Does there exist a ring with $5$ elements with ${\rm Char}(R)=2$? Definition of a ring needs not to have $1$, nor does the definition of characteristic depends on it.
Question: Does there exist a ring with $5$ elements with ${\rm Char}(R)=2$?
Let say the definition of ${\rm Char}(R)$ is the smallest positive integer $n$ such that $nr=0$ for all $r\in R$.
I really have not got much of a clue for this question and this is the best I can think of:
Call the ring $R$, consider a non-trivial element $x\in R$. Then $\{0, x\}$ is a subgroup of $R$, where $R$ is treated as an additive subgroup. Since $2$ does not divide $5$, this violates Lagrange's Theorem and so such $R$ does not exist.
Would this even be correct? Is there a way of showing this without the usage of Lagrange's Theorem? (This is a Rings past paper problem and so ideally I hope this can be solved using rings content)
 A: You have indeed shown that such a ring (or rather, the additive group of such a ring) would violate Lagrange's theorem, so it cannot exist. I think that's good enough.
Alternately, you can look at the additive group, see how many elements it has, and conclude what that group must be. It is not a group that allows for characteristic $2$.
Note that the characteristic of a ring is an inherently additive property, so not using the multiplicative structure here is completely natural. I wouldn't worry too much about that.
A: No.  By the definition of characteristic, there would be a nontrivial ring homomorphism $h:\Bbb Z/2\Bbb Z\to\mathcal R$.  Then the image $\operatorname {im}h$ is a $2$ element subring.  Now apply Lagrange to the additive groups, to get that the order of $\mathcal R$ is even, or infinite.
A: No, because $5$ is not a power of $2$, and the order of any finite dimensional vector space over the field with $2$ elements is a power of $2$, with the exponent being the dimension of the vector space.
