For what prime numbers $p$ there exists a ring with identity and exactly $p$ invertible elements ?
REMARK It can be shown that for $p=5$ there is no such ring, so I am wondering for what values of $p$ such a ring exists.
1 Answer
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For $p=2$, $\mathbb F_3$ is an example.
Otherwise, $p$ is odd, so $-1$ is a unit of order $2$ unless it is equal to $1$, so $1=-1$, so $2=0$. We can form the ring:
$\mathbb Z[x]/(2,x^p-1)= \mathbb F_2[x]/(x^p-1)$. Any ring with exactly $p$ invertible elements admits a map from this universal ring whose image is the subring generated by the invertible elements, which must also have exactly $p$ invertible elements, so if there is any example, one example is quotient of this ring.
This ring is a product of finite fields of characteristic $2$, since $x^p-1$ has distinct roots over $\mathbb F_2$. So any quotient must be a product of finite fields of characteristic $2$. Since the unit group of a product is the product of the unit group, and the unit group, being cyclic, does not decompose as a product, it must be the unit group of a single field of characteristic $2$.
So if $p\neq 2$, it must be of the form $2^n-1$ - a Mersenne prime.
answered Aug 4, 2013 at 19:25
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$\begingroup$ Send $x$ to a generator of the group of units of $R$. $\endgroup$ Commented Aug 4, 2013 at 20:26
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$\begingroup$ @Sally: Will isn't claiming that the kernel is $(x^p - 1)$. He only needs that the map is surjective, from which it follows that $R$ is a quotient of $\mathbb{F}_2[x]/(x^p - 1)$, and then he classifies all such quotients. $\endgroup$ Commented Aug 4, 2013 at 21:53
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$\begingroup$ @QiaochuYuan $R$ is a quotient of $\mathbb{F}_2[x]/(x^p - 1)$. How do you drive that $\mathbb{F}_2[x]/(x^p - 1)$ has exactly $p$ invertible elements ? $\endgroup$– dimoCommented Aug 5, 2013 at 8:13
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$\begingroup$ @dimo: Will isn't claiming that either. The claim is that some direct factor of $\mathbb{F}_2[x](x^p - 1)$ has exactly $p$ invertible elements. $\endgroup$ Commented Aug 5, 2013 at 13:28