Past exam of the rings and modules course of the University of Oxford / 2017 / 1 / b

Take $R=\mathbb{Z}/4\mathbb{Z}$, a quotient ring with four residue classes. The class $\overline{2}$ is nonzero but $\overline{2}\overline{2}=\overline{0}$. Thus $R$ has a zero divisor and is not an integral domain.

No such ring exists. Characteristic two gives $r+r=0$ for every $r$. Choose a nonzero element $a$, which exists in a five-element set. Translation $r\mapsto r+a$ is an involution without fixed points: applying it twice gives $r$, while $r+a=r$ would force $a=0$ by additive cancellation. Its disjoint pairs would partition all five elements, which is impossible because five is odd.

This argument applies even to a nonunital ring. With the unital convention, it is also the case $n=2$ of finite ring characteristic divides cardinality.

No such ring exists: every finite integral domain is a field. Here is the proof. For $a\ne 0$, the map $L_a:R\to R$ defined by $L_a(x)=ax$ is injective, since $ax=ay$ gives $a(x-y)=0$ and therefore $x=y$. An injective self-map of a finite set is surjective, so some $x$ satisfies $ax=1$. Since an integral domain is commutative, $xa=1$ as well. Thus every nonzero element has a multiplicative inverse, making $R$ a field.

The conclusion still holds if an identity was not initially required. Surjectivity first gives an element $e$ with $ae=a$. Then $a(ex-x)=0$ implies $ex=x$ for every $x$, and commutativity makes $e$ a two-sided identity. Applying the preceding argument with $e$ supplies all inverses.

A suitable field with four elements isWrite $\alpha$ for the class of $t$. Monic polynomial division over a ring gives a unique representative $a+b\alpha$, with $a,b\in {\mathbb{F}}_2$: existence follows by division, and uniqueness follows because a nonzero multiple of the degree-two modulus cannot have degree less than two. The four elements are therefore $0,1,\alpha ,\alpha +1$.

The defining relation gives ${\alpha }^2=\alpha +1$, and hence $\alpha (\alpha +1)=1$. Thus the nonzero elements have inversesThe quotient ring is commutative and every nonzero element is a unit, so it is a field. This verifies the field property directly, without quoting a theorem about irreducible-polynomial quotients.

No field has six elements. A finite field has positive characteristic: the sequence $0,1,2\cdot 1,\dots$ must repeat. By part (a), this characteristic is a prime number $p$. Translation by $1$ partitions the field into cycles of length $p$, since returning to a starting point after $k$ steps is equivalent to $k1=0$. Thus $p$ divides six and $p=2$ or $p=3$.

The multiples of $1$ form a copy of the prime field ${\mathbb{F}}_p$. For completeness, the inverse of a nonzero residue $a$ lies among these multiples: the integer Bezout identity $ab+cp=1$ gives $(a1)(b1)=1$. The entire field is a vector space over this subfield. A maximal linearly independent set is finite and spans, so it is a basis of some positive size $d$. Unique basis coordinates give exactly $p^d$ elements. This proves finite field cardinality is a prime power here. But six is neither a power of two nor a power of three, a contradiction.

Under the unital ring convention used elsewhere in this wiki, no such pair exists. In a two-element ring the elements are $0$ and $1$. Additive cancellation forces $1+1=0$, and the identity axiom forces $1\cdot 1=1$. The distributive property forces products involving zero to vanish. Every operation is therefore determined, giving the prime field ${\mathbb{F}}_2$ as the unique possibility up to ring isomorphism.

If the word ring allows a nonunital ring, there are instead two non-isomorphic examples: ${\mathbb{F}}_2$ and the zero-multiplication rng on the additive group $\mathbb{Z}/2\mathbb{Z}$. In the second example the unique nonzero element $u$ satisfies $u+u=0$ and $u^2=0$; every product is zero, so associativity and both distributive laws hold. An isomorphism would have to send $1$ to $u$ because it preserves zero and is bijective, but then it would send $1^2=1$ to both $u^2=0$ and $u$, which is impossible. These exhaust the rngs with two elements, since $u^2$ can only be $0$ or $u$. Specifying the identity convention is essential for this part.