Ring whose identity is $42$?

Question

I am looking to find an example of a ring, in which the identity is $42$.

My initial thoughts are, well would $\mathbb{Z}_{84}$ work?

Answer 1

How about $\Bbb Z/41\Bbb Z$? A field even!

Answer 2

Take any ring $X$, quotient by the ideal generated by $41$.

Answer 3

Define a ring structure on $\mathbb{Z}$ with addition $\oplus$ and multiplication $\odot$ as follows: $$x\oplus y:=(x-41)+(y-41)+41$$ and $$x\odot y:=(x-41)(y-41)+41$$ for all $x,y\in\mathbb{Z}$. Then, $(\mathbb{Z},\oplus,\odot)$ has $42$ as the multiplicative identity.

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If you want $42$ to be the additive identity, then you can do something similar: $$x \boxplus y:=(x-42)+(y-42)+42$$ and $$x\boxdot y:=(x-42)(y-42)+42$$ for all $x,y\in\mathbb{Z}$. Then, $(\mathbb{Z},\boxplus,\boxdot)$ is a ring with $42$ as the additive identity.

Answer 4

How about the ring $R=\{0,42\}\subset\mathbb{Z}$ (subset as sets only, not a subring), where we define $0+x=x+0=x$, $0\cdot x=x\cdot0=0$, $42+42=0$, and $42\cdot42=42$? In other words, this is just $\mathbb{Z}_2$, but using "$42$" as the symbol for $\overline{1}$.