Past exam of the rings and modules course of the University of Oxford / 2021 / 1 / b

All principal ideals in this part are taken inside the subring $T$, rather than inside $\mathbb{R}[X]$. First, $T$ is a subring: sums and additive inverses retain zero coefficient of $X$, and the coefficient of $X$ in a product $uv$ is $u(0)[X]v+v(0)[X]u=0$. It contains $1$.

The principal ideal $I=(X^3{)}_T=X^{3}T$ is preciselyIndeed, multiplying an element of $T$ by $X^3$ gives a possible term of polynomial degree $3$ and terms of polynomial degrees at least $5$, with no $X^4$ term. Conversely, dividing any displayed polynomial by $X^3$ gives a polynomial with no $X$ term, hence an element of $T$.

Given $p\in T$, let $a,b,c$ be its coefficients of $1,X^2,X^4$, respectively. Subtracting these terms leaves an element of $I$, soFor uniqueness, if two such representatives differ by an element of $I$, their difference has possible monomials only $1,X^2,X^4$, whereas every element of $I$ has zero coefficients at all three. Comparing coefficients gives $a=a^{\prime }$, $b=b^{\prime }$, $c=c^{\prime }$.

In particular, $X^4\notin I$. Treating $I$ as the principal ideal of $\mathbb{R}[X]$ instead would incorrectly discard this surviving term.

Set $I=(X^3{)}_T$ and $J=(Y^3{)}_{\mathbb{R}[Y]}$. Part (b)(i) gives a unique representative $a+b{X}^2+c{X}^4$ for each element of the quotient ring $T/I$. By polynomial division, each element of the quotient ring $\mathbb{R}[Y]/J$ likewise has a unique representative $a+bY+c{Y}^2$. Thus the proposed map $\Phi$ is well-defined and a bijection.

It preserves addition and $1$. To verify multiplication, all terms of polynomial degree at least $5$ lie in $I$. HenceIn the other quotient ring, $Y^3=0$, giving exactly the same coefficient formula:Thereforeis a ring isomorphism. Both sides are examples of a truncated polynomial ring; the generator on the left is the class of $X^2$, whose square is nonzero and whose cube vanishes.

No such property exists. The ring isomorphism in part (b)(ii) identifies quotient rings by elements having different irreducible element behaviour.

The element $X^3$ is an irreducible element of $T$. It is nonzero and a nonunit. Every nonzero nonunit of $T$ has polynomial degree at least $2$: the only possible polynomials of smaller polynomial degree are constant polynomials, and every nonzero constant polynomial is a unit. If $X^3=uv$ with $u,v\in T$ both nonunits, additivity of polynomial degree in the integral domain $\mathbb{R}[X]$ would givewhich is impossible. Thus one factor must be a unit.

In $\mathbb{R}[Y]$, however, $Y^3=Y\cdot Y^2$ is a product of two nonunits, so $Y^3$ is not an irreducible element. If a ring isomorphism invariant $P$ characterized irreducible elements by the indicated quotient ring, the irreducible element $X^3\in T$ would imply that $T/(X^3{)}_T$ has $P$. Its isomorphic quotient ring $\mathbb{R}[Y]/(Y^3)$ would then have $P$, incorrectly making $Y^3\in \mathbb{R}[Y]$ an irreducible element.

Consequently,

The required property isFor a nonzero element $x$ of an integral domain $R$, this follows fromIndeed, $\overline{a}\overline{b}=0$ in the quotient ring exactly when $x\mid ab$. The prime element condition says this forces $x\mid a$ or $x\mid b$, exactly the absence of zero divisors in the quotient ring. The nonunit condition makes the quotient ring nonzero. Being an integral domain is preserved by every ring isomorphism.

With the usual definition of a prime element, $x\ne 0$ is understood. The prime ideal formulation works for every ideal of any commutative ring, including the zero ideal; it avoids this convention about elements.