Strongly non-Ramsey order type in polarized partition problems It is known (a theorem of Komjáth and Hajnal) that it is consistent (by adding a Cohen real to a universe where there is no Suslin line) that there exists an order type $\theta$ such that for any other order type $\eta$, there exists a bad coloring $f: [\eta]^2 \to \omega$ such that for any order preserving embedding $g: \theta \to \eta$, $f''[g''\theta]^2=\omega$, i.e. $\eta \not \rightarrow [\theta]_\omega^2$, in particular $\eta \not \rightarrow (\theta)_\omega^2$ (there are more results but enough for now). 
I'm wondering if there are references about similar phenomenon for polarized partition relation, more concretely: do there exist order types $\theta_0, \theta_1$ and cardinal $\gamma$ such that for any order types $\eta_0, \eta_1$, there exists $f: \eta_0\times \eta_1 \to \gamma$ such that for all order-preserving embeddings $g_0: \theta_0 \to \eta_0, g_1: \theta_1\to \eta_1$, $f''[g_0''\theta_0\times g_1''\theta_1]=\omega$ (or just not constant), i.e. ${\eta_0 \choose \eta_1} \not \rightarrow {\theta_0\choose \theta_1}_\gamma^{1,1}$. Any pointers would be appreciated. Thanks in advance. 
 A: We can't find such pair of bad order types, aka there indeed is some Erdös-Rado phenomenon in the polarized partitions with respect to linear orderings. Indeed given $\gamma$ number of colors and $\theta$, $\psi$ we can find large saturated dense linear orders $\alpha, \beta$ such that for any $f: \alpha\times \beta \to \gamma$, there exists $X\subset \alpha, Y\subset \beta$ such that $tp(X)=tp(\alpha), tp(Y)=tp(\beta)$ such that $f''X\times Y$ is constant.
To elaborate: Given $\gamma$ the number of colors and order types $\theta, \psi$, given let $\lambda$ be regular a lot larger than $|\theta|, |\psi|, \gamma$, let $\beta$ be a $\lambda$-saturated dense linear order. Let $\lambda'$ be a lot larger than $2^{|\beta|}, \gamma$ and let $\alpha$ be $\lambda'$-saturated dense linear ordering, we claim that ${\alpha\choose \beta}\to {\theta\choose \psi}^{1,1}_{\gamma}$. Given a coloring $f: \alpha \times \beta\to \gamma$, for each $i\in \alpha$, consider $f_i: \beta\to \gamma$. 

There exists a monochromatic under $f_i$ sub order of $\beta$ that is $\lambda$-saturated.
  Why? Suppose not, then induct on $i\in \gamma$ to build a cut. Stage 0, there exists $\eta_0,\eta_1<\lambda$ and $A_0=\{a_j: j<\eta_0\}<B_0=\{b_j: j<\eta_1\}$ such that there is no $c\in \beta$ that has color 0 such that $A_0<c<B_0$. Note that $(A_0,B_0)=\{c\in \beta: A_0<c<B_0\}$ is still $\lambda$-saturated. Inductively at stage $\alpha$, suppose we have $A_0<A_1<\cdots<A_\beta<\cdots < B_\beta<\cdots <B_0$ (i.e. $\forall \beta<\gamma<\alpha \ A_\beta<A_\gamma<B_\gamma<B_\beta$). Since $\beta$ is $\lambda$-saturated, let $A'_\alpha= \bigcup_{i<\alpha}A_i, B'_\alpha=\bigcup_{i<\alpha}B_i$, we know $(A'_\alpha, B'_\alpha)$ is still $\lambda$-saturated. By hypothesis, we can find $A_\alpha, B_\alpha$ such that $A'_\alpha<A_\alpha<B_\alpha<B'_\alpha$ such that no element in $(A_\alpha, B_\alpha)$ has color $\alpha$. This can proceed in $\gamma$ steps since $\gamma<\lambda$ but absurdity arises since $\beta$ is saturated and every element gets colored.

Denote the monochromatic $\lambda$-saturated suborder of $\beta$ under $f_i$ $\beta_i$, and the color be $\gamma_i$. Now define a coloring $g: \alpha\to 2^{\beta}\times \gamma$ such that $g(i)=(\beta_i,\gamma_i)$. By the similar argument above we can get a suborder of $\alpha$, say $\alpha'$ that is monochromatic and $\lambda'$-saturated. Let $(\beta',\gamma')$ be the color $g''\alpha'$, then $f''\alpha'\times \beta'=\gamma'$, as $\alpha'$ and $\beta'$ are sufficiently saturated we can embed $\theta$ and $\psi$ into them respectively.
