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Is there any commutative ring that satisfies all of the following:

1) when two numbers are multiplied, the resulting product can only be dissected into these two numbers or their "prime factor" multiplication form or multiplicative identity times the product. ( 1) excludes the case when multiplicative identity and some number are multiplied.)

2) when two numbers are added, the resulting sum can only be dissected into these two numbers or the sum of additive identity and the resulting sum. ( 2) excludes the case when additive identity and some number are added.)

Edit-3) For the aforementioned two cases, the following construction is considered equivalent: $(x+y)+z = x+(y+z)$ and commutative multiplication cases.

Edit-optional-4) Multiplicative identity and additive identity do not have to exist in the commutative ring. In this case, by "commutative ring", I mean commutative ring minus multiplicative identity and additive identity.

I assume non-trivial case.

Edit: I also assume that the number of non-equivalent elements are at least countable infinite.

Edit: What happens if we remove the assumption of multiplicative identity and additive identity from commutative ring?

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    $\begingroup$ If there is a nonzero element $x$ in the ring, then $2x+0=x+x$ with $\{x\}\ne\{0,2x\}$, contra #2. $\endgroup$ – anon Apr 9 '13 at 8:46
  • $\begingroup$ @anon I guess my edit will fix that problem. $\endgroup$ – Zeus Apr 9 '13 at 8:53
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The trivial ring $R=\{0\}$, whose addition and multiplication are defined via $$\begin{array}{c|c|} + & 0\\\hline 0 & 0\\\hline \end{array}\qquad \begin{array}{c|c|} \cdot & 0\\\hline 0 & 0\\\hline \end{array}$$


A non-trivial example is the field of two elements $\mathbb{F}_2=\{0,1\}$, whose addition and multiplication are defined via $$\begin{array}{c|c|c|} + & 0 & 1\\\hline 0 & 0 & 1\\\hline 1 & 1 & 0\\\hline \end{array}\qquad \begin{array}{c|c|c|} \cdot & 0 & 1\\\hline 0 & 0 & 0\\\hline 1 & 0 & 1\\\hline \end{array}$$

Since we may disregard any cases where the additive identity is being added to something, the only additive case to look at is $1+1=0$. The only additive "dissections" of $0$ are $1+1$ (the original sum) or $0+0$ (the sum plus the additive identity).

Since we may disregard any cases where the multiplicative identity is being multiplied by something, the only multiplicative case to look at is $0\cdot 0=0$. The only multiplicative "dissections" of $0$ are $0\cdot 0$ (the original product) or $0\cdot 1$ (the product times the multiplicative identity).

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Hint $\ $ In any ring with $> 3$ elements, $\rm\,\ 1\!+\!(x\!-\!1) = (1\!+\!x)\!-\!1,\,\ x\ne 0,\pm1\:$ is a counterexample if $-1\ne 1.\:$ Else if $-1 = 1,\:$ so too is $\rm\:x\!+\!(y\!+\!z) = (x\!+\!y)\!+\!z\:$ for distinct nonzero $\rm\,x,y,z.$

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  • $\begingroup$ Edited the question. What happens now? $\endgroup$ – Zeus Apr 9 '13 at 15:13

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