Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 3 / 10E (Groups, Rings and Modules)

Suppose $I$ is prime. Ifthen $ab\in I$, so $a\in I$ or $b\in I$. Thus one factor is zero and $R/I$ is an integral domain.

Conversely, if $R/I$ is an integral domain and $ab\in I$, thenso $a+I=0+I$ or $b+I=0+I$. Therefore $a\in I$ or $b\in I$, and $I$ is prime. This proves the prime ideal quotient criterion.

In a Boolean ring, $(r+1{)}^2=r+1$. Using $r^2=r$ and $1^2=1$ givesso

If $R$ is also a nonzero integral domain, thenforces every $r$ to equal $0$ or $1$. Hence $R\cong {\mathbb{F}}_2$.

If $I$ is prime in a Boolean ring, then $R/I$ is a nonzero Boolean integral domain by part (i), and hence is ${\mathbb{F}}_2$. Since the quotient is a field,This is the prime ideals of a Boolean ring are maximal property.

Reduce coefficients modulo $I$:This is a surjective ring homomorphism and its kernel consists exactly of polynomials all of whose coefficients lie in $I$, namely $I[X]$. The first isomorphism theorem gives the coefficientwise quotient of a polynomial ringIf $I$ is prime, then $R/I$ is an integral domain, so $(R/I)[X]$ is an integral domain. The quotient criterion therefore shows that $I[X]$ is prime in $R[X]$.

Take $R=\mathbb{Z}$ and $I=(2)$. The ideal $I$ is maximal, butis not a field, since $X$ is nonzero and not invertible. Therefore

Let $G={\mathbb{F}}_p^{\times }$, and let $m$ be the least common multiple of the orders of its elements. For every prime power $q^a$ dividing $m$, some element of $G$ has order divisible by $q^a$, and a suitable power of it has order exactly $q^a$. Multiplying these elements over the distinct primes produces, because their orders are coprime, an element of order $m$.

Every element of $G$ is a root of $X^m-1$. A nonzero polynomial of degree $m$ over a field has at most $m$ roots, so $p-1\le m$. On the other hand, every element order divides $p-1$ by Lagrange's theorem, so $m\le p-1$. Hence $m=p-1$, and

The squaring homomorphism has kernel $\{1,-1\}$ because $p$ is odd. Its image $H$ therefore has order $(p-1)/2$ and index two. If $p=3$, then $3=0$ is already a square. Otherwise $2,3\in G$. In the two-element quotient $G/H$, either $2H=H$, or $3H=H$, or both are the nontrivial coset, in which case $6H=H$. Thus one of $2,3,6$ is a square modulo $p$.

Finally,Whichever of $2,3,6$ is a square supplies a root, soThis is the index-two square-class argument for three related residues.