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It's well-known that if you take the definition of surreal multiplication and attempts to generalize it to all games, the result is not well-defined, in that it does not respect equivalence of games.

What I'm wondering is this: What if we consider games that are equivalent to numbers, such as $\{*|*\}$, which is equivalent to $0$? Or to put it differently, what if we consider surreal multiplication, but we allow representing our numbers in non-numeric ways, such as representing a $0$ by $\{*|*\}$?

Is multiplication well-defined (equivalence-respecting) in this intermediate setting?

(I ask because I noticed that, although multiplication is well-defined for impartial games, it's not well-defined for games equivalent to impartial games. I'm wondering if something similar happens with numbers, or whether it remains well-defined; I haven't been able to find a counterexample.)

Thank you!

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(the counterexample has been worked out in collaboration with Harry Altman. All errors are mine)

The counterexample provides a game $K$ such that $K=0$ but $K^2\neq 0$. Here $\cong$ is identity of games, $=$ is Conway equivalence and juxtaposition (or sometimes $\cdot$) denote Conway product, and $K^2$ is $K \cdot K$.

Consider $G \cong^{def} \{-1,0|0,1\}$ and $H \cong^{def} * \cong^{def} \{0|0\}$.

$G$ and $H$ are Conway equivalent, and moreover $*+*=0$, hence $K\cong^{def} G + H =0$.

Let's compute

$* \cdot * \cong *$

$GH \cong G \cdot * \cong \{*|*\} =0 $

$K^2 \cong (G+H)(G+H) = G^2 + GH + HG + H^2 = G^2 + \{0|0\}$.

Considering the left options $-1$, $0$ in the definition of the Conway product $G \cdot G$ we get that

$ G \cdot (-1) + 0 \cdot G - 0 \cdot (-1) = -G = G = * $

is a left option of $G^2$; similarly

$G =*$ is a right option of $G^2$.

There is no need to fully compute $G^2$ to see that $G^2 + \{0|0\}$ is a first-winner game (hence not $=0$), since the first player can always move in $G^2$ choosing the option equivalent to $*$. The second player is left to play with a game equivalent to $*+*$, so he loses.

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