If $\kappa \rightarrow (\alpha)^r_2$ holds for every $r\in \omega$, then is $\kappa$ an $\alpha$-Erdős cardinal?

Question

If $\kappa \rightarrow (\alpha)^r_2$ holds for every $r\in \omega$, then is $\kappa$ an $\alpha$-Erdős cardinal? (or rather, does $\kappa \rightarrow (\alpha)^{<\omega}_2$ hold?)

$\kappa \rightarrow (\alpha)^r_2$ means that $\forall f: [\kappa]^r \rightarrow 2$, $\exists A\in[\kappa]^\alpha$ such that $|f"[A]^r|=1$.

$\kappa \rightarrow (\alpha)^{<\omega}_2$ is $\forall f: [\kappa]^{<\omega} \rightarrow 2$, $\exists A\in[\kappa]^\alpha$ such that $\forall l \in \omega$, $|f"[A]^l|=1$.

Clearly if $\kappa \rightarrow (\alpha)^r_2$ holds for every $r\in \omega$, then splitting $f:[\kappa]^{<\omega}\rightarrow 2$ into functions $f_r:[\kappa]^r\rightarrow 2$ and applying the partition relations, we get that for each $l\in \omega$ we can find an $A_l\in[\kappa]^\alpha$ where $|f"[A_l]^l|=1$. But we can't simply take the intersection of these $A_l$ as this may not have size $\alpha$. (It could easily be empty!)

We can also find the $A_l$ such that $A_i\supseteq A_{i+1}$ for all $i$, by restricting the $f_r:[A_{r-1}]^r\rightarrow 2$ and finding homogenous sets for these. But again, the intersection of all of these may not have size $\alpha$.

In particular, does this hold when $\alpha=\kappa$?

What about if $\kappa \rightarrow (\alpha)^r_\gamma$ holds for every $r\in \omega$? Then does $\kappa \rightarrow (\alpha)^{<\omega}_\gamma$ hold?

If this is not the case, then is there a name for these $\kappa$ where $\kappa \rightarrow (\alpha)^r_2$ holds for every $r\in \omega$?

EDIT: In view of the answers; $\kappa$ is weakly compact implies $\kappa\rightarrow (\kappa)^r_\gamma$ for any $r<\omega$, and $\gamma<\kappa$. But does $\forall \alpha<\kappa$, $\kappa \rightarrow (\alpha)^2_2$ imply $\forall \alpha<\kappa,r<\omega, \gamma<\kappa, \kappa\rightarrow (\alpha)^r_\gamma$?

Answer 1

The two properties are not the same.

If $\kappa$ is weakly compact, then $\kappa\rightarrow(\kappa)^r_\lambda$ for any $r<\omega$ and $\lambda<\kappa$. These cardinals are compatible with $V=L$.

A cardinal $\kappa$ for which $\kappa\rightarrow(\kappa)^{<\omega}_2$ is called a Ramsey cardinal, and these cardinals imply that $0^\sharp$ exists.

Answer 2

Unfortunately $\kappa\rightarrow(\alpha)^r_2$ for every $r$ is much weaker than $\kappa\rightarrow(\alpha)^{<\omega}_2$. In fact, the least $\kappa$ for which $\kappa\rightarrow(\alpha)^r_2$ holds is $\exp_{r-1}(|\alpha|)^+$ (the general Erdos-Rado theorem), and so the least $\kappa$ for which this holds for all $r$, is $\exp_\omega(|\alpha|)$, a singular cardinal. Here $\exp_r(\tau)$ is the $r$-fold iterated power of $\tau$.

If $\alpha$ is limit, then the least $\kappa$ for which $\kappa\rightarrow(\alpha)^{<\omega}_2$ holds, is strongly inaccessible.