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Some clues of mine:

Also I'm wondering whether a nontrivial immediate extension of $K$ gives a nontrivial extension of $\mathfrak{m}$ by $R$. Or whether an unsolvable system of pairwise solvable congruence equations gives rise to a nontrivial extension.

Definition: A valued field $K$ is called spherically complete if for any chain of balls $B_1\supset B_2\supset B_3\supset \dots$, we have $\bigcap\limits_{i=1}^\infty B_i\neq\emptyset$. An immediate extension of a valued field $L$ is a field extension $F/L$ such that $F$ shares the same value group and residue field with $L$. A maximally complete field is a valued field with no non-trivial immediate extension.

The valuation ring $R$ of a valued field $L$ is call maximal if any system of pairwised intersecting sets $\{a_i+L_i|a_i\in R,L_i\text{ an ideal of }R\}$ has a non-empty intersection $\bigcap\limits_{i}(a_i+L_i)\neq\emptyset$.

Some clues of mine:

Also I'm wondering whether a nontrivial immediate extension of $K$ gives a nontrivial extension of $\mathfrak{m}$ by $R$. Or whether an unsolvable system of pairwise solvable congruence equations gives rise to a nontrivial extension. If $K\subset L$ is an immediete extension, $R\subset T$ the corresponding valuation rings, $\mathfrak{m}\subset\mathfrak{M}$ the maximal ideals, then every element $y\in T$ has the form $y=u\cdot x$ where $x\in R, u$ is a unit of $T$.

In B. Bhatt's lecture notes[1], Remark 4.2.5 says

... $\operatorname{Ext}_R^2(k,R)$ is non-zero if $K$ is not spherically complete.

which amounts to the following pure algebraic question.

Statement I want to prove: Let $K$ be a non-Archimedean complete valued field with rank 1 valuation (i.e. the value group is an ordered subgroup of $\mathbb{R}$ or equivalently its absolute value on $K$ is real-valued function, see notes[2] ). $R$ be its valuation ring with $R:=\{x\in K, |x|\leq1\}$, and $\mathfrak{m}$ be the maximal ideal, $k:=R/\mathfrak{m}$ is the residue field. The above assertion becomes if $K$ is not spherically complete, then the group $\operatorname{Ext}_R^2(k,R)\neq0$.

I'm wondering how to show the above proposition. Obviously from the long exact sequence obtained from $0\to\mathfrak{m}\to R\to k\to0$ one may identify $\operatorname{Ext}_R^2(k,R)=\operatorname{Ext}_R^1(\mathfrak{m},R)$ where the later corresponds to the equivalent classes of extensions of $\mathfrak{m}$ by $R$ (i.e. short exact sequence $0\to R\to X\to\mathfrak{m}\to0$ with $X$ an $R$-module to be determined).

Some clues of mine:

If R is a discrete valuation ring then complete=spherically complete there's nothing to prove. So it remains only the case where $\mathfrak{m}$ is infinitely generated, $R$ is non-noetherian.

In this survey[3] (see 6.1-6.3) and reference there, one knows for a rank 1 complete valuation ring, spherically complete=maximally complete=maximal=almost maximal=linearly compact.

And in this paper[4], theorem 9 establishes the equivalence of the following two conditions:

1. $R$ is a maximal valuation ring.
2. $\operatorname{Ext}_R^1(A,S)=0$ for $A$ any torsion-free module and $S$ any torsion-free module of rank 1 (hence is isomorphic to $R$ as a module).

So if $R$ is maximal valuation ring, we have $\operatorname{Ext}_R^1(\mathfrak{m},R)=0$ for free by condition 2. While I'm trying to figure out if $\operatorname{Ext}_R^1(\mathfrak{m},R)=0$ implies $\operatorname{Ext}_R^1(A,R)=0$ for any torsion-free module $A$ but have trouble getting over it.

Also I'm wondering whether a nontrivial immediate extension of $K$ gives a nontrivial extension of $\mathfrak{m}$ by $R$. Or whether an unsolvable system of pairwise solvable congruence equations gives rise to a nontrivial extension.

[1]Bhatt. Lecture notes for perfectoid spaces, 2017.

[2]Conrad. Lecture notes on space of valuations.

[3]Angel Barría Comicheo, Khodr Shamseddine. Summary on non-Archimedean valued fields, Contemp. Math., 704 , 1–36. 2018.

[4]Matlis, E. Injective Modules Over Prufer Rings. Nagoya Mathematical Journal, 15, 57-69. 1959.

In B. Bhatt's lecture notes[1], Remark 4.2.5 says

... $\operatorname{Ext}_R^2(k,R)$ is non-zero if $K$ is not spherically complete.

which amounts to the following pure algebraic question.

Statement I want to prove: Let $K$ be a non-Archimedean complete valued field with rank 1 valuation (i.e. the value group is an ordered subgroup of $\mathbb{R}$ or equivalently its absolute value on $K$ is real-valued function, see notes[2] ). $R$ be its valuation ring with $R:=\{x\in K, |x|\leq1\}$, and $\mathfrak{m}$ be the maximal ideal, $k:=R/\mathfrak{m}$ is the residue field. The above assertion becomes if $K$ is not spherically complete, then the group $\operatorname{Ext}_R^2(k,R)\neq0$.

I'm wondering how to show the above proposition. Obviously from the long exact sequence obtained from $0\to\mathfrak{m}\to R\to k\to0$ one may identify $\operatorname{Ext}_R^2(k,R)=\operatorname{Ext}_R^1(\mathfrak{m},R)$ where the later corresponds to the equivalent classes of extensions of $\mathfrak{m}$ by $R$ (i.e. short exact sequence $0\to R\to X\to\mathfrak{m}\to0$ with $X$ an $R$-module to be determined).

Some clues of mine:

If R is a discrete valuation ring then complete=spherically complete there's nothing to prove. So it remains only the case where $\mathfrak{m}$ is infinitely generated, $R$ is non-noetherian.

In this survey[3] (see 6.1-6.3) and reference there, one knows for a rank 1 complete valuation ring, spherically complete=maximally complete=maximal=almost maximal=linearly compact.

And in this paper[4], theorem 9 establishes the equivalence of the following two conditions (assuming $R$ is an integral domain):

1. $R$ is a maximal valuation ring.
2. $\operatorname{Ext}_R^1(A,S)=0$ for $A$ any torsion-free module and $S$ any torsion-free module of rank 1.

So if $R$ is maximal valuation ring, we have $\operatorname{Ext}_R^1(\mathfrak{m},R)=0$ for free by condition 2. While I'm trying to figure out if $\operatorname{Ext}_R^1(\mathfrak{m},R)=0$ implies $\operatorname{Ext}_R^1(A,S)=0$ for any torsion-free module $A$ and rank 1 torsion free $S$ but have trouble getting over it. (At first I made a mistake here as Moret-Bailly has commented by confusing rank of a module as the minimal cardinality of generators. But this is wrong, and the right definition is rank=$\dim_KS\otimes_RK$, $K$ fraction field)

Also I'm wondering whether a nontrivial immediate extension of $K$ gives a nontrivial extension of $\mathfrak{m}$ by $R$. Or whether an unsolvable system of pairwise solvable congruence equations gives rise to a nontrivial extension.

[1]Bhatt. Lecture notes for perfectoid spaces, 2017.

[2]Conrad. Lecture notes on space of valuations.

[3]Angel Barría Comicheo, Khodr Shamseddine. Summary on non-Archimedean valued fields, Contemp. Math., 704 , 1–36. 2018.

[4]Matlis, E. Injective Modules Over Prufer Rings. Nagoya Mathematical Journal, 15, 57-69. 1959.

In B. Bhatt's lecture notes[1], Remark 4.2.5 says

... $\operatorname{Ext}_R^2(k,R)$ is non-zero if $K$ is not spherically complete.

which amounts to the following pure algebraic question.

Statement I want to prove: Let $K$ be a non-Archimedean complete valued field with rank 1 valuation (i.e. the value group is an ordered subgroup of $\mathbb{R}$ or equivalently its absolute value on $K$ is real-valued function, see notes[2] ). $R$ be its valuation ring with $R:=\{x\in K, |x|\leq1\}$, and $\mathfrak{m}$ be the maximal ideal, $k:=R/\mathfrak{m}$ is the residue field. The above assertion becomes if $K$ is not spherically complete, then the group $\operatorname{Ext}_R^2(k,R)\neq0$.

I'm wondering how to show the above proposition. Obviously from the long exact sequence obtained from $0\to\mathfrak{m}\to R\to k\to0$ one may identify $\operatorname{Ext}_R^2(k,R)=\operatorname{Ext}_R^1(\mathfrak{m},R)$ where the later corresponds to the equivalent classes of extensions of $\mathfrak{m}$ by $R$ (i.e. short exact sequence $0\to R\to X\to\mathfrak{m}\to0$ with $X$ an $R$-module to be determined).

Some clues of mine:

If R is a discrete valuation ring then complete=spherically complete there's nothing to prove. So it remains only the case where $\mathfrak{m}$ is infinitely generated, $R$ is non-noetherian.

In this survey[3] (see 6.1-6.3) and reference there, one knows for a rank 1 complete valuation ring, spherically complete=maximally complete=maximal=almost maximal=linearly compact.

And in this paper[4], theorem 9 establishes the equivalence of the following two conditions:

1. $R$ is a maximal valuation ring.
2. $\operatorname{Ext}_R^1(A,S)$ for $A$ any torsion-free module and $S$ any torsion-free module of rank 1 (hence is isomorphic to $R$ as a module).

So if $R$ is maximal valuation ring, we have $\operatorname{Ext}_R^1(\mathfrak{m},R)=0$ for free by condition 2. While I'm trying to figure out if $\operatorname{Ext}_R^1(\mathfrak{m},R)=0$ implies $\operatorname{Ext}_R^1(A,R)$ for any torsion-free module $A$ but have trouble getting over it.

Also I'm wondering whether a nontrivial immediate extension of $K$ gives a nontrivial extension of $\mathfrak{m}$ by $R$. Or whether a unsolvable pairwise solvable congruence equations gives rise to a nontrivial extension.

[1]Bhatt. Lecture notes for perfectoid spaces, 2017.

[2]Conrad. Lecture notes on space of valuations.

[3]Angel Barría Comicheo, Khodr Shamseddine. Summary on non-Archimedean valued fields, Contemp. Math., 704 , 1–36. 2018.

[4]Matlis, E. Injective Modules Over Prufer Rings. Nagoya Mathematical Journal, 15, 57-69. 1959.

In B. Bhatt's lecture notes[1], Remark 4.2.5 says

... $\operatorname{Ext}_R^2(k,R)$ is non-zero if $K$ is not spherically complete.

which amounts to the following pure algebraic question.

Statement I want to prove: Let $K$ be a non-Archimedean complete valued field with rank 1 valuation (i.e. the value group is an ordered subgroup of $\mathbb{R}$ or equivalently its absolute value on $K$ is real-valued function, see notes[2] ). $R$ be its valuation ring with $R:=\{x\in K, |x|\leq1\}$, and $\mathfrak{m}$ be the maximal ideal, $k:=R/\mathfrak{m}$ is the residue field. The above assertion becomes if $K$ is not spherically complete, then the group $\operatorname{Ext}_R^2(k,R)\neq0$.

I'm wondering how to show the above proposition. Obviously from the long exact sequence obtained from $0\to\mathfrak{m}\to R\to k\to0$ one may identify $\operatorname{Ext}_R^2(k,R)=\operatorname{Ext}_R^1(\mathfrak{m},R)$ where the later corresponds to the equivalent classes of extensions of $\mathfrak{m}$ by $R$ (i.e. short exact sequence $0\to R\to X\to\mathfrak{m}\to0$ with $X$ an $R$-module to be determined).

Some clues of mine:

If R is a discrete valuation ring then complete=spherically complete there's nothing to prove. So it remains only the case where $\mathfrak{m}$ is infinitely generated, $R$ is non-noetherian.

In this survey[3] (see 6.1-6.3) and reference there, one knows for a rank 1 complete valuation ring, spherically complete=maximally complete=maximal=almost maximal=linearly compact.

And in this paper[4], theorem 9 establishes the equivalence of the following two conditions:

1. $R$ is a maximal valuation ring.
2. $\operatorname{Ext}_R^1(A,S)=0$ for $A$ any torsion-free module and $S$ any torsion-free module of rank 1 (hence is isomorphic to $R$ as a module).

So if $R$ is maximal valuation ring, we have $\operatorname{Ext}_R^1(\mathfrak{m},R)=0$ for free by condition 2. While I'm trying to figure out if $\operatorname{Ext}_R^1(\mathfrak{m},R)=0$ implies $\operatorname{Ext}_R^1(A,R)=0$ for any torsion-free module $A$ but have trouble getting over it.

Also I'm wondering whether a nontrivial immediate extension of $K$ gives a nontrivial extension of $\mathfrak{m}$ by $R$. Or whether an unsolvable system of pairwise solvable congruence equations gives rise to a nontrivial extension.

[1]Bhatt. Lecture notes for perfectoid spaces, 2017.

[2]Conrad. Lecture notes on space of valuations.

[3]Angel Barría Comicheo, Khodr Shamseddine. Summary on non-Archimedean valued fields, Contemp. Math., 704 , 1–36. 2018.

[4]Matlis, E. Injective Modules Over Prufer Rings. Nagoya Mathematical Journal, 15, 57-69. 1959.