Past exam of the algebra 2 course of the University of Oxford / 2014 / 1

A prime ideal is a proper ideal $I\ne R$ such that $ab\in I$ implies $a\in I$ or $b\in I$, for every $a,b\in R$.

Suppose first that $I$ is prime. Since $I$ is proper, $1+I\ne 0+I$ in the quotient ring. If $(a+I)(b+I)=0+I$, then $ab\in I$, so primeness gives $a+I=0+I$ or $b+I=0+I$. Thus the commutative quotient ring has a nonzero identity and no nonzero zero divisors, hence is an integral domain.

Conversely, if $R/I$ is an integral domain, its identity is nonzero, so $I\ne R$. If $ab\in I$, then $(a+I)(b+I)=0$ in $R/I$. Absence of nonzero zero divisors implies $a+I=0$ or $b+I=0$, namely $a\in I$ or $b\in I$. ThereforeThe properness condition matters: taking $I=R$ would produce the zero ring, which is not an integral domain.

Suppose, for a contradiction, that the monic irreducible polynomials in the polynomial ring $k[t]$ form a finite list $p_1,\dots ,p_r$. The list is nonempty because $t$ itself is monic and irreducible. PutThis has positive degree, so it is a nonzero nonunit. Every such polynomial over a field has an irreducible polynomial as a factor: repeatedly factor a reducible polynomial, with the positive degrees strictly decreasing until an irreducible factor is reached. Dividing that factor by its leading coefficient makes it monic; call it $q$.

By the assumed completeness of the list, $q=p_j$ for some $j$. But then $p_j$ divides both $F$ and $p_1\cdots p_r$, hence their difference $1$. A positive-degree polynomial cannot divide $1$. This contradiction proves there are infinitely many monic irreducible polynomials over every field. This is the polynomial version of Euclid's theorem; it does not require the field itself to be infinite.

Take a reduced fraction $a/b\in R$, with $b\ne 0$. Since $\gcd (a,b)=1$, Bézout's identity supplies integers $u,v$ with $ua+vb=1$. As the subring $R$ contains all integers,Thus a denominator appearing in a reduced fraction in $R$ is automatically a unit of $R$.

Now let $I\cap \mathbb{Z}=d\mathbb{Z}$. Because $d\in I$, the principal ideal $dR$ is contained in $I$. For the reverse inclusion, take $x\in I$ and write $x=a/b$ in lowest terms. Multiplying by $b\in R$ gives $a=bx\in I\cap \mathbb{Z}=d\mathbb{Z}$, so $a=dc$ for some integer $c$. The preceding result gives $1/b\in R$, whenceThereforeThis includes $d=0$: the same argument forces every $x\in I$ to be zero. This is the ideals of subrings of the rational numbers property; every such ring is a principal ideal domain.

In this localization of a ring, a nonzero fraction $a/b$ with odd $b$ is a unit exactly when $a$ is odd: its reciprocal then has odd denominator; if $a$ is even, an inverse would equate an even integer with an odd one after clearing denominators. Consequently every nonzero element has a unique formHere $r$ is the exponent of $2$ in the numerator of the reduced fraction, its nonnegative 2-adic valuation.

Every nonzero ideal has the form $dR$ by part (c). Write the nonzero integer $d=2^{r}v$ with $v$ odd; since $v$ is a unit, $dR=2^{r}R$. Hence the nonzero proper ideals are exactlyThe zero ideal and $R=2^{0}R$ complete the full list. These ideals are distinct: $2^r$ does not belong to $2^{r+1}R$, since $1/2\notin R$.

The reduction map $a/b\mapsto a\mathrm{mod}2$ is a well-defined surjective ring homomorphism $R\to {\mathbb{F}}_2$ with kernel of a ring homomorphism $2R$; odd denominators reduce to $1$. Thus $R/2R\cong {\mathbb{F}}_2$ is a field, so $2R$ is a prime ideal and $2$ is a prime element. Every nonzero nonunit is associated to $2^r$ with $r\ge 1$. If $r\ge 2$, the factorization $2^r=2\cdot 2^{r-1}$ is into two nonunits, so it is not an irreducible element and therefore not a prime element. We concludeThe failed step in the argument of part (b) is the guarantee that the product plus one is a nonunit and hence has an irreducible factor. Here every prime element is an even fraction, so a product of prime elements plus $1$ is an odd fraction, therefore a unit. Already $2+1=3$ has no prime divisor in this ring. In $k[t]$, positive degree prevented that failure.