Construct a ring containing $16$ element where EVERY element $r\neq 0$,$1$ is a zero divisor

Question

This is a practice question to prepare me for my final exam in abstract algebra.

Construct a ring containing $16$ elements where EVERY element $r\neq 0$,$1$ is a zero divisor

I'm having a hard time starting this out. Especially, how can I ensure a ring I create has 16 elements. My first thought it to create a quotient ring R/I.

Since we're restricting our ring to have exactly 16 elements, it, of course, has to be finite.

I'm not the best at abstract algebra so this may not be even close, but here is one I thought would work:

Let $R = \{a \in \mathbb{Z}_{32} |$ a is even, $a > 0\}$ and let $I = 2$.

So then $R = \{2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32\}$ which has 16 elements, and since we're modding out by 2, they're all zero divisors.

Any help would be greatly appreciated.

Answer 1

I suggest $(\Bbb{Z}/2\Bbb{Z})^4$. We can identify this ring with the collection of subsets of $X=\{1,2,3,4\}$ by using indicator functions. Then addition is symmetric difference and multiplication is intersection. The null set is $0$, and $X$ is the identity. For any subset $S\subseteq X$, we then have $S\cdot S^C=0$. Thus either $S=\varnothing=0$, $S=X=1$, or $S$ is a zero divisor.

Edit

The ring I suggest is the following. It is the set of 4-tuples of whose entries are either $0$ or $1$.

Some example elements of this ring are $(0,0,0,0)$, $(0,1,1,0)$, $(1,0,1,1)$, and $(1,1,1,1)$. Since there are two choices for each entry, there are 16 total elements.

The ring operations are pointwise addition and multiplication where $1+1=0$ and otherwise addition and multiplication behave normally (i.e. $0\cdot 1=0$, $1\cdot 1=1$, $0\cdot 0=0$, and $0+0=0$, and $1+0=1$). As examples, $$(0,1,1,0)+(1,0,1,1)=(0+1,1+0,1+1,0+1)=(1,1,0,1),$$ and $$(0,1,1,0)\cdot (1,0,1,1)=(0\cdot 1,1\cdot 0, 1\cdot 1,0\cdot 1)=(0,0,1,0).$$

You can check that $(0,0,0,0)$ is the additive identity and $(1,1,1,1)$ is the multiplicative identity.

Then if $(a,b,c,d)$ is a tuple that is not one of these, if we define a new tuple $(\bar{a},\bar{b},\bar{c},\bar{d})$ by making $\bar{0}=1$ and $\bar{1}=0$, then $0\cdot \bar{0}=0\cdot 1=0$ and $1\cdot \bar{1}=1\cdot 0=0$, so $$(a,b,c,d)\cdot (\bar{a},\bar{b},\bar{c},\bar{d})= (a\bar{a},b\bar{b},c\bar{c},d\bar{d}) =(0,0,0,0).$$ Thus all nonzero elements are zero divisors in this ring.

Answer 2

In a finite commutative ring $R$, every noninvertible element is a zero divisor.

Indeed, if $r$ is not invertible, the map $x\mapsto rx$ is not surjective (because $1$ is missed), so it is not injective. Being a group homomorphism with respect to addition, its kernel is nonzero, which proves the claim.

Your ring must have a single invertible element, hence its characteristic needs to be $2$: indeed, $-1$ is invertible, so we need $-1=1$.

Thus $R$ has to be a four dimensional vector space over $\mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}$ and it's quite natural to look for a ring structure on $\mathbb{F}_2^4$.

One possibility is the product ring, which indeed satisfies the requirement, because an element of the form $a=(a_1,a_2,a_3,a_4)$ where one of the entries is zero is surely a zero divisor: if $e_i$ denotes the $4$-tuple with $1$ at the $i$-th place an zero elsewhere, then $a_i=0$ implies $ae_i=0$.