Example of inequivalent valuations inducing the same topology

Question

Few definitions:

1. Let $F$ be a field. We define a valuation on $F$ as a function $v:K\to G\cup \infty$ (where $(G,+,<)$ is an ordered abelian group) that satisfy:

$v(a)=\infty$ iff $a=0_F$.

$v(ab)=v(a)+v(b)$ for all $a,b\in F^*$.

$v(a+b)\geq min\{v(a),v(b)\}$.

2. The valuation group of $(F,v)$ is $G_v:=v(F^*)\subset G$.

3. Let $v:F\to G_1\cup \infty, w:F\to G_2\cup \infty$ valuations on $F$. Then they are equivalent if there exists an isomorphism of ordered groups $a:(G_v,+,<) \to (G_w,+,<)$ such that $w(x)=a(v(x))$ for all $x\in F^*$

4. Given valuations: $v:F\to G_v, w:F\to G_w$ then $v,w$ are equivalent iff $O_v=O_w$, where $O_v: \{x\in F: v(x)\geq 0\}$.

Question (edited) ((My main concern is in parts 3, 4):

Let $F$ be a field with a valuation $v$ and an ordered abelian group $G_v=(Z^2,+,<_l)$. Let $p_1:Z^2\to Z$ be the map $p_1(a,b)=a$. Show that:

1. $v'(x)=p_1(v(x))$ ($v'(0)=\infty$) is a valuation on $F$ with $G_{v'}=Z$.
2. $v'$ induces the same topology as $v$ on $F$.
3. If two valuations induce the same topology on a field $F$, they do not need to be equivalent.
4. Let $F$ be a field with a valuation $v$ with valuation group $G$ and $G'$ is an orderded abelian group. Let $p:G\to G'$ an order preserving surjective homomorphism. Show that $v_2(a)=p(v(a))$ (with $v_2(0)=\infty$), is a valuation on $F$ that induces the same topology as $v$ on $F$.

Here $<_l$ is the order defined by: $(a_1,b_1)<_l (a_2,b_2)$ if $a_1<a_2$ or $a_1=a_2, b_1<b_2$.

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My trial:

In 1.

--we're given that $v'(0)=\infty$.

-- if $x,y \in F^*$ then $v(x), v(y)\in Z^2$ so I can write $v(x)=(a_1,b_1), v(y)=(a_2,b_2)$ (where $a_1, b_1, a_2, b_2 \in Z$), then $p_1(v(x))=p_1(a_1,b_1)=a_1$ and $p_1(a_2,b_2)=a_2$.

So, $v'(xy)=p_1(v(xy))=p_1(v(x)+v(y))=p_1((a_1,b_1)+(a_2,b_2))=p_1((a_1+a_2,b_1+b_2))=a_1+a_2=..=v'(x)+v'(y)$ (since $v$ is a valuation).

--If $x,y\in F$, we need to show $v'(x+y)\geq min\{ v'(x),v'(y)\}$. If $(a_1,b_1)\leq_l (a_2,b_2)$ then $a_1\leq_l a_2$. So, if $v(x+y)\geq min\{v(x),v(y)\}$ and we assume that $v(x)\leq v(y)$ then $v'(x+y)\geq p_1(v(x))=v'(x)$ and if we assume that $v(y)\leq v(x) $ by symmetry we get that $v'(x+y)\geq v'(y)$, therefore $v'(x+y)\geq min\{v'(x),v'(y)\}$.

And, $G_{v'}=v'(F^*)=p_1(v(F^*))=p_1(Z^2)=Z$.

In 2. The topology induced by $v$ is given by the basis $\{B_g(a): a\in F, g\in G_v\}$ where $B_{g}^v(a)=\{x\in F: v(x-a)>g\}$. So, if I want to see that $v, v'$ induce the same topology on $F$, it is enough to show that a $v'$-neighborhood of any point $x_0\in F$ is contained in the $v$-neighborhood of $x_0$ (i.e the open ball in the $v'$ topology, around $x_0$ is contained in the open ball in the $v$ topology around $x_0$) and vise versa.

-- If $x\in B_{(a, b)}^v(x_0)$ then $v(x-x_0)>(a,b)$, write $v(x-x_0)=(c,d)$ so $c>a$ or $c=a, d=b$. Therefore, $v'(x-x_0)=p_1(v(x-x_0))=p_1(c,d)>c\geq a$ So $x\in B_{a-1}^{v'}(x_0)$.

-- if $x\in B_{a-1}^{v'}(x_0)$ then $v'(x-x_0)>a-1$ but this the same as $p_1(v(x-x_0))>a-1$. Write $v(x-x_0)=(c,d)$ so $p_1(c,d)>a-1$ then $c>a-1$. Then, $(c,d)>_l (a-1,b)$ therefore $v(x-x_0)>(a-1,b)$(for some $b,d$). This shows that $x\in B_{(a-1,b)}^v$. In total we have: $B_{(a, b)}^v(x_0) \subset B_{a-1}^{v'}(x_0) \subset B_{(a-1,b)}^v$. Does this look fine?

In 3. The value groups of $v,v'$ are $Z^2, Z$ (respectively) which are not isomorphic as groups. But what a simple example of such valuations can I consider?

In 4. First, $v_2$ is a valuation:

--we're given that $v_2(0)=\infty$.

-- let $a,b\in F^*$ then $v_2(ab)=p(v(ab))=$(v is a valuation) $p(v(a)+v(b))=$($p$ is a homomorphism)= $p(v(a))+p(v(b))=v_2(a)+v_2(b)$.

--Let $a,b\in F$ then $v_2(a+b)=p(v(a+b))\geq$ ($p$ is order preserving) $p(min\{v(a),v(b)\})=$ (assume $v(a)\leq v(b)$) $p(v(a))=v_2(a)$ and similarly, if $v(b)\leq v(a)$ then $v_2(a+b)\geq v_2(b)$, so, $v_2(a+b)\geq min\{v_2(a),v_2(b)\}$.

But I did not use that $p$ is surjective in the above calculation!

To show that both valuation induce the same topology:

We find elements of the base such that,

$B_{g_1}^{v}(x_0) \subseteq B_g^{v_2}(x_0) \subseteq B_{g_2}^{v}(x_0)$. ($x_0, g$ are arbitrary).

If $x\in B_{a}^v(x_0)$ then $v(x-x_0)=c>a$. Then $v_2(x-x_0)=p(v(x-x_0))=p(c)>p(a)$ ($p$ is order preserving), but can I say now that $x\in B_{p(a)}^{v_2}$ and so $B_{a}^v(x_0)\subset B_{p(a)}^{v_2}(x_0)$?

Now, let $x\in B_{a}^{v_2}(x_0)$ so $v_2(x-x_0)>a$ but $p(v(x-x_0))=v_2(x-x_0)>a$ then how one can say that $v(x-x_0)>..$?

It would be highly appreciated if you could help me improve my way of thinking and finish the points that I did not succeed to deal with.

Answer 1

For $v_2$ it is not true that it is a valuation.

If $v(a)=(0,0)$, $v(c)=(1,-1)$, $b=c-a$ then $v(c)>v(-a)$ gives that $v(b) =v(-a)=(0,0)$

so $v_2(a+b) = v_2(c)=-1$ is not $\ge \min(v_2(a),v_2(b))=0$