Choice principles and Ramsey's theorem. I was reading Ramsey theory and wanted to know which choice principle was equivalent to infinite Ramsey theorem (by which I mean "For all $n,k \in \mathbb N$ and for all infinite sets $Z$, every $n$-colouring of $k$-tuples of $Z$ has an infinite homogeneous subset.") .
I know that countable choice implies Ramsey's theorem, and even Ramsey's theorem for pairs in two colours implies Kőnig's lemma (and hence countable choice for finite sets) over ZF, but I do not know about the other directions.
There is a question here that claims Kőnig's lemma implies Ramsey's theorem, but I am not sure if it is correct, thank you.
 A: Andreas Blass pointed in the comments to his paper

Blass, Andreas, Ramsey’s theorem in the hierarchy of choice principles, J. Symb. Log. 42(1977), 387-390 (1978). ZBL0374.02037.

In that paper he shows that in Cohen's first model, if $A$ is the set of generic Cohen reals, which is Dedekind-finite there, then $F({a,b})=\min(a\mathop{\triangle}b)\pmod 2$ is a colouring without an infinite homogeneous subset.
Indeed, if $B$ was homogeneous, let $\dot B$ be a name for it, and $E$ a finite support for $\dot B$ and for some $p$ such that $p\Vdash``\dot B$ is homogeneous for $\dot F"$. Extend $p$ to $q$ which decides that some $\dot a_n\in\dot B$ for some $n\notin E$, let $m$ be an integer not mentioned in $q$ or $E$.
We consider $q'$ to be $q$ with the information that $q'(m,i)=q(n,i)$ for all $i$ on which $q(n,i)$ was defined. It is easy to see that the $2$-cycle $(m\ n)$ is a permutation of $\omega$ which fixes $E$ pointwise, and indeed $q'$ is not moved. Therefore, $q'\Vdash\dot a_m\in\dot B$ as well. But since we did not decide yet the value of $\dot F(\{\dot a_m,\dot a_n\}^\bullet)$, we can throw a wrench into the whole $\dot B$ is homogeneous.
Why did we do all of that? Well, in Cohen's first model every set can be linearly ordered. So given a finitely branching tree of height $\omega$, by induction each level is finite, linearly ordering the underlying set we get a uniform enumeration of the levels, and therefore a proof that the tree is countable. Therefore, it has a branch.
It is worth noting that Kőnig's lemma is equivalent to any of the four formulations found in "Every countable/infinite family of finite sets admits a [partial] choice function" (the partial means an infinite subfamily with a choice function).

Now, it seems that at the same time as Andreas was working on his paper, Gabriele Lolli published

Lolli, Gabriele, On Ramsey’s theorem and the axiom of choice, Notre Dame J. Formal Logic 18, 599-601 (1977). ZBL0351.02041.

Where he incorrectly claims that the equivalence does in fact hold. To quote the Math Review by Robert Cowen, "the proofs given are not convincing."
This is further addressed by Eleftherios Tachtsis in his paper. Which also contain some relevant results.

Tachtsis, Eleftherios, On Ramsey’s theorem and the existence of infinite chains or infinite anti-chains in infinite posets, J. Symb. Log. 81, No. 1, 384-394 (2016). ZBL1342.03036.

Finally, John Truss in a paper that followed by a paper joint with Thomas Forester studied Kőnig's lemma and Ramsey theorem, and it seems that they should be mentioned here.

Truss, J., Some cases of Konig’s lemma, Set Theory Hierarchy Theory, Mem. Tribute A. Mostowski, Bierutowice 1975, Lect. Notes Math. 537, 273-284 (1976). ZBL0338.02039.
Forster, T. E.; Truss, J. K., Ramsey’s theorem and König’s lemma, Arch. Math. Logic 46, No. 1, 37-42 (2007). ZBL1114.03038.

