I was reading the book On Number And Games and I have some question. In this book Conway constructed the set of "games" with a addition and a multiplication. I understand that the surreal numbers are a particular case of such games. My question is if it games form a commutative ring. And if it is true, does the ring games form a integral domain? has finite Krull dimension? Can one describe its maximal spectrum?. And finally, is there any book about these questions?.
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2$\begingroup$ One obstruction to games being a ring: multplication of games (as opposed to numbers) is dependent on the form and not just the value of the game; in particular, IIRC, $\{-1\mid1\}\cdot G$ is not necessarily equal to $0\cdot G$. This gets described in either ONAG or Winning Ways. $\endgroup$– Steven StadnickiCommented Mar 6, 2018 at 0:06
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1$\begingroup$ "Conmutative", short for Conway-commutative? $\endgroup$– Gerald EdgarCommented Mar 6, 2018 at 0:32
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1$\begingroup$ @GeraldEdgar no, just a mistake in Spanish$\to$English translation $\endgroup$– YCorCommented Mar 6, 2018 at 0:55
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1$\begingroup$ You might want to look at this question (which I phrased in an admittedly very awkward way, so skip over most of the definitions). $\endgroup$– Gro-TsenCommented Mar 12, 2018 at 6:39
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1 Answer
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See the correction below.
The class of games does indeed constitute a commutative ring under Conway's definitions of sums and products. This is already in Conway's book, but more explicitly stated and proved in Combinatorial Game Theory by Aaron N. Siegel, American Mathematical Society, 2013. On the other hand, the theorem that ensures that the cancellation law holds for multiplication is explicitly formulated and proved for surreal numbers (Theorem 8, pp.19-20 for Conway; Proposition 2.3, p.413 for Siegel). I suspect it fails for games.
Correction (1/13/22): (Siegel p. 413) establishes the following for the multiplication of long games. For all long games $x,y,z$: (a) $x0 \cong 0$, (b) $xy \cong yx$, (c) $x1 \cong x$, (d) $(xy)z=x(yz)$, (e) $(x+y)z=xz +yz$. Siegel's long games are Conway's games.
However, as David E Speyer observes in his comments, contrary to what I previously claimed this is not enough to show that the class of games constitutes a commutative ring under Conway's definitions of sums and products. See Speyer's comments for details.
answered Mar 6, 2018 at 1:54
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$\begingroup$ I am confused as to how this can be right. As usual, define $0 = \{\ |\ \}$, $1 = \{0|\ \}$, $2 = \{ 1|\ \} = \{ 0, 1|\ \}$, $\tfrac{1}{2} = \{ 0| 1 \}$ and $\ast = \{ 0,0 \}$. Then $2 = 1+1$. It is well known that $\tfrac{1}{2}+\tfrac{1}{2} = 1$ and $\ast + \ast = 0$. So, if we had distributive multiplication, we would have $\tfrac{1}{2} \cdot 2 = 1$ and $2 \cdot \ast = 0$. But then $( \tfrac{1}{2}\cdot 2 ) \cdot \ast \neq \tfrac{1}{2}\cdot (2 \ \cdot \ast)$. $\endgroup$ Commented Jan 13, 2022 at 11:31
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1$\begingroup$ I guess this answer may be referring to multiplying games, not multiplying "games modulo equivalence". But if we are working with games not modulo equivalence, then we don't have negatives, so we only get a commutative rig. $\endgroup$ Commented Jan 13, 2022 at 12:26
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$\begingroup$ @ David E Speyer What is established in (Siegel p. 413) for the multiplication of long games is the following. For all long games $x,y,z$: (a) $x0 \cong 0$, (b) $xy \cong yx$, (c) $x1 \cong x$, (d) $(xy)z=x(yz)$, (e) $(x+y)z=xz +yz$. Siegel's long games are Conway's games. $\endgroup$ Commented Jan 13, 2022 at 17:05
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1$\begingroup$ So, there are two options: (1) Considering games up to $=$ and (2) considering games up to $\cong$.(I find it very confusing that $=$ is the stricter of the two relations, but this seems to be Siegel's notation.) If we do (1), then I don't see a statement in Siegel that multiplication is a well defined operation. If we do (2), then we don't have additive inverses. This doesn't seem to contradict Prop 2.2, but it means that there is no commutative ring to be studied. $\endgroup$ Commented Jan 13, 2022 at 17:34
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$\begingroup$ Siegel does show that multiplication is a well-defined operation on surreal numbers (2.3.(b)) , but not for $=$-classes of games. $\endgroup$ Commented Jan 13, 2022 at 17:41