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I tried to come up with an ideal generated by two elements but additively generated by three elements and came up with the following ideal in the following ring.

However, this ring feels too complicated for what I set out to do. Is there an easier example of an ideal that's generated by two elements but additively generated by three elements?

First, let me define my ring $A$ as $\mathbb{Z}$ together with $\boxed{2}, \boxed{3}, \boxed{4}, \boxed{8}$, with $\boxed{1}$ being the same as $1$.

Let me also define the map $f : A \to \mathbb{Z}$ as $k\boxed{m} \mapsto km$, e.g. $f(4\boxed{3} + 2\boxed{8} + 7) = 35$.

Let me define multiplication of the new unknowns in the following way:

  1. $\boxed{m}\boxed{n} = \frac{mn}{2}\boxed{2}$ if $mn$ is even.
  2. $\boxed{m}\boxed{n} = \frac{mn}{3}\boxed{3}$ otherwise.

So $\boxed{3}\boxed{8} = 12\boxed{2}$ and $\boxed{3}\boxed{3} = 3\boxed{3}$.

I argue that this multiplication rule is associative by noting that $\boxed{m}\boxed{n}\boxed{r} = \frac{mnr}{2}\boxed{2}$ when $mnr$ is even or $\frac{mnr}{3}\boxed{3}$ when $mnr$ is odd.

Let's examine the ideal $I = (\boxed{3}, \boxed{4})$.

$I$ is not principal.

If $I$ were principal, it would need to be generated by some $a$ such that $f(a) = 1$, since $\boxed{4}-\boxed{3}$ is in $I$.

And thus an element of $I$ would have the form:

$$ b_0 + b_1a + b_2aa + b_3a^3 + \cdots $$

However, any power of $a^n$ for $n \neq 1$ does not contain $\boxed{4}$ when written in normal form. Therefore $a$ must contain $\boxed{4}$ when written in normal form. Therefore $a$ is of the form $w + k \boxed{4}$. $f$ is multiplicative, so $f(a)$ must equal $1$. Thus $w=1$ and $k=0$ and thus $a = 0$, so $I$ must be the unit ideal which is a contradiction.

Thus, by contradiction $I$ is not principal.

I is also additively generated by $\{2\boxed{2}, \boxed{3}, \boxed{4}\}$.

I claim that $I$ cannot be additively generated by two elements.

As proof I note that $(2 \boxed{2}, \boxed{3}, \boxed{4})$ is isomorphic to a free Abelian group on three generators. This argument is kind of sketchy, I'm not sure how to make it better.

Thus $I$, an ideal of $A$, is generated by two elements but additively generated by three elements.

Martin Sleziak
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Greg Nisbet
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1 Answers1

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(I have not verified that your construction works.)

Here's an easier construction (inspired by standard examples in algebraic number theory, but you don't need algebraic number theory to understand it.) Let $R=\Bbb Z[\sqrt{-7}]$ and $I=(2,1+\sqrt{-7})$. Then $I$ is not prinicipal, as shown here. $I$ is clearly generated by two elements as an ideal. $I$ is also generated by two elements as an abelian group. To see this, note that additively $\Bbb Z[\sqrt{-7}]$ is a free abelian group of rank two and $I$ is a subgroup, so $I$ is a free abelian group of rank at most two. If $I$ was of rank one as an abelian group, then it would also be generated by a single element as an ideal, which is not the case.

Now consider the ideal $\Bbb Z \times I$ inside $\Bbb Z \times \Bbb Z[\sqrt{-7}]$. As an additive group, this is free of rank three, because $I$ is free of rank two, as explained above. $\Bbb Z\times I$ can be generated by two elements as an ideal. For example $(1,2)$ and $(0,1+\sqrt{-7})$. If $\Bbb Z \times I$, was prinicipal, then also $I$ would be prinicipal by applying the projection $\Bbb Z \times \Bbb Z[\sqrt{-7}] \to \Bbb Z[\sqrt{-7}]$ which is a surjective ring homomorphism that maps $\Bbb Z \times I$ to $I$. Thus $\Bbb Z \times I$ is an example of what you want.

Lukas Heger
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