Past exam of the rings and modules course of the University of Oxford / 2023 / 1

Multiplication in a direct product of rings is coordinatewise. Thus $(r_1,\dots ,r_n)$ has a two-sided inverse exactly when each $r_j$ has a two-sided inverse in $R_j$, and thenConsequently its unit group is the direct product of groupsThe identification preserves multiplication because both sides use coordinatewise multiplication. This is the unit group of a direct product of rings formula.

Write $T={\prod }_{j=1}^n{R}_j$, and let ${\rho }_j:T\to R_j$ be the coordinate ring homomorphism. Set $I_j={\rho }_j(I)$. Each $I_j$ is an ideal: it is an additive subgroup, and multiplying a lift in $I$ on either side by a tuple with prescribed $j$th coordinate proves closure under left and right multiplication by $R_j$.

Certainly $I\subseteq {\prod }_j{I}_j$. Conversely, let $a=(a_1,\dots ,a_n)\in {\prod }_j{I}_j$. For each $j$, choose $b^{(j)}\in I$ with $j$th coordinate $a_j$, and let $e_j$ have $1_{R_j}$ in coordinate $j$ and zero elsewhere. Since $I$ is an ideal, $e_j{b}^{(j)}\in I$, andThereforeThe finite sum is essential to this proof of ideals of a finite product of rings.

The element $-1_R$ lies in the unit group and satisfies $(-1_R{)}^2=1_R$. Its order of a group element divides two and, by Lagrange's theorem, divides the odd number $|U(R)|$. Its order is therefore one, so $-1_R=1_R$ and $2\cdot 1_R=0$.

It follows thatis a well-defined unital ring homomorphism. For a nonzero ring, this says its characteristic of a ring is two. If the zero ring is permitted, the same map still exists. This is the odd unit group forces characteristic two argument, with that harmless exception to the characteristic wording.

Since the irreducible element $x$ is not a unit, the principal ideal $(x)$ is proper. Suppose $(x)\subseteq J\subseteq R$. As $R$ is a principal ideal domain, write $J=(d)$. Then $x=dr$ for some $r\in R$. The definition of an irreducible element says that $d$ or $r$ is a unit.

If $d$ is a unit, then $J=R$. If $r$ is a unit, then $d$ and $x$ are associate elements, so $J=(x)$. Thus $(x)$ is a maximal ideal. By the permitted criterion that a quotient ring by a maximal ideal is a field,

Every multiple of $x_1\cdots x_r$ is a multiple of each $x_j$, giving one inclusion. For the reverse inclusion, a principal ideal domain is a unique factorization domain, so each irreducible element $x_j$ is a prime element.

Let $a\in {\bigcap }_{j=1}^r(x_j)$. The case $a=0$ is immediate. Otherwise, write $a=x_{1}b$. Since $x_2\mid a$ and the prime element $x_2$ cannot divide the non-associate irreducible element $x_1$, it divides $b$. Write $a=x_1{x}_2{b}_2$, and repeat. At stage $j$, the prime element $x_j$ divides none of $x_1,\dots ,x_{j-1}$, so it divides the remaining factor. Hence $x_1\cdots x_r\mid a$, proving

The polynomial ring ${\mathbb{F}}_2[X]$ is a euclidean domain under polynomial degree, hence a principal ideal domain. By part (b)(i), eachis a field. By polynomial division, every residue class has exactly one representative of polynomial degree less than $n_j$. Thus $1,\overline{X},\dots ,{\overline{X}}^{n_j-1}$ form a basis over ${\mathbb{F}}_2$, and $|K_j|=2^{n_j}$.

For $j\ne k$, the ideals $(f_j)$ and $(f_k)$ are distinct maximal ideals, so they are comaximal ideals: their sum is an ideal properly containing $(f_j)$ and must be the whole ring. The Chinese remainder theorem for ideals and part (b)(ii) therefore give the ring isomorphismIn particular, this is a product of finite fields of the required sizes. The map preserves multiplication as well as addition and is surjective by the Chinese remainder theorem for ideals.

Choose a generator $u$ of the cyclic group $U(R)$, so $u^p=1$. Part (a)(iii) gives a unital ring homomorphism ${\mathbb{F}}_2\to R$. Its image lies in the centre of $R$, so even if $R$ is not a commutative ring, evaluating a polynomial at $u$ defines a ring homomorphismBecause $\varphi (X^p-1)=0$, it factors through the quotient ring to giveIts image is the subring $S={\mathbb{F}}_2[u]$. Every unit of $S$ is a unit of $R$, because an inverse in $S$ also lies in $R$. Conversely, every member of $U(R)$ is $u^j$ for some $j$, and both $u^j$ and its inverse $u^{p-j}$ lie in $S$. HenceThe inverse argument matters: merely belonging to a subring would not by itself make an ambient unit invertible there.

Using the assumed square-free polynomial property, factorinto distinct monic polynomials that are irreducible polynomials, with $n_j=\mathrm{deg}{f}_j$. Part (b)(iii) identifies $T={\mathbb{F}}_2[X]/(X^p-1)$ with ${\prod }_j{K}_j$, where $|K_j|=2^{n_j}$.

By the first isomorphism theorem for rings, the image $S$ from part (c)(i) is $T/J$, where $J$ is the kernel of a ring homomorphism. By part (a)(ii), $J={\prod }_j{J}_j$ for ideals $J_j\subseteq K_j$. A field has only the zero ideal and the whole field as ideals, so each $K_j/J_j$ is either $K_j$ or the zero ring. Consequently, for a subset $D\subseteq \{1,\dots ,r\}$,This is the quotient of a finite product of fields principle. The subset is nonempty because $|U(S)|=p>2$, whereas the zero ring has only one unit. By part (a)(i),Each factor is a positive integer. Since $p$ is a prime number, exactly one factor is $p$ and all remaining factors are one. Thusso $p$ is a Mersenne prime. If $n=ab$ with $a,b>1$, then $2^a-1$ would be a proper divisor of $2^{ab}-1$; thus $n$ is itself a prime number, although only the displayed form is needed.

The assumed square-free polynomial property can also be checked directly: in characteristic of a ring two, the formal derivative of $X^p-1$ is $X^{p-1}$ because $p$ is odd. These two polynomials have greatest common divisor one, since $X$ does not divide $X^p-1$, so no repeated irreducible polynomial factor occurs.