0) Every nonzero element of a finite ring is either a zero divisor or a unit. This is proved in Every nonzero element in a finite ring is either a unit or a zero divisor

1) If a ring R satisfies the condition that "every nonzero element is either a zero divisor or a unit", must R be finite? If not, can you please give at least two non-isomorphic counterexamples?

2) If a ring R satisfies the condition that "every nonzero element is a unit", will R be finite or infinite? If both cases are possible, can you please give at least two non-isomorphic examples in both cases?

3) Does there exist finite/infinite rings such that "every nonzero nonidentity element is a zero divisor"? If both answers are yes, can you please give at least two non-isomorphic examples in both cases?

Edit: 4) If a ring R has an element that is neither a unit nor a zero divisor, then R must be infinite. Now will R be countable or uncountable? Can you please give examples (especially if both cases are possible)?------(Ok I know $\mathbb{Z}$ is a countable example. Does there exist uncountable examples?)

P.S. I am self-learning undergraduate level mathematics. Sorry if the question is trivial or stupid.

Edit: After reading the answers, I find that doing textbook excises is still not enough to be proficient in this subject. I need to learn to think in more various ways.

• For 1, just start by looking through a few of the rings you know for counterexamples. You should find one or two very soon. For 3, a stupid (but fun!) counterexample would be for example if the set of non-unit, non-zero elements were empty... – Circonflexe Apr 30 '15 at 17:01
• Ok for 1), $\mathbb{Z}$ and $\mathbb{R}$ are non-isomorphic examples. – willhuang00 Apr 30 '15 at 17:27
• counterexample for 1: $\mathbb Z$. 3: every non-unit is in maximal ideal. – user 1 Apr 30 '15 at 17:28
• @willhuang00 $\Bbb{Z}$ is no-good for 1, because it has no zero divisors, but the only units are $\pm 1$. – A.P. Apr 30 '15 at 17:28
• Yes, you can wonder. I fear that many people (myself included), especially among the oldest users, have been bittered by the too many people resorting to MSE as a homework-solving mechanical Turk. I'm curious, though: if you did know that $2 \times 2$ matrices over $\Bbb{R}$ are a counterexample for 1), why didn't you try to see if $3 \times 3$ (or even $n \times n$) matrices worked, too? – A.P. Apr 30 '15 at 18:13

1) This is true for all fields, like $\mathbb{R}$, $\mathbb{C}$, $\mathbb{Q}$ and $\mathbb{Z}/2\mathbb{Z}$ and all finite-dimensional algebras over these fields. Finite dimensional algebras covers matrix rings, and lots more besides.

2) This is exactly the condition that defines division rings, and the zero ring which is not a division ring. A division ring with a commutative product is called a field and I gave plenty of examples of fields above.

3) $(\mathbb{Z}/2\mathbb{Z})$ and $(\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z})$ and $(\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z})$ etc. I think that these and the zero ring are the only finite rings that meet this condition.

Response to edit

4) The ring of polynomials over the real numbers. Also the ring of continuous function from the real numbers to the real numbers; the function $x \rightarrow x$ is not a unit and not a zero divisor.

• Technically the class of cardinals doesn't count since a ring must be a SET with binary operations... – Nishant Apr 30 '15 at 23:18
• @Nishant I added this – jkabrg Apr 30 '15 at 23:22
• More problematically, no cardinal besides $0$ has an additive inverse. – Nishant Apr 30 '15 at 23:27

The rings $\mathbb{Q}$, $\mathbb{R}$,$\mathbb{C}$ (these are also fields) and $\mathbb{H}$ (the quaternions) are infinite examples of 2).

For 1) and 3) you can consider the subrings of $M(2,\mathbb{R})$ formed by the matrices of the form: $$D=\begin{bmatrix}a&0\\0&b\end{bmatrix} \qquad T=\begin{bmatrix}a&c\\0&b\end{bmatrix} \qquad U=\begin{bmatrix}0&c\\0&0\end{bmatrix}$$ The matrices $D$ and $T$ form two non isomorphic rings, The matrices $U$ are a ring without unity that is not isomorphic to $D$-ring and $T$-ring.

You can see that matrices of the form $\begin{bmatrix}0&0\\0&b\end{bmatrix}$ are zero divisors in $D$-ring and in the $T$-ring and you can easely see that there are invertible elements in these rings.

The ring of matrices $U$ has no invertible elements, all its elements are zero divisors.

Finally, You can construct similar exemples in $M(n;\mathbb{K})$ for any field $\mathbb{K}$, and all are not isomorphic.

1/2) Every field has this property; as part of the definition of a field every non-zero element is a unit. $\mathbb{Q}[x]/(p_n(x))$ where $p_n(x)$ is an irreducible polynomial of degree $n$ gives an infinite non-isomorphic family. As an example of an infinite ring where nonzero elements are zero divisors or units and where both occur take $n \times n$ matrices over some infinite field ($\mathbb{Q},\mathbb{R},\mathbb{C}$ for example).*

3) Check out http://en.wikipedia.org/wiki/Boolean_ring. These have the property that all elements $x$ are idempotent meaning $x^2=x$. But then $x(x-1)=0$ so every element besides $1$ is a zero divisor.

• * to explain this example, take some matrix $x \not \in \{0,1\}$. Either it's invertible (a unit) or has a nonzero kernel. If there's a nonzero kernel pick some other matrix which which maps a basis vector into the kernel of $x$ and sends all other basis vectors to 0. Then the product of this with $x$ will be 0. – rVitale Apr 30 '15 at 17:32
• Even (arguably) more easily: the ring of $n \times n$ matrices on a field $K$ is a vector space on $K$, thus it has property 1 because it is a product of fields. All products of fields have property 1 because multiplication is defined component-wise, so the $0$ divisors are exactly the elements with $0$ in at least one component, while every other element has an inverse because each of its components has one. – A.P. Apr 30 '15 at 17:45
• As a remark for the OP our examples are not isomorphic as rings. Matrix multiplication in the sense of composition of linear maps and component-wise multiplication give different ring structures. – rVitale Apr 30 '15 at 20:12

Define a ring R such that R has underlying group structure Z/(nZ), where n is an arbitrary integer greater or equal to 2 but define multiplication as xy=0 for all x,y in R. Then R is a ring and R satisfies part (3).