Example of irreducible ideal which is not strongly irreducible

Question

I have read a paper with title Ideal Theory in Commutative Semirings by Reza Ebrahimi Atani and Shahabaddin Ebrahimi Atani. In this paper we have the following definitions:

An ideal I is irreducible if and only if A∩B=I for two ideals A, B implies A = I or B = I.

An ideal I is strongly irreducible if and only if A∩B⊆I for two ideals A, B implies A⊆I or B⊆I.

Theorem 7. If I is strongly irreducible, then I is irreducible.

Can you help me to find an example of ideal I where I is irreducible but not strongly irreducible?

Thank you.

Answer 1

For future readers, let me give a simpler example of an irreducible ideal, which is not strongly irreducible, even in the context of rings. I claim that $\langle X^2,Y^2\rangle\subseteq K[X,Y]$, where $K$ is a field, is one such example.

To see why, I will say more. First of all, notice how the strongly irreducible ideals are related to the other well-known classes of ideals:

Proposition $0$. Any prime ideal is strongly irreducible, and any strongly irreducible ideal is irreducible.

Next, some useful characterizations and examples, left as exercises to the reader:

Proposition $1$. $a)$ Let $R$ be a commutative ring, $I\unlhd R$ a proper ideal. Then $I$ is strongly irreducible if and only if: $\forall a,b\in R,\langle a\rangle\cap\langle b\rangle\subseteq I\Rightarrow a\in I\lor b\in I$. (hence, enough to test it for principal ideals)
$b)$ If $R$ is a GCD-domain, then $I$ is strongly irreducible if and only if: $\forall a,b\in R, {\rm lcm}(a,b)\in I\Rightarrow a\in I\lor b\in I$.
$c)$ If $R$ is a UFD, then $I$ is strongly irreducible if and only if: $\forall a,b\in R^\ast\setminus U(R)$ coprime, $ab\in I\Rightarrow a\in I\lor b\in I$.
$d)$ Let $R$ be a UFD, $p\in R$ prime element, $r\in\mathbb{N}^\ast$. Then $\langle p^r\rangle\subseteq R$ is strongly irreducible.

It will be challenging to find an irreducible ideal which is not strongly, because of this:

Proposition $2$. Let $R$ be a PID. Then the primary ideals are precisely the irreducible ideals, which are precisely the strongly irreducible ideals, which are: $\langle 0\rangle$ and $\langle p^r\rangle$ for $p\in R$ prime and $r\in\mathbb{N}^\ast$.

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We can have a better chance, if we investigate the polynomial rings over a field $K[X_1,...,X_n]$. In this ring, we have special ideals, the so-called monomial ideals, i.e. ideals generated by monomials. A standard commutative algebra result tells us the following (a proof can be found here):

Theorem $2$. $a)$ The prime monomial ideals of $K[X_1,...,X_n]$ are precisely the monomial ideals generated by subsets of variables.
$b)$ The irreducible monomial ideals of $K[X_1,...,X_n]$ are precisely the monomial ideals generated by pure powers of variables.

My initial example will be completely justified, if we are able to prove the following:

Theorem $3$. The strongly irreducible monomial ideals of $K[X_1,...,X_n]$ are precisely the monomial ideals generated by subsets of variables or by a power of a single variable.

Proof: $(\Leftarrow)$ If $I$ is a monomial ideal generated by subsets of variables, then by [Theorem $2,a$)] $I$ is prime, hence strongly irreducible. If $I$ is a monomial ideal generated by a power of a single variable, then $I$ is strongly irreducible, due to [Proposition $1,d)$].

$(\Rightarrow)$ Let $I\unlhd K[X_1,...,X_n]$ be a strongly irreducible monomial ideal. Then by [Proposition $0$], $I$ is irreducible. The zero ideal is prime (because we are in a domain), so it is strongly irreducible. From now on, to avoid trivialities, assume that $I\neq \langle 0\rangle$.

By [Theorem $2,b)$] we must have $I=\langle X_{i_1}^{a_1},...X_{i_k}^{a_k}\rangle\unlhd K[X_1,...,X_n]$, for some $1\leq k\leq n;1\leq i_1<...<i_k\leq n; a_1,...,a_k\in\mathbb{N}^\ast$. $k=1$ is fine, because then $I=\langle X_{i_1}^{a_1}\rangle$, has the desired form. Suppose from now on that $k>1$.

If $\forall i\in\overline{1,k},a_i=1$, then $I=\langle X_{i_1},...,X_{i_k}\rangle$, again has the desired form. Suppose now the contrary, that some exponent is greater than $1$, say $a_j>1$, for some $j\in\overline{1,n}$. It remains to prove that, under these conditions, $I$ is not strongly irreducible. For this, since $k>1,\exists m\neq j,m\in\overline{1,k}$.

Look at the polynomials $f:=X_{i_j}$ and $g:=X_{i_j}^{a_j-1}+X_{i_m}^{a_m}$. Clearly, $f,g\notin I=\langle X_{i_1}^{a_1},...X_{i_k}^{a_k}\rangle$, and ${\rm gcd}(f,g)=1$, they are coprime. However, $fg=X_{i_j}^{a_j}+X_{i_j}X_{i_m}^{a_m}\in I$. Hence, by [Proposition $1,c)$], the ideal $I$ cannot be strongly irreducible, and we are done. $\Box$