There is a unique up to isomorphism algebraically closed field of characteristic 0 and cardinality of the continuum. Let's call it $K$. We usually call it $\mathbb{C}$, but by this we impose a topological structure* on $K$. In fact there are many other possible ring topologies on $K$ (by a ring topology we mean a topology for which the addition and multiplication are continuous maps $K\times K \to K$). There seems to be a zoo of those**. I care more for better behaved ones.

It is natural to consider separable topologies (ie admitting a countable dense subset) such that the uniform structure imposed by the additive group structure is complete. In fact, I care mostly for Polish topologies (separable and defined by a complete metric). For these the uniform structure is automatically complete.

A famous Polish topology on $K$ is $\mathbb{C}_p$ which is obtained by completing the algebraic closure of $\mathbb{Q}_p$ with respect to the unique absolute value on it extending the $p$-adic absolute value on $\mathbb{Q}_p$ (absolute value = multiplicative norm). Here is another example: consider $k$, a countable field (eg $\mathbb{Q}$ or its algebraic closure), and take the degree valuation on $k(t)$. Complete to $k((t))$, extend (uniquely) the absolute value to the algebraic closure and complete again (use Krasner's Lemma). Get a Polish topology on $K$ for which $k$ is discrete.

All the examples of topologies on $K$ above, and all the examples I am aware of are defined by absolute values. Hence the question in the title: Is every Polish ring topology on $\mathbb{C}$ defined by an absolute value?

Remark: Given a topology on $K$ there is a unique minimal closed, algebraically closed subfield. I am willing to assume $K$ is minimal if it helps.

Note: for locally compact fields it is always the case that the topology is defined by an absolute value. Indeed, one can use the modular function applied to multiplication operator on the additive group in order to construct an absolute value on the field.

$*$ There are plenty of ways to identify $K$ with $\mathbb{C}$. Given a topology on $K$ we can conjugate it by a field automorphism and define another. We consider here topologies up to conjugation.

$**$ related: http://mathoverflow.net/a/106454/89334

There is a unique up to isomorphism algebraically closed field of characteristic 0 and cardinality of the continuum. Let's call it $K$. We usually call it $\mathbb{C}$, but by this we impose a topological structure* on $K$. In fact there are many other possible ring topologies on $K$ (by a ring topology we mean a topology for which the addition and multiplication are continuous maps $K\times K \to K$). There seems to be a zoo of those**. I care more for better behaved ones.

It is natural to consider separable topologies (ie admitting a countable dense subset) such that the uniform structure imposed by the additive group structure is complete. In fact, I care mostly for Polish topologies (separable and defined by a complete metric). For these the uniform structure is automatically complete.

A famous Polish topology on $K$ is $\mathbb{C}_p$ which is obtained by completing the algebraic closure of $\mathbb{Q}_p$ with respect to the unique absolute value on it extending the $p$-adic absolute value on $\mathbb{Q}_p$ (absolute value = multiplicative norm). Here is another example: consider $k$, a countable field (eg $\mathbb{Q}$ or its algebraic closure), and take the degree valuation on $k(t)$. Complete to $k((t))$, extend (uniquely) the absolute value to the algebraic closure and complete again (use Krasner's Lemma). Get a Polish topology on $K$ for which $k$ is discrete.

All the examples of topologies on $K$ above, and all the examples I am aware of are defined by absolute values. Hence the question in the title: Is every Polish ring topology on $\mathbb{C}$ defined by an absolute value?

Remark: Given a topology on $K$ there is a unique minimal closed, algebraically closed subfield. I am willing to assume $K$ is minimal if it helps.

Note: for locally compact fields it is always the case that the topology is defined by an absolute value. Indeed, one can use the modular function applied to multiplication operator on the additive group in order to construct an absolute value on the field.

$*$ There are plenty of ways to identify $K$ with $\mathbb{C}$. Given a topology on $K$ we can conjugate it by a field automorphism and define another. We consider here topologies up to conjugation.

$**$ related: https://mathoverflow.net/a/106454/89334

There is a unique up to isomorphism algebraically closed field of characteristic 0 and cardinality of the continuoum. Let's call it $K$. We usually call it $\mathbb{C}$, but by this we impose a topological structure on $K$. In fact there are many other possible ring topologies on $K$ (by a ring topology we mean a topology for which the adition and multiplication are continuous maps $K\times K \to K$). There seems to be a zoo of those*. I care more for better behaved ones.

It is natural to consider separable topologies (ie admiting a countable dense subset) which are complete with respect to the uniform structure imposed by the additive group structure. In fact, I care mostly for Polish topologies (separable which admit a complete compatible metric). These are automatically complete.

A famous Polish topology on $K$ is $\mathbb{C}_p$ which is obtained by completing the algebraic closure of $\mathbb{Q}_p$. Here is another example: consider $k$, a countable subfield (eg $\mathbb{Q}$ or its algebraic closure), and take the degree valuation on $k(t)$. Complete to $k((t))$, extend (uniquely) the absulote value to the algebraic closure and complete again (use Krasner's Lemma). Get a Polish topology on $K$ for which $k$ is discrete.

All the examples above, and all the examples I am aware of come from absolute values. Hence the question in the title: Does every Polish ring topology on $K$ come from an absolute value?

Remark: Given a topology on $K$ there is a unique minimal closed, algebrically closed subfield. I am willing to assume $K$ is minimal if it helps.

Note: for locally compact fields the topology alway come from an absolute value, given by the modular function applied to multiplication operators on the additive group.

$*$ related: http://mathoverflow.net/a/106454/89334

There is a unique up to isomorphism algebraically closed field of characteristic 0 and cardinality of the continuum. Let's call it $K$. We usually call it $\mathbb{C}$, but by this we impose a topological structure* on $K$. In fact there are many other possible ring topologies on $K$ (by a ring topology we mean a topology for which the addition and multiplication are continuous maps $K\times K \to K$). There seems to be a zoo of those**. I care more for better behaved ones.

It is natural to consider separable topologies (ie admitting a countable dense subset) such that the uniform structure imposed by the additive group structure is complete. In fact, I care mostly for Polish topologies (separable and defined by a complete metric). For these the uniform structure is automatically complete.

A famous Polish topology on $K$ is $\mathbb{C}_p$ which is obtained by completing the algebraic closure of $\mathbb{Q}_p$ with respect to the unique absolute value on it extending the $p$-adic absolute value on $\mathbb{Q}_p$ (absolute value = multiplicative norm). Here is another example: consider $k$, a countable field (eg $\mathbb{Q}$ or its algebraic closure), and take the degree valuation on $k(t)$. Complete to $k((t))$, extend (uniquely) the absolute value to the algebraic closure and complete again (use Krasner's Lemma). Get a Polish topology on $K$ for which $k$ is discrete.

All the examples of topologies on $K$ above, and all the examples I am aware of are defined by absolute values. Hence the question in the title: Is every Polish ring topology on $\mathbb{C}$ defined by an absolute value?

Remark: Given a topology on $K$ there is a unique minimal closed, algebraically closed subfield. I am willing to assume $K$ is minimal if it helps.

Note: for locally compact fields it is always the case that the topology is defined by an absolute value. Indeed, one can use the modular function applied to multiplication operator on the additive group in order to construct an absolute value on the field.

$*$ There are plenty of ways to identify $K$ with $\mathbb{C}$. Given a topology on $K$ we can conjugate it by a field automorphism and define another. We consider here topologies up to conjugation.

$**$ related: http://mathoverflow.net/a/106454/89334