What is the minimal number of generators of the ideal $(6x, 10x^2, 15x^3)$ in $\Bbb Z[x]$?

I know that the ideal $J=(6x, 10x^2, 15x^3) \trianglelefteq \Bbb Z[x]$ is not principal – I give the proof below. But can it be generated by two polynomials?

I believe that the answer is no. I wrote the corresponding $5$ equations , but it seems quite long, so I would like some help.

More generally, if I consider the $n$-th first prime numbers $p_1,...,p_n$, then I think that the minimal number of generators of the ideal $I_n = (q_1x, \dots, q_nx^n)\trianglelefteq \Bbb Z[x]$ is $n$, where $q_i = \prod\limits_{j\neq i} p_j$. Do you think this is correct? The ideal $J$ above corresponds to $I_3$ — I'm trying to start with an easy case.

I don't really know the notions of dimension, of height, etc. in commutative algebra, but answers introducing these notions would be appreciated.

If $(6x, 10x^2, 15x^3)=(f)$ were principal, then we could write $$f(x)=6xP(x)+10x^2Q(x)+15x^3R(x) \qquad 6x=a(x)f(x)$$ so that $6=a(x) \;[ 6P(x)+10xQ(x)+15x^2R(x) ]$ would yield $a(x)q(x)=0=a(x)r(x)$ and $p(x)a(x)=1$. Therefore $q=r=0$ and $f(x)=6xp(x)$, so that $10x^2=b(x)f(x)=6xb(x)p(x)$ which is impossible when evaluated at $1$, since all the polynomials have integer coefficients.

Solution 1:

I have been working hard for a week on this question, and I found that matters around minimal number of generators are really delicate. The final answer is that the present case is not correct. It would be enough to show a counterexample, but I previously wrote a wrong positive answer based on localization techniques: you are interested in, so I will leave paragraphs about them. To see how much questions on number of generators are delicate, you can try to show with local methods that $(12x,10x^2,15x^3)$ cannot be generated by two elements.

I introduce you to these techniques because I find them really useful in a variety of cases. I try to be concise even if there is a lot to say (it could be a half of a basic course in commutative algebra). Then, I give you the counterexample.

Modules. Given a ring $A$, an $A$-module $M$ is an (additive) abelian group, with a linear action of $A$. What I mean? The linear endomorphisms of $M$ $$End(M) = \{f: M \to M \ :\ f(x+y) = f(x)+f(y) \ \forall \ x,y \in M\}$$ form a ring, where the sum is $(f+g)(x) := f(x)+g(x)$ and the product is the composition. Well, a linear action of $A$ is a ring-homomorphism $\varphi:A \to End(M)$: you can 'multiply' by elements of $A$ by $a \cdot m := \varphi(a)(m)$, and the multiplication is linear in both entries $a,m$.

Examples (try to understand why these statements hold)

• If $k$ is a field, a $k$-module is a $k$-vector space.
• A $\mathbb{Z}$ module is just an abelian group.
• An ideal $I \subset A$ is an $A$-module.
• A quotient $A/I$ of a ring by an ideal is an $A$-module.
• If $M$ is an $A$-module, $I$ an ideal of $A$, then $IM=\{im: i \in I, m \in M\}$ is an $A$-module.
• If $M$ is an $A$-module, $M/IM$ is an $A/I$- module (the quotient $M/IM$ is a quotient of abelian groups).

Localization. Given a ring $A$ and a prime ideal $P \subset A$, you can form $A_P$, the localized ring at $P$(see here ), in the following way. Take $B=\{a/s: a \in A, s \not \in P\}$ the ring of formal quotients with denominator not in $P$; the operations are given as if they were rational numbers (like $a/b+c/d = (ad+bc)/bd$). Then $A_P=B/\sim$, where $\sim$ is given by $a/b \sim a'/b' \ \ \Leftrightarrow \ \exists \ s \not \in P: s(ab'-a'b) = 0$. This is weird, but note that if $A$ is a domain the condition reduces to the prettier $ab'=a'b$. Check these statements:

• $\sim$ is an equivalence relation (this is very boring, but you must do it once in your life).
• There is a morphism $\varphi_p:A \to A_P$ which sends $a \to a/1$ and it is injective when $A$ is a domain.
• There is a corrispondence of ideals between $$ \{\text{Ideals of }A\text{ contained in }P\} \ \leftrightarrow \ \{\text{Ideals of }A_P\} $$ which explicitly sends $I \subseteq A$ in $I_P = I A_P = \{x/s: x \in I\}$. This corrispondence preserve inclusions and prime ideals.
• $A_P$ is a local ring, i.e. it has just one maximal ideal $\mathfrak{m}=PA_P$. In particular, $A_P/PA_P$ is a field.
• If $A$ is a local ring with unique maximal ideal $\mathfrak{m}$, then $A_{\mathfrak{m}}\simeq A$.
• If $I$ is an ideal of $A$,$P$ a prime ideal, then $(A/I)_P \simeq A_P/I_P$. In particular, using the two previous facts, show $A_P/PA_P \simeq k(A/P)$ (his field of fractions).

Localized modules. Similarly, given a domain $A$, a module $M$ and a prime $P$, you can form the $A_P$ module $M_P$ in the following way. Take $N=\{m/s, m \in M, s \not \in P\}$ the module of formal quotient with denominator not in $P$. Then $M_P = N/\sim$, where $\sim$ is given by $m/s \sim m'/s'$ iff $\exists \ t \not \in P:\ t(s'm-sm') = 0$. The action of $A_p$ is given by $a/s \cdot m/t := a \cdot m / st$ Check these statements to make practice:

• The action is well defined (both $a/s, m/t$ can be replaced by equivalent elements: the result must be equivalent).

• If $M_P=0$ for every prime $P$, then $M=0$ (and viceversa).

• If $I$ is an ideal of $A$, then $I_P$ is exactly what we described before.

• $M_P/PM_P$ is a vector space (over..?).

Dimension. Given a ring $A$, we can form 'chains of primes' $P_0= \subsetneq \ldots \subsetneq P_n$. We call this a chain of length $n$. Then we define $dim A = sup \{ n: \text{ exist a prime chain of length }n\}$.To get an idea of why this is called dimension, you have to look at algebraic geometry. If $P \subset \mathbb{C}[x_1, \ldots, x_n]$ is a a prime ideal, you can look at the algebraic variety $V(P)=\{(x_1, \ldots, x_n) \in \mathbb{C}^n: f(x_1, \ldots, x_n)=0 \ \forall f \in P\}$. It turns out that the dimension of $P$ as a geometric locus (line, surface,..) is exactly the dimension of $A/P$ as a ring. To make practice, show that

• A product of fields $k_1 \times \ldots \times k_n$ has dimension 0.
• If A is PID, then $dimA =1$. In particular, $dim \mathbb{Z} = 1$.

Number of generators by local reasoning. And now we approach the initial question. We have a domain $A$, an $A$ module $M$ and we wonder: how many elements at least we need to generate $M$? We call this number $\mu(M)$. Localizations give bounds which are easier to calculate.

Given $P$ a prime ideal, we can form the field $A_P/PA_P$ and the corresponding vector space $M_P/PM_P$. We know exactly how any generators we need here: the dimension of $M_P/PM_P$ over $A_P/PA_P$. We call this number $\mu_P(M)$.

Lower bound. If $m_1, \ldots, m_k$ generate $M$, then $m_1/1, \ldots, m_k/1$ generates $M_P/PA_P$ (why?). So $\mu(M) \ge \mu_P(M)$. We get the lower bound $$\mu(M) \ge \max_P \mu_P(M)$$

Upper bound. This is known as Forster Swan theorem, and it states that $$\mu(M) \le \max_P \mu_P(M) + \dim A/P$$

Present case. There is a classification of primes in $\mathbb{Z}[x]$. So i tried all the localizations (with some tricks I can't reveal here), but unfortunately the lower bound is $2$ and the upper bound is $3$ (localizing at $(0)$). If you want to know more about trying 'all' localization, see primary decomposition ( you can verify that $J=(2,x^2) \cap (3,x)$ is its primary decomposition). Let's see the counterexample.

First step. $J=(6,10x,15x^2)=(6,2x,x^2)=:I$

Of course $J \subseteq I$ because every generator of $J$ is a multiple of a generator of $I$. Conversely, $2x=-10x+2x\cdot6$ and $x^2=-15x^2+10x\cdot x + 6 \cdot x^2$, so $I \subseteq J$.

Second step. $I=(6,2x,x^2) = (6,2x+3x^2)=:K$.

Again, it is easy to see that $K \subseteq I$. Conversely, we have $$(2x+3x^2) (3+2x)-6(x+2x^2+x^3) = (6x+4x^2+9x^2+6x^3)-6x-12x^2-6x^3 = x^2 \in K$$ And consequently also $2x= (2x+3x^2)-3 \cdot x^2 \in K$.

If you have further questions, do not hesitate to ask.

Edit: Indeed, I found a little refinement of local inequalities which solves the present case. Here Mohan says that in the case of polynomial rings, we can not to take into account primes such that the height of $P$ is 0 in forster swan inequality (this should be due to eisenbud and evans). In this case, we can exclude $p=(0)$. Let's examine the other cases.

(1) If $P \neq (2,x),(3,x)$, we claim that $JA_P = A_P$. Note that $\sqrt{J} = \sqrt{(6,2x,x^2)} = (6,x) = (2,x) \cap (3,x)$. Otherwise, one should have $J \subseteq P$, then $(2,x) \cap (3,x) = \sqrt{J} \subseteq \sqrt{P} = P$ because a prime is radical, then $(2,x) \subseteq P$ or $(3,x) \subseteq P$, then $(2,x)=P$ or $(3,x) =P$ by maximality of $(2,x), (3,x)$. On balance, in this case $\mu_P(J)=1$.

(2) If $P=(2,x), (3,x)$, then $\mu_P(J) = 2$. Infact, getting rid of invertibles from generators:

$P=(2,x)$: $JA_P = (6,2x,x^2)A_P = (2,2x,x^2)A_P=(2,x^2)A_P$

So $\mu_P(J) = dim(\mathbb{Z}[x]/(2,x), (2,x^2)_P/P(2,x^2) ) = dim(\mathbb{Z}/2\mathbb{Z}, [(2,x^2)/(4,2x,x^3)]_P ) = 2$

$P=(3,x)$: $JA_P = (6,2x,x^2)A_P = (3,x,x^2)A_P = (3,x)A_P$

So $\mu_P(J) = dim(\mathbb{Z}/3\mathbb{Z}, [(3,x)/(9,3x,x^2)]_P ) = 2$.

(3) Putting all together, $\mu_P(J) + dim(A/P)$ equals $1+1=2$ for primes of height 1 , and it is $\le 2+0 = 2$ for primes of height 2. We can ignore primes of height 0, so forster swan gives $\mu(M) \le 2$. Of course, we have $\mu(M) \ge \mu_{(3,x)}(M) = 2$, so we conclude that $\mu(M)=2$.