Past exam of the algebra 2 course of the University of Oxford / 2014 (A3, 4322)

Source: Mathematics Part A release, member FOI-20210708-1 - Data/4322.pdf, dated 26 June 2014. All thirteen source leaf parts are answered below.

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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.

For a nonzero integer-coefficient polynomial $F$, its polynomial content $c(F)$ is the positive greatest common divisor of all its coefficients; one may set $c(0)=0$. A primitive polynomial has content one.

For $f\in \mathbb{Q}[t]\setminus \{0\}$, choose a positive integer $m$ clearing its denominators, so $F=mf\in \mathbb{Z}[t]$. Put $d=c(F)>0$. Thenwhere $f_1=F/d$ is an integer-coefficient primitive polynomial. This proves existence with $\alpha =d/m>0$.

For uniqueness, suppose $\alpha f_1=\beta g_1$, where both integer-coefficient polynomials are primitive and $\alpha ,\beta >0$. Write $\alpha /\beta =a/b$ in lowest terms with $a,b$ positive. Then $g_1=(a/b)f_1$ has integer coefficients. Coprimality forces $b$ to divide every coefficient of $f_1$, hence $b=1$. Now $g_1=a{f}_1$ has content $a$, so $a=1$. Therefore $\alpha =\beta$ and $f_1=g_1$.

Define the rational polynomial content by this unique positive multiplier. If $f=\alpha f_1$ and $g=\beta g_1$, Gauss lemma for polynomials says that $f_1{g}_1$ is primitive. ThusThe uniqueness argument also shows that the construction is independent of the chosen denominator-clearing integer $m$.

Write a rational root of a polynomial as $\gamma =a/b$ in lowest terms with $b>0$. Substituting into the monic integer-coefficient polynomial and multiplying by $b^d$ givesEvery term other than $a^d$ is divisible by $b$, so $b\mid a^d$. But $\gcd (a,b)=1$ implies $\gcd (a^d,b)=1$, and hence $b=1$. ThusThis is the monic case of the rational root theorem. The monic polynomial assumption is essential because it removes any leading coefficient that could supply prime factors of the denominator.

Use the reduction criterion: a primitive integer-coefficient polynomial whose leading coefficient remains nonzero modulo a prime number $p$ is irreducible over $\mathbb{Q}$ if its reduction is irreducible over ${\mathbb{F}}_p$. This follows from Gauss lemma for polynomials, since a nontrivial factorization over the integers would reduce to a nontrivial factorization of the same degrees. For a monic polynomial the factors may be taken monic, so neither degree can drop.

Modulo $2$ the polynomial is $\overline{f}=t^4+t+1$. Its values at $0$ and $1$ are both $1$, so it has no linear polynomial factor. A reducible quartic polynomial without a linear polynomial factor must factor into two irreducible quadratic polynomials. The only monic irreducible quadratic polynomial over ${\mathbb{F}}_2$ is $q=t^2+t+1$: among monic quadratic polynomials a nonzero constant is necessary, and $t^2+1$ has a root, while $q$ has none.

But $t^2\equiv t+1(\mathrm{mod}q)$, so $t^3\equiv 1$ and $t^4\equiv t$. ConsequentlyThus $q$ does not divide $\overline{f}$, excluding the remaining possible factorization. We obtain

Eisenstein's criterion states that an integer-coefficient polynomial is irreducible over $\mathbb{Q}$ if some prime number divides every nonleading coefficient, does not divide the leading coefficient, and its square does not divide the constant coefficient. Equivalently apply this to the primitive part when phrased for primitive polynomials.

Take $p=2$. Every nonleading coefficient of $t^5+4t+2$ is divisible by $2$ (including the zero coefficients), its leading coefficient is $1$, and its constant coefficient $2$ is not divisible by $4$. Therefore

A cubic polynomial over a field is reducible if and only if it has a root in that field: a nontrivial factorization of degree three has a linear polynomial factor, and conversely a root supplies one by the factor theorem.

The rational root theorem, or part (b), shows that a rational root of this monic polynomial must be an integer. Such an integer must divide the constant coefficient $4$, so the only candidates are $\pm 1,\pm 2,\pm 4$. Direct evaluation givesNone vanishes. Hence there is no rational root, provingThe same no-root argument alone would not decide a quartic polynomial, which could factor as two quadratic polynomials; its sufficiency here uses degree three.

The structure theorem for finitely generated modules over a principal ideal domain applies to a Euclidean domain because every Euclidean domain is a principal ideal domain. In invariant-factor form, every finitely generated module $M$ has a decompositionwhere $r\ge 0$, and each $d_i$ is a nonzero nonunit. The rank of a module $r$ is unique and the invariant factors of a finitely generated module $d_i$ are unique up to multiplication by units. The zero module is allowed by taking $r=s=0$.

Every abelian group is a module over $\mathbb{Z}$, and a finite abelian group is finitely generated. Since $\mathbb{Z}$ is a Euclidean domain, the theorem applies. A nonzero free module summand would be infinite, so a finite group has $r=0$. Choosing positive integer representatives gives the finite abelian classificationThe ordered integers $d_i$ are uniquely determined. Splitting each $d_i$ into prime powers using the Chinese remainder theorem gives the equivalent elementary-divisor form as a direct sum of cyclic prime-power groups.

Use the Chinese remainder theorem to separate relatively prime factors:These five cyclic prime-power summands are the elementary-divisor form. Leave one $\mathbb{Z}/3\mathbb{Z}$ summand aside and combine the factors of orders $4,3,5,7$, which are pairwise coprime. Their direct sum of modules is cyclic of order $420$. The invariant-factor canonical form is thereforeIts group order is $3\cdot 420=1260=60\cdot 21$, and its exponent of a finite group is $420=\mathrm{lcm}(60,21)$, providing independent checks on the decomposition. Since its exponent is smaller than its order, $A$ itself is not cyclic.

We prove the unique-subgroup criterion for a finite abelian group to be cyclic using the invariant-factor classification from part (a). If $A$ is nontrivial, writewhere $C_m$ denotes a cyclic group of order $m$. If $s\ge 2$, choose a prime number $p$ dividing $d_1$; then $p$ also divides $d_2$. In the first cyclic factor, the element represented by $d_1/p$ has order of a group element $p$, and the corresponding element in the second factor has order $p$ as well.

The two elements supported in their separate coordinates generate distinct subgroups of order $p$: one has zero second coordinate, and the other has zero first coordinate. Both are subgroups of $A$, and $p\mid |A|$. This contradicts the assumed uniqueness for that divisor. Hence $s\le 1$, andIf $A$ is the trivial group, it is already cyclic, generated by its identity. Thus the argument covers every finite order.

The multiplicative group of a field is abelian, so its finite subgroup $G$ is a finite abelian group. Let $d\mid |G|$, and suppose $H$ is a subgroup of order $d$. By Lagrange's theorem, the order of each $h\in H$ divides $d$, hence $h^d=1$. Thus $H$ is contained in the set of roots in $k$ of $t^d-1$.

This nonzero polynomial has degree $d$, so it has at most $d$ roots. Since the $d$ distinct elements of $H$ are already roots, they comprise the whole root set. Any second subgroup $K$ of order $d$ would also consist of $d$ roots of the same polynomial, forcing $K=H$. Therefore there is at most one subgroup of each order dividing $|G|$.

Part (c) now applies to giveThe root bound remains valid in positive characteristic, even if the formal derivative of $t^d-1$ vanishes: the argument counts distinct roots, so no separability assumption is needed.