Past exam of the algebra 2 course of the University of Oxford / 2015 / 1

If $a=ub$ with $u$ a unit in a ring, then $a$ lies in the principal ideal generated by $b$. This gives $(a)\subseteq (b)$. Since $u$ has a multiplicative inverse, $b=u^{-1}a$, giving the reverse inclusion. ThereforeThus associate elements always generate the same principal ideal; the associator ring condition asks whether equality of these ideals also forces the generators to differ by a unit.

Suppose $R$ is an integral domain and $(a)=(b)$. If $a=0$, then $(b)=0$, so $b=0$ and $a=1\cdot b$.

Otherwise membership in the equal principal ideals supplies $r,s\in R$ with $a=rb$ and $b=sa$. Substituting gives $a=rsa$, or $(1-rs)a=0$. There are no nonzero zero divisors in an integral domain, and $a\ne 0$, so $rs=1$. Hence $r$ is a unit with inverse $s$, and $a=rb$ makes $a$ and $b$ associate elements. We have proved

The ideal correspondence theorem identifies the ideals of $\mathbb{Z}/4\mathbb{Z}$ with the ideals of $\mathbb{Z}$ containing $4\mathbb{Z}$. The possible positive generators divide $4$, giving exactlyThe units are $\overline{1}$ and $\overline{3}$. The generators of the three displayed ideals are, respectively,Within the last pair, multiplication by the unit $\overline{3}$ exchanges the generators. The other two sets are singletons, whose members are associate elements to themselves by multiplication by $\overline{1}$. Thus equal principal ideals always have associate generators: this is an associator ring.

However, $\overline{2}\ne \overline{0}$ and ${\overline{2}}^2=\overline{0}$, so $\overline{2}$ is a nonzero zero divisor. Therefore

For two factors, the source's $R_1\oplus R_2$ is the direct product of rings with componentwise operations. Its principal ideals satisfyOne inclusion follows by projection to the factors. For the reverse inclusion, multiplication by $(1,0)$ and $(0,1)$ puts $(a_1,0)$ and $(0,a_2)$ in the generated ideal; taking ideal combinations separately produces every pair in $(a_1)\times (a_2)$.

Suppose $(a_1,a_2)$ and $(b_1,b_2)$ generate the same principal ideal. Projecting to the factors gives $(a_i)=(b_i)$ for $i=1,2$. Since each factor is an associator ring, there are units $u_i\in R_i$ with $a_i=u_i{b}_i$. The pair $(u_1,u_2)$ is a unit of the direct product of rings, with inverse $(u_1^{-1},u_2^{-1})$, andHenceRepeating this argument proves the same assertion for every finite direct product of rings.

Put $A=R/(p^k)$. A principal ideal domain is a unique factorization domain, so every nonzero element of this quotient ring can be writtenIndeed, remove the largest power of the prime element $p$ from a representative that is not divisible by $p^k$. Since $s$ and $p^k$ have greatest common divisor $1$ up to a unit, Bézout's identity in the principal ideal domain gives $xs+y{p}^k=1$. Thus $\overline{s}$ is a unit in $A$.

Consequently every nonzero principal ideal in $A$ is generated by one of $1,\overline{p},\dots ,{\overline{p}}^{k-1}$. Include the zero ideal by assigning it exponent $e=k$. These ideals are distinct: their inverse images in $R$ are $(1),(p),\dots ,(p^k)$, and equality of two different powers would, by cancellation in the integral domain, make the nonunit $p$ a unit.

Now suppose $(\overline{a})=(\overline{b})$. If this is the zero ideal, both generators are zero and are associate elements. Otherwise their exponents agree, so writewith $\overline{s},\overline{t}$ both units. Thenand the multiplier is a unit. ThusThe exponent detects the principal ideal, while its remaining factor is invertible; this is why zero divisors cause no obstruction here.

By the first isomorphism theorem for rings, the given surjection identifies $S$ with $R/\ker \varphi$. Since $R$ is a principal ideal domain, write $\ker \varphi =(a)$.

If $a=0$, the quotient ring is $R$, an integral domain, so part (a)(ii) applies. If $a$ is a unit, the quotient ring is the zero ring; its unique element is associate to itself by multiplication by its identity, so it too is an associator ring.

In the remaining case, unique factorization giveswhere $\epsilon$ is a unit, the $p_i$ are pairwise nonassociate prime elements, and $k_i\ge 1$. The principal ideals $(p_i^{k_i})$ are pairwise comaximal: different prime factors have greatest common divisor $1$, and Bézout's identity gives a linear combination equal to $1$. The Chinese remainder theorem therefore gives a ring isomorphismEvery factor is an associator ring by part (c), and their finite direct product of rings is an associator ring by part (b). A ring isomorphism preserves units and principal ideals, so the property transfers to $S$. Hence