Exercise about Ideals, Generators, principal ideals and prime ideals

Question

Recently I started learning about ring theory and wanted to try some examples to see if I did understand everything correct. I would like to know If my solution (especially Nr. 4) is okay.

Define $A := \left\{ f \in \mathbb{Q}[X] \mathop| f(0)=f'(0)=0 \right\}$

1. Show that $A$ is an Ideal
2. Find a generator of $A$
3. Is $A$ principal ideal?
4. Is $A$ prime ideal?

About 1): Considering $f(0)=f'(0)=0$ the elements of $A$ are of the form $a_2X²+a_3X³+...+a_nX^n$

$(A,+)$ is a subgroup of $(\mathbb{Q}[X],+)$ and for $a \in A$, $q \in \mathbb{Q}[X] \Rightarrow aq \in A$ Thus $A$ is an Ideal

About 2):

Since $\mathbb{Q}[X]$ commutative and has a $1$, I can use the following theorem:

Let $R$ be a commutative Ring with $1$. Then for $M \subset R$:

$$ (M)= \left\{ r_1 a_1+r_2 a_2+...+r_n a_n \mathop| r_i \in R, a_i \in M \right\} \tag{*} $$

In my case is $(M)=A$, and since the representation $(*)$ is very similar to polynomial I get, $M= \{ a_2,a_3,...a_n\} $ to ba a generator of the Ideal $A$.

About 3):

Because $M$ has more than one element $\Rightarrow A$ is not a principal Ideal

About 4):

Let $R$ be a commutative Ring. An Ideal $I \neq R$ is calle prime ideal if:

$a,b \in R$ and $ab \in I \Rightarrow a\in I$ or $b \in I $

If I take $a=b=a_1X \in \mathbb{Q}[X]$, then $ab=a_1^2X^2 \in A$ but $a=b \notin A$. Thus, A is not a prime ideal.

Question: As stated above, I am new to ring theory (algebra in general) and would like to know if my solution is correct (especially Nr.4)

Edit: In 2) I meant $M= \{ X^2,X^3,...X^n\} $

Answer 1

You seem to be using $M$, $A$, and $I$ for the same set here. Stick with $A$ as given in the original problem. Edit Now I see that $M$ stands for something different; a set of generators.

Your answer for 2) isn't a solution. You haven't said what $a_1,\dots,a_n$ are. You need to be explicit here.

For 3), you haven't said what the generators are, but even if you have found a way to generate the set with more than one element, that doesn't mean it's not also generateable with a single element. For instance $A = (2X^2, X^2 + X^4)$, but $A = (X^2)$ too.

Your answer to 4) is correct! To simplify, I would suggest $a=b=X$ (i.e., set $a_1 = 1$).

Your answer to 1) has no mistakes in it, but it is rather terse. If I were grading it, I would wonder if you could supply the missing steps. Based on your answer to 2) and 3) I would have doubts.

Answer 2

I'll start with correct proofs of each of the claims (assuming I didn't make a mistake), and then comment on each section of your question.

I'm putting my answers first because that's the order in which I wrote this answer. It's hard to review stuff without trying to solve it yourself first.

My answers

Lemma 51: $p \in A$ if and only if the constant term and linear term of $p$ are zero.

Suppose $p$ is in $A$. Then $p(0) = 0$; thus the constant term is $0$. Suppose that $p'(0)$ is $0$, this means that the linear term of $p$ is now the constant term of $p'$ and must also be $0$.

Suppose $p$ has zero constant term and zero linear term. Thus $p(0) = 0$. Thus $p'(0) = 0$.

End of proof of Lemma 51.

Part 1

Let $f$ be an element of $A$ and $r$ an arbitrary element of $\mathbb{Q}[x]$.

• $(rf)(0) = 0$
• $(rf)'(0) = (rf')(0) + (r'f)(0) = 0 + 0 = 0$

Thus $A$ is closed under multiplication by an arbitrary ring element.

• $(f+g)(0) = f(0)+g(0) = 0 + 0 = 0$
• $(f+g)'(0) = f'(0) + g'(0) = 0 + 0 = 0$

Thus $A$ is closed under addition.

Thus $A$ is an ideal.

Part 2

I choose $x^2$ as my generator. By multiplying by a monomial $kx^n$, I can produce monomials of the form $kx^{(n+2)}$ and by adding monomials of this form together, I can produce an arbitrary $\sum_i a_ix^i$ where $a_0 = a_1 = 0$. By Lemma 51, the elements of $A$ are precisely the elements of $\mathbb{Q}[x]$ of that form.

Part 3

Yes. It is generated by $x^2$.

Part 4

No. An ideal is prime if and only if its complement is multiplicatively closed.

$A^c$ is not multiplicatively closed.

For example, $x$ is in $A^c$, but $xx = x^2$ is not in $A^c$.

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Review of question

Part 1

Your argument about the shape of an element of $A$ is correct. It is a bit too abbreviated in my opinion. Feel free to ignore this if you're relying on a previously proven lemma.

Your argument about $A$ being a subgroup and multiplicatively closed, in my opinion, skips too much. You did not demonstrate that $a+b$ or $aq$ satisfy either the initial conditions $f(0) = f'(0) = 0$ or lack linear and constant terms. You did correctly assert that they're in the ideal, though, and that's the condition you need to check.

Part 2

First, as an aside, there's a theorem you can invoke to show that every ideal is principal. $\mathbb{Q}$ is a field and thus a principal ideal domain. Therefore $\mathbb{Q}[x]$ is a principal ideal domain. Therefore any ideal of $\mathbb{Q}[x]$ is principal. See this answer for an example.

You are exploiting the structure theorem that you have about elements of $A$ to get a generating set for $A$.

Your generating set is $\{x^2, x^3, x^4 \cdots\}$.

This set does indeed generate $A$.

However, it is not minimal.

For example, $x^3$ is $x\,x^2$. $x$ is not in $A$, but $x$ is in $\mathbb{Q}[x]$, so we're allowed to multiply by it when trying to build elements of $A$ from our generator.

Remember that an ideal is not a subring. It is closed under multiplication by arbitrary ring elements, not just those inside the ideal.

Part 3

Part 3 is wrong. A principal ideal is defined as one that is generated by a single element, not one that contains a single element.

Interestingly, there is exactly one ideal that has a single element. The ideal $(0) = \{0\}$. Also, this prime if and only if there are no zero divisors, which is kinda fun.

Part 4

Your argument is correct. $ax$ is outside the ideal, $bx^2$ is inside the ideal. Since all nonzero coefficients are units (because $\mathbb{Q}$ is a field) it doesn't matter what the coefficient is.