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

For a ring $R$, its ring characteristic is the least positive integer $n$ with $n{1}_R=0$, or zero if no positive such integer exists. Equivalently it is the least positive integer satisfying $nr=0$ for every $r\in R$, because $nr=(n{1}_R)r$. This additive formulation also defines the characteristic of a rng when an identity is not assumed.

Let $R$ be an integral domain. Since $1_R\ne 0$, its characteristic is not one. If its positive characteristic were a composite integer $n=ab$ with $1<a,b<n$, minimality would give $a{1}_R\ne 0$ and $b{1}_R\ne 0$, butThis contradicts the absence of zero divisors. ThereforeThis is the characteristic of an integral domain theorem. If an integral domain is defined without requiring an identity, the same conclusion follows from the additive definition: for a hypothetical composite characteristic $n=ab$, choose $r$ with $ar\ne 0$, which exists by minimality of $n$. Then $(ar)(bs)=n(rs)=0$ for every $s$, forcing $bs=0$ for every $s$ and contradicting minimality again.

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.