One way of viewing a field is just as a ring where all but one of its elements (namely $0$) is a unit. I'm looking for rings (commutative with a 1) where all but two of its elements are units. I found one fairly trivial example, $\mathbb Z/4 \mathbb Z,$ and I think I may have a proof that it's the only example of such a ring that's finite:

If $R$ is some finite ring, then considering $R$ as an additive group and applying the structure theorem for finite abelian groups, we get that $R$ is isomorphic to a direct product of groups of the form $\mathbb Z/n_i \mathbb Z,$ and the $n_i$ are coprime. I believe that this means that $R$ is also isomorphic to this as a ring (is this correct?). For each $n_i$ the size of $\left(\mathbb Z/n_i \mathbb Z\right)^*$ is $\phi(n_i).$ Let $n=|R|$. Since $\phi$ is multiplicative over coprime elements, we have $n=\Pi_i n_i$ and $|R^*|=n-2=\Pi_i \phi(n_i)$. But $\phi(n_i)\leq n_i-1,$ so obviously there can only be one factor $\mathbb Z/n_1 \mathbb Z$. Then $n_1=n$ and all we need is that $\phi(n)=n-2$. Writing $n$ as a product of primes $n=2^{r_2}3^{r_3}5^{r_5}\cdots$ we get $$\phi(n)= n-2=\phi(2^{r_2})\phi(3^{r_3}) \phi(5^{r_5})\cdots\\ =(2^{r_2}-2^{r_2-1})(3^{r_3}-3^{r_3-1})(5^{r_5}-5^{r_5-1})\cdots $$ Again, it's clear that there can only be one factor as otherwise the expression for $\phi(n)$ in the final line would be too small. Say $n=p^{r_p}$ and $p^{r_p}-2=p^{r_p}-p^{r_p-1}$ Then $p^{r_p-1}=2$. Hence $p=2$ and $r_p=2$ i.e. $n=4.$

Can anyone tell me whether this proof is correct or not.

My main question: are there any more examples with infinite rings? I feel like the answer is no because it seems very hard to have so many units, but I have no idea how to start to prove this in general. I tried local rings: If $R$ was a local ring, with unique maximal ideal $I$, then $R=I\sqcup R^*$ so I need a maximal ideal containing all but two elements which seems impossible, but again I don't know how to prove it. Most of my intuition for this being false is coming from Lagrange I think. Lagrange says that subgroups can't get too large, in particular that if $|G|=n$ then any proper subgroup $H$ must have $|H| \leq n/2$. But Lagrange doesn't work for infinite groups, so I don't know if this intuition is at all valid.

  • $\begingroup$ $R$ might be isomorphic to $\mathbb Z/4\mathbb Z\oplus\mathbb Z/8\mathbb Z,$ and $4$ and $8$ are not coprime, so your structure theorem might need some modifications. For an example, look here. $\endgroup$ – awllower Apr 20 '15 at 11:13
  • $\begingroup$ Whoops! Yes you're right, I misremembered the result!! $\endgroup$ – James Apr 20 '15 at 11:16
  • $\begingroup$ But if there are more than two direct summands, then there are more than two non-units $(1, 0, ...),$ and $(0, 1, ...),$ thus, after some modifications your proof still works. $\endgroup$ – awllower Apr 20 '15 at 11:27
  • $\begingroup$ @James : There is only a unique unital ring structure on $\mathbb Z/n\mathbb Z$ because of the distributivity law : if you require $1 \cdot 1 = 1$, then $n \cdot 1 = n$ and $n \cdot m = nm$ by induction (where the LHS is ring multiplication and the RHS is integer multiplication). However, if you don't ask $1 \cdot 1 = 1$, the multiplication $x\cdot y = 0$ for all $x,y$ in your ring is a perfectly fine non-unital ring structure. But since you are speaking of units, assuming the context of unital rings makes sense. $\endgroup$ – Patrick Da Silva Apr 20 '15 at 15:40

Below is an alternative discussion.
Suppose $R$ is such a ring, and $\{0, x\}$ is the set of non-units of $R.$ Then any ideal that contains a unit element is the unit ideal $R,$ thus $R$ has exactly $1$ non-trivial ideal, i.e. $(x).$ (This shows that $R$ is a local artinien principal ideal ring.)
Now the nilradical of $R$ is equal to $\bigcap_{\mathfrak p\in\text{Spec} R}\mathfrak p=(x),$ thus $x^n=0,$ for some $n,$ and hence $x^2$ is a non-unit that cannot equal $x,$ as that would imply that $x^n=x\not=0, \forall n\in \mathbb N-\{0\},$ i.e. $x^2=0.$
If $2x\not=0,$ then $2x$ is a non-unit that is different from $0$ and $x$ ($2x=x$ implies that $x=0$), a contradiction. Hence $2x=0.$ So $2$ is a non-unit, and hence $2=0$ or $2=x.$

The case $2=0$ gives us a new example: $\mathbb F_4:=\mathbb Z_2[x]/(x^2),$ where $\mathbb Z_2:=\mathbb Z/2\mathbb Z.$
The case $2=x$ gives us the example: $\mathbb Z/4\mathbb Z.$

Thanks to @user73985's arguments, these two examples are the only two rings with this property.

If I misunderstood at some point, or if I made some mistakes, please tell me, thanks in advance.

  • $\begingroup$ Sorry, but since I'm just an undergraduate, this is starting to go beyond my current level of knowledge, albeit slightly. What is the nilradical? Is it just the intersection of the prime ideals of a ring? How do you get that $x^n=0$ from this? Does the 'nil' in 'nilradical' mean nilpotent? $\endgroup$ – James Apr 20 '15 at 16:41
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    $\begingroup$ Elementary: the only proper ideal is in fact $\{0,x\}$. Then $x+x=0$, $x^2=0$, and so on. $\endgroup$ – user26857 Apr 20 '15 at 18:00
  • $\begingroup$ @James Yes, the nilradical is just the intersection of the prime ideals, and it also equals $\{x\in R| x^n=0 \text{ for some }n\}.$ (You can try to prove this or find this statement here) And yes, the "nil" means "nilpotent." $\endgroup$ – awllower Apr 20 '15 at 23:46
  • $\begingroup$ @user26857 But how do you know that $x^2=0$ instead of $x^2=x?$ $\endgroup$ – awllower Apr 20 '15 at 23:47
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    $\begingroup$ If $x^2=x$ then $x(x-1)=0$. But $x-1\notin\{0,x\}$, so it's invertible, etc. (Maybe you want to say "how do you know that $x^2=0$".) $\endgroup$ – user26857 Apr 20 '15 at 23:49

If $a \neq 1$ is a unit, and $b \neq 0$ is not, then $ab$ is not a unit either; further, we cannot have $ab = 0$, thus $ab = b$. But then, since $a \neq 1$, $a - 1$ is a zero divisor, so not a unit, so $a - 1 = b$. So there can be at most one unit which is not equal to $1$. Thus if $R$ is a ring with precisely two nonunits then $3 \leq |R| \leq 4$. When combined with awllower's answer we see that the two examples in that answer are the only two, even including infinite rings.

  • $\begingroup$ Thanks for this simple and ingenious argument, which clarifies the matter a lot. :) $\endgroup$ – awllower Apr 20 '15 at 23:49

A suggestion: Look for a commutative inverse semigroup with two identities. Then Take some Z_n and form the semigroup ring.


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