1,b1) <_lex (a2,b2) if a1< a2 or
a1=a2 and b1<b2.

Again, let x, y be in K.
Denote v(x)=(a1, b1) and v(y)=(a2, b2).

We need to prove that
w_1(x+y) \geq min{w_1(v(x)),w_1(y)}.

I know that v(x+y) \geq min{v(x),v(y)}.

In addition, do we need to explain why Γ_w ?
v: K –> Z^2 and p_i: Z^2 –> Z  inf.
So I think that Γ_w must be Z as w_i is the composition of p_i and v.

But it's not clear to me how to use this in the calculations.

Any guide would be appreciated!

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