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Let $X$ be an algebraic complex K3 surface, we know that $X$ is deformation equivalent to a smooth quartic surface or more generally a K3 surface with Picard number $1$ (a very general K3 surface in the family of K3 surfaces). It means that there exists a proper morphism $\mathcal{X}\rightarrow S$ with each fibre a K3 surface, $\mathcal{X}_{\eta}=X$ ($\eta$ a general point on $S$) and $\rho(\mathcal{X}_t)=1$ for some closed $t\in S$.

I want to show that $S$ can be a smooth algebraic curve or better be $\mathbb{A}^1$.

Thought: Consider the universal family $M$ of algebraic complex K3 surfaces, the family of K3 surfaces with $\rho>1$ is a countable union of dimension $18$ family $D_i$ of subvarieties of $M$. Then we only need to pick a curve outside all of the diviors $D_i$, how can one explicitly do it?

Let $X$ be an algebraic complex K3 surface, we know that $X$ is deformation equivalent to a smooth quartic surface or more generally a K3 surface with Picard number $1$ (a very general K3 surface in the family of K3 surfaces). It means that there exists a proper morphism $\mathcal{X}\rightarrow S$ with each fibre a K3 surface, $\mathcal{X}_{\eta}=X$ ($\eta$ a general point on $S$) and $\rho(\mathcal{X}_t)=1$ for some closed $t\in S$.

I want to show that $S$ can be a smooth algebraic curve or better be $\mathbb{A}^1$.

Thought: Consider the universal family $M$ of algebraic complex K3 surfaces, the family of K3 surfaces with $\rho>1$ is a countable union of dimension $18$ family $D_i$ of subvarieties of $M$. Then we only need to pick a curve outside all of the diviors $D_i$, how can one explicitly do it?

Let $X$ be an algebraic complex K3 surface, we know that $X$ is deformation equivalent to a smooth quartic surface or more generally a K3 surface with Picard number $1$ (a very general K3 surface in the family of K3 surfaces). It means that there exists a proper morphism $\mathcal{X}\rightarrow S$ with each fibre a K3 surface, $\mathcal{X}_{\eta}=X$ ($\eta$ a general point on $S$) and $\rho(\mathcal{X}_t)=1$ for some closed $t\in S$.

I want to show that $S$ can be a smooth algebraic curve or better be $\mathbb{A}^1$.

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user494851
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Let $X$ be an algebraic complex K3 surface, we know that $X$ is deformation equivalent to a smooth quartic surface or more generally a K3 surface with Picard number $1$ (a very general K3 surface in the family of K3 surfaces). It means that there exists a proper morphism $\mathcal{X}\rightarrow S$ with each fibre a K3 surface, $\mathcal{X}_{\eta}=X$ ($\eta$ the generica general point on $S$) and $\rho(\mathcal{X}_t)=1$ for some closed $t\in S$.

I want to show that $S$ can be a smooth algebraic curve or better be $\mathbb{A}^1$.

Thought: Consider the universal family $M$ of algebraic complex K3 surfaces, the family of K3 surfaces with $\rho>1$ is a countable union of dimension $18$ family $D_i$ of subvarieties of $M$. Then we only need to pick a curve outside all of the diviors $D_i$, how can one explicitly do it?

Let $X$ be an algebraic complex K3 surface, we know that $X$ is deformation equivalent to a smooth quartic surface or more generally a K3 surface with Picard number $1$ (a very general K3 surface in the family of K3 surfaces). It means that there exists a proper morphism $\mathcal{X}\rightarrow S$ with each fibre a K3 surface, $\mathcal{X}_{\eta}=X$ ($\eta$ the generic point on $S$) and $\rho(\mathcal{X}_t)=1$ for some closed $t\in S$.

I want to show that $S$ can be a smooth algebraic curve or better be $\mathbb{A}^1$.

Thought: Consider the universal family $M$ of algebraic complex K3 surfaces, the family of K3 surfaces with $\rho>1$ is a countable union of dimension $18$ family $D_i$ of subvarieties of $M$. Then we only need to pick a curve outside all of the diviors $D_i$, how can one explicitly do it?

Let $X$ be an algebraic complex K3 surface, we know that $X$ is deformation equivalent to a smooth quartic surface or more generally a K3 surface with Picard number $1$ (a very general K3 surface in the family of K3 surfaces). It means that there exists a proper morphism $\mathcal{X}\rightarrow S$ with each fibre a K3 surface, $\mathcal{X}_{\eta}=X$ ($\eta$ a general point on $S$) and $\rho(\mathcal{X}_t)=1$ for some closed $t\in S$.

I want to show that $S$ can be a smooth algebraic curve or better be $\mathbb{A}^1$.

Thought: Consider the universal family $M$ of algebraic complex K3 surfaces, the family of K3 surfaces with $\rho>1$ is a countable union of dimension $18$ family $D_i$ of subvarieties of $M$. Then we only need to pick a curve outside all of the diviors $D_i$, how can one explicitly do it?

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one One-dimensional family of complex algebraic K3 surfaces

Let $X$ be aan algebraic complex K3 surface, we know that $X$ is deformation equivalent to a smooth quartic surface or more generally a K3 surface with Picard number $1$ (a very general K3 surface in the family of K3 surfaces). It means that there exists a proper morphism $\mathcal{X}\rightarrow S$ with each fibre a K3 surface, $\mathcal{X}_{\eta}=X$ ($\eta$ the generic point on $S$) and $\rho(\mathcal{X}_t)=1$ for some closed $t\in S$.

I want to show that $S$ can be a smooth algebraic curve or better be $\mathbb{A}^1$.

Thought: Consider the universal family $M$ of algebraic complex K3 surfaces, the family of K3 surfaces with $\rho>1$ is a countable union of dimension $18$ family $D_i$ of subvarieties of $M$. Then we only need to pick a curve outside all of the diviors $D_i$, how can one explicitly do it?

one-dimensional family of complex algebraic K3 surfaces

Let $X$ be a algebraic complex K3 surface, we know that $X$ is deformation equivalent to a smooth quartic surface or more generally a K3 surface with Picard number $1$ (a very general K3 surface in the family of K3 surfaces). It means that there exists a proper morphism $\mathcal{X}\rightarrow S$ with each fibre a K3 surface, $\mathcal{X}_{\eta}=X$ ($\eta$ the generic point on $S$) and $\rho(\mathcal{X}_t)=1$ for some closed $t\in S$.

I want to show that $S$ can be a smooth algebraic curve or better be $\mathbb{A}^1$.

Thought: Consider the universal family $M$ of algebraic complex K3 surfaces, the family of K3 surfaces with $\rho>1$ is a countable union of dimension $18$ family $D_i$ of subvarieties of $M$. Then we only need to pick a curve outside all of the diviors $D_i$, how can one explicitly do it?

One-dimensional family of complex algebraic K3 surfaces

Let $X$ be an algebraic complex K3 surface, we know that $X$ is deformation equivalent to a smooth quartic surface or more generally a K3 surface with Picard number $1$ (a very general K3 surface in the family of K3 surfaces). It means that there exists a proper morphism $\mathcal{X}\rightarrow S$ with each fibre a K3 surface, $\mathcal{X}_{\eta}=X$ ($\eta$ the generic point on $S$) and $\rho(\mathcal{X}_t)=1$ for some closed $t\in S$.

I want to show that $S$ can be a smooth algebraic curve or better be $\mathbb{A}^1$.

Thought: Consider the universal family $M$ of algebraic complex K3 surfaces, the family of K3 surfaces with $\rho>1$ is a countable union of dimension $18$ family $D_i$ of subvarieties of $M$. Then we only need to pick a curve outside all of the diviors $D_i$, how can one explicitly do it?

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