The Lumenary · First-year graduate

Commutative algebra

Name: ______________________

Date: ______________________

You will learn to

Work with prime and maximal ideals, the spectrum $\operatorname{Spec} R$ and its Zariski topology, and the nilradical and Jacobson radical
Localise rings and modules, describe the primes of $S^{-1}R$ , and use localisation to reduce questions to local rings
Prove that Noetherian means "every ideal is finitely generated", prove Hilbert's basis theorem, and show that $k[x_1, x_2, \dots]$ is not Noetherian
Prove Nakayama's lemma by the determinant trick and use it for minimal generators over a local ring
Prove the Nullstellensatz over $\C$ and use it to match radical ideals with algebraic sets; state the dimension theorem $\dim k[x_1, \dots, x_n] = n$

1.
Which of these ideals are prime? (Choose all that apply.)
- A. $(0) \subseteq \mathbb{Z}$
- B. $(6) \subseteq \mathbb{Z}$
- C. $(7) \subseteq \mathbb{Z}$
- D. $(x) \subseteq k[x, y]$
- E. $(xy) \subseteq k[x, y]$
- F. $(x^2) \subseteq k[x]$
2.
What is the Krull dimension of $\mathbb{Z}[x]$ ?
3.
How many elements does the localisation $(\mathbb{Z}/360)_{(3)}$ have? (Here $(3)$ is the prime ideal of $\mathbb{Z}/360$ generated by $3$ .)
4.
In the proof of Hilbert's basis theorem, $L_d \subseteq R$ is the ideal of leading coefficients of degree- $d$ elements of $I \subseteq R[x]$ . Why is there an $N$ with $L_N = L_{N+1} = L_{N+2} = \cdots$ ?
- A. $L_0 \subseteq L_1 \subseteq \cdots$ is an ascending chain of ideals of the Noetherian ring $R$
- B. $I$ is finitely generated, so its degrees are bounded
- C.Each $L_d$ is a principal ideal
- D.Leading coefficients can be normalised to $1$
5.
Let $R = \mathbb{Z}_{(p)}$ , $\mathfrak{m} = p\mathbb{Z}_{(p)}$ and $M = \mathbb{Q}$ . Then $\mathfrak{m}M = M$ but $M \neq 0$ . Which hypothesis of Nakayama's lemma fails?
- A. $M$ is not finitely generated over $R$
- B. $\mathfrak{m}$ is not contained in the Jacobson radical
- C. $R$ is not a local ring
- D. $R$ is not Noetherian
6.
Let $I = (x^2 + y^2 - 1,\; x - 1) \subseteq \mathbb{C}[x, y]$ : the unit circle and its tangent line at $(1, 0)$ . Which of these polynomials lie in $I$ itself? (Choose all that apply.)
- A. $y$
- B. $y^2$
- C. $x - 1$
- D. $x^2 - 1$
- E. $y^3$
- F. $x$
7.
Which Lean statement formalises “the polynomial ring in infinitely many variables over a field $k$ is not Noetherian”?
- A.¬ IsNoetherianRing (MvPolynomial ℕ k)
- B.¬ IsNoetherianRing (MvPolynomial (Fin n) k)
- C.∀ I : Ideal (MvPolynomial ℕ k), ¬ I.FG
- D.IsNoetherianRing (Polynomial k)
8.
The ideals of $\mathbb{Z}/n$ are $d\mathbb{Z}/n\mathbb{Z}$ for the divisors $d$ of $n$ . Write prime_ideals(n) (for $n \ge 2$ ) returning the sorted list of divisors $d$ of $n$ for which the ideal generated by $d$ in $\mathbb{Z}/n$ is prime, checking the definition directly: proper, and $ab \in I \Rightarrow a \in I$ or $b \in I$ for all residues $a, b$ .
Computational — Practice

Answer key — Commutative algebra

1.
Prime: $(0)$ and $(7)$ in $\mathbb{Z}$ , and $(x)$ in $k[x, y]$ (quotients $\mathbb{Z}$ , $\mathbb{F}_7$ , $k[y]$ are domains). Not prime: $(6)$ ( $2 \cdot 3$ ), $(xy)$ ( $x \cdot y$ ) and $(x^2)$ ( $x \cdot x$ ). Of the primes, only $(7)$ is maximal.
2.
The chain $(0) \subsetneq (x) \subsetneq (2, x)$ consists of primes (quotients $\mathbb{Z}[x]$ , $\mathbb{Z}$ , $\mathbb{F}_2$ ), so $\dim \geq 2$ . Since $\mathbb{Z}$ is Noetherian of dimension $1$ , $\dim \mathbb{Z}[x] = 1 + 1 = 2$ . Geometrically $\operatorname{Spec}\mathbb{Z}[x]$ is an “arithmetic surface”: a family of lines $\operatorname{Spec}\mathbb{F}_p[x]$ over the points of $\operatorname{Spec}\mathbb{Z}$ .
3.
By the Chinese remainder theorem $\mathbb{Z}/360 \cong \mathbb{Z}/9 \times \mathbb{Z}/40$ . The element $(1, 0)$ lies outside the prime $(3)$ , and it annihilates $\mathbb{Z}/40$ , so that factor dies; on $\mathbb{Z}/9$ every element not divisible by $3$ is already a unit. Hence $(\mathbb{Z}/360)_{(3)} \cong \mathbb{Z}/9$ , with $9$ elements. The experiment commutative_algebra_spec_zmod checks this rule for all $n \le 60$ .
4.
If $f \in I$ has degree $d$ and leading coefficient $a$ , then $xf \in I$ has degree $d + 1$ and the same leading coefficient, so $L_d \subseteq L_{d+1}$ . Since $R$ is Noetherian, this ascending chain of ideals stabilises at some $L_N$ . The proof then uses finitely many generators of $L_0, \dots, L_N$ to generate $I$ .
5.
$\mathbb{Z}_{(p)}$ is a local (Noetherian) ring with maximal ideal $\mathfrak{m} = J(R)$ , and $p\mathbb{Q} = \mathbb{Q}$ because $p$ is invertible in $\mathbb{Q}$ . The failing hypothesis is finite generation: finitely many fractions $a_i/b_i$ generate a submodule of $\tfrac{1}{d}\mathbb{Z}_{(p)}$ , where $d = \prod b_i$ , and $1/(pd)$ is not in it. The determinant trick needs a finite generating set.
6.
Setting $x = 1$ identifies $\mathbb{C}[x, y]/I$ with $\mathbb{C}[y]/(y^2)$ . A polynomial $f$ lies in $I$ iff $f(1, y)$ is divisible by $y^2$ . So $y^2$ , $y^3$ , $x - 1$ and $x^2 - 1$ lie in $I$ ; $y$ does not ( $f(1, y) = y$ ), although it lies in $\sqrt{I} = I(V(I)) = (x - 1, y)$ , as the Nullstellensatz predicts; $x$ does not even vanish at the point. The first experiment finds the same: least $k$ with $y^k \in I$ is $2$ .
7.
MvPolynomial σ k has one variable for each element of σ; infinitely many variables means σ = ℕ, and the statement is ¬ IsNoetherianRing (MvPolynomial ℕ k). The proof exhibits the chain $(x_0) \subsetneq (x_0, x_1) \subsetneq \cdots$ and an evaluation map showing $x_n \notin (x_0, \dots, x_{n-1})$ .

def prime_ideals(n):
    result = []
    for d in range(1, n + 1):
        if n % d:
            continue
        ideal = {d * k % n for k in range(n)}
        if len(ideal) == n:
            continue            # the whole ring is not prime
        if all(a in ideal or b in ideal
               for a in range(n) for b in range(n)
               if a * b % n in ideal):
            result.append(d)
    return result

The answer is always the list of primes $p \mid n$ — and, when $n$ itself is prime, that is $d = n$ , the zero ideal of the field $\mathbb{Z}/n$ . Since $\mathbb{Z}/n$ is finite, its prime ideals are maximal: $\operatorname{Spec}\mathbb{Z}/n$ is a finite discrete space with one point for each prime factor of $n$ .