Show that $U(A)\cup\{0\}$ is a field and $|A|\equiv1\pmod p$

Question

Let $p$ be a prime number and $A$ a finite ring in which the group $U(A)$ of the invertible elements has order $p$. If there is an element $a\in U(A)$ such that $1-a\in U(A)$, show that $U(A)\cup\{0\}$ is a field and $|A|\equiv1\mod p$.

The first seems pretty trivial as $U(A)$ is a field as $U(A)$ is a group with the multiplication. But the second reminds me of Wilson`s theorem, but I get nowhere. Any help?

Since all finite fields have order $q^k$ for some prime $q$, this forces $p+1=q^k$, so either $p=2$ or $q=2$ (and $p$ is a Mersenne prime). — Robert Shore, Mar 11 '19 at 20:48
Re: the "pretty trivial" comment: is it clear that $U(A)\cup{0}$ is closed under addition? — Greg Martin, Mar 11 '19 at 22:21
Agree with @GregMartin. Proving that $U(A)\cup{0}$ is a field with respect to the addition and multiplication operations of $A$ is the hard part. — Jyrki Lahtonen, Mar 12 '19 at 06:26
@Jyrki See also A finite ring is a field if its units $\cup\ {0}$ comprise a field of characteristic $\ne 2$ $\ \ $ — Bill Dubuque, Mar 15 '19 at 20:50
A nice find @BillDubuque, and surely useful reading (even though not a duplicate). The vibe I get from this question is that it is designed to be done with relatively elementary tools (e.g. those in my answer). IIRC Romanians have contests for students at approximately this level, and this would fit. — Jyrki Lahtonen, Mar 15 '19 at 21:07
As observed by Robert Shore, here the field is most likely to be of characteristic two. — Jyrki Lahtonen, Mar 15 '19 at 21:26

Jyrki Lahtonen · Answer 1 · 2019-03-15T21:25:05.697

Let $a$ be the unit with the property that $1-a$ is also a unit.

Because $U(A)$ has prime order, it must be a cyclic group. Hence every non-identity element is a generator. In particular $a$ is a generator of $U(A)$.
Let $a^m\in U(A)\setminus\{1\}$ be arbitrary. Because $a^m$ is then also a generator of $U(A)$ there exists an integer $n$ such that $(a^m)^n=a$.
With $m$ and $n$ as in the previous bullet, the polynomial factorization $$1-x^n=(1-x)(1+x+x^2+\cdots+x^{n-1})$$ applied to $x=a^m$ implies that $1-a^m$ is a factor of $1-(a^m)^n=1-a$. As a factor of a unit $1-a^m$ is a unit itself. Observe that the inverse of $1-a$ is a power of $a$. All the integral linear combinations of powers of $a$ commute with each other, so any one-sided inverse found here is automatically a two-sided inverse, hence an element of $U(A)$.
Let $k,\ell$ be distinct integers such that $a^k\neq a^\ell$. Then $a^k-a^\ell=a^\ell(1-a^{\ell-k})$ is a unit as a product of units.
By the result of the preceding bullet the set $k=U(A)\cup\{0\}$ is closed under differences. By the familiar subgroup criterion it is then a subgroup of the additive group of $A$. It is trivially closed under products, so it is a subring. The non-zero elements of $k$ form a commutative group implying that $k$ is a field.
$A$ is a finite dimensional vector space over $k$. Therefore $|A|=|k|^n$. As $|k|\equiv1\pmod p$ we also have $|A|=|k|^n\equiv1\pmod p$.

Observe that I am not assuming $A$ to be commutative. Powers of $a$ automatically commute with each other, and so do their linear combinations with integer coefficients. — Jyrki Lahtonen, Mar 12 '19 at 07:24
I think that $a$ is a generator because $a^2-a^3$ is in $U(A)$ too. Could you give more indications? — , Mar 15 '19 at 19:29
@SeptimiuCristian What does Lagrange's theorem tell you about generators of $U(A)$? — Jyrki Lahtonen, Mar 15 '19 at 20:31

Show that $U(A)\cup\{0\}$ is a field and $|A|\equiv1\pmod p$

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