# How does prime factorization fail for "nice" non-Dedekind Domains like $\mathbb Z[x]$?

$$\mathbb Z[x]$$ is "close" to being a Dedekind Domain: It is Noetherian and an UFD (and therefore integrally closed). However it has a few non-$$0$$ prime ideals that are not maximal. This made me wonder how unique factorization into prime ideals fails for it. More specifically, since $$(x)$$ is a prime ideal that isn't maximal, i'd expect there to be some ideal that has two different factorizations, at least one of them having $$(x)$$ in it. But i wasn't able to find any.. Is there one or am i missing something?

• I've often found $\Bbb Z[X]$ to be mean. Commented Aug 8 at 10:59
• $\mathbb Z[x]$ has plenty of non-zero prime ideals that are not maximal. See math.stackexchange.com/questions/174595/…
– lhf
Commented Aug 8 at 11:06
• @lhf I know? I even wrote a specific one in my question. Commented Aug 8 at 11:06
• I think that @lhf's comment was in response to you writing "a few" in the question, which suggests "not many". lhf countered that there are "plenty", which means "many". Commented Aug 8 at 13:24
• @OlivierRoche Is there any reason why you comment that? I think everyone here knows $\mathbb Z$ is not a field. Commented Aug 10 at 12:50

The only prime ideal that contains the non-prime ideal $$(4,x)$$ is $$(2,x)$$. But $$(2,x)^2=(4,2x,x^2)$$ does not contain $$x$$, so $$(4,x)$$ can't have a factorization into prime ideals.

This is an example where no prime factorization exists. Moreover, we have $$(2,x)^3=(2,x)(4,x^2)$$ where all ideals can't be factored further, so factorizations are also not unique. Since $$\mathbb Z[x]$$ is a Noetherian UFD, I believe it's not possible to have two different factorizations where all factors are prime ideals, but I don't see how to prove this right now.

• So what fails here is that that ideal doesn't even have a prime factorization. Is there any ideal of $\mathbb Z[x]$ with multiple prime factorizations tho? Commented Aug 8 at 11:19
• A problem is that factorization as product of ideals is the wrong concept. The correct way to generalize the FTA (valid in $\mathbb{Z}$ and in PIDs) is to work with intersections of ideals. In general $I\cap J\neq IJ$. Note that the decomposition as intersection of primes doesn't hold even in $\mathbb{Z}$. Commented Aug 10 at 10:54
• @AndreaMori I agree, and the other answers explain this well. But the OP asked for examples where unique factorization into products fails, so I provided some. Commented Aug 10 at 11:54
• @AndreaMori That makes so much more sense! I was so confused why my book talked about factorization using products of ideals when intersection seems to be much more natural. It did because it was talking about Dedekind Domains where that works, of course. But its nice to know my intuition was correct. Commented Aug 10 at 13:32

My algebra is so rusty, I cannot do this with any confidence, and have already gone back to fix a number of mistakes, but here is at least some ideas for others to check or work on with.

The core idea is to show that factorization into prime ideals is unique if it exists. As has already been pointed out, factorization into prime ideals does not exist for all ideals. However, I suspect the minimal/irredundant primary decomposition, even if not unique, can be reduced to a unique choice for ideals in $$\mathbb{Z}[x]$$.

So, given an ideal $$I\subset\mathbb{Z}[x]$$, there is a minimal primary decomposition $$I=\cap_i\mathfrak{q}_i$$ with associated prime ideals $$\mathfrak{p}_i=r(\mathfrak{q}_i)$$. The set of associated prime ideals is uniquely determined.

Actually, for minimal primes $$\mathfrak{p}_i$$ in the set, the corresponding $$\mathfrak{q}_i$$ are also uniquely determined. The principal prime ideals $$(p)$$ and $$(f(x))$$ for $$p$$ prime and $$f(x)$$ irreducible would have to be minimal if present, corresponding to primary ideals $$(p^n)$$ and $$(f(x)^n)$$ for $$n\ge1$$, ie also principal ideals. We can get rid of all the principal primary ideals $$\mathfrak{q}_i$$ since the product/intersection of these is again a principal ideal: ie the largest principal ideal $$(h(x))$$ for which $$I\subset(h(x))$$. The ideal $$J=I:(h(x))=\{f(x)\mid f(x)h(x)\in I\}$$ then has a minimal primary decomposition consisting of non-principal primary ideals, ie those with associated prime ideals $$\mathfrak{p}_i=(p_i,f_i(x))$$ for $$p_i$$ prime and $$f_i(x)$$ irreducible modulo $$p_i$$.

The above argument could actually have been reduced to taking the greatest common divisor of all elements in $$I$$, which would be the $$h(x)$$ in the previous paragraph, and divide all elements by it to obtain $$J$$.

An irreducible primary ideal over $$(p,f(x))$$, if I'm not mistaken, takes the form $$(p^{n_0},p^{n_1}f(x)^{m_i},\ldots,f(x)^{m_K})$$ where $$n_k,m_k>0$$. For this to be a prime power $$(p,f(x))^N$$, it would require $$\{(n_k,m_k)\}=\{(N,0),(N-1,1),\ldots,(0,N)\}$$.

So far I'm reasonably confident about my reasoning. The next steps, less so, so please verify.

Let $$J=\cap\mathfrak{q}_j$$ with associated prime ideals $$\mathfrak{p}_j=(p_j,f_j(x))$$. My suspicion is that, not only are the prime ideals $$\mathfrak{p}_j$$ uniquely determined, but so are the primary ideals $$\mathfrak{q}_j$$. Note that this depends on having gotten rid of the minimal/principal ideals, so for the remaining primary ideals, there are no $$\mathfrak{p}_i\subset\mathfrak{p}_j$$ with $$i\not=j$$ for associated prime ideals.

And wouldn't this also make $$J=\cap\mathfrak{q}_j=\prod\mathfrak{q}_j$$?

Can't the primary ideals $$\mathfrak{q}_j$$ be retrieved simply as the image of the ideal $$J$$ in the localizations to each of the associated prime ideals: $$I\mapsto I_{\mathfrak{p}_j}\subset\mathbb{Z}[x]_{\mathfrak{p}_j}$$?

I'm sure there are alternative constructions that have the same effect. Eg if we take $$n_j$$ be a power such that $$\mathfrak{p}_j^{n_j}\subset\mathfrak{q}_j$$, wouldn't $$\mathfrak{q}_j = J + \mathfrak{p}_j^{n_j}$$?

Upon revisiting the argument, after removing the principal ideals to obtain $$J=\cap\mathfrak{q}_j$$, the associated prime ideals $$\mathfrak{p}_j$$ will all be minimal within the set, and so the corresponding primary ideals will be unique (the proof of which is by the localization explained above).

It is a general result in Commutative Algebra that in a noetherian ring $$R$$ every (proper) ideal admits a primary decomposition. That means that if $$I\subset R$$ is an ideal, then $$I=\bigcap_{j=1}^mQ_j.\qquad\qquad(\ast)$$ where each $$Q_j$$ is primary (I will assume that the definition of primary ideal is known).

The decomposition is in general not unique. Given a primary ideal $$Q$$, its radical $$P=\sqrt{Q}$$ is a prime ideal. Given the decomposition $$(\ast)$$ consider the set $$\Sigma=\{P_i=\sqrt{Q_i}\}.$$ A prime $$P\in\Sigma$$ is called isolated if it doesn't contain properly any other prime ideal in $$\Sigma$$. Then the general result (that holds also for ideals admitting a primary decomposition in a non-noetherian ring) is that the primary ideals in $$(\ast)$$ corresponding to isolated primes in $$\Sigma$$ are uniquely determined.

I believe that the confusion comes from thinking of the decomposition of integers as a product of primes (Fundamental Theorem of Arithmetic). When we transalte that into a property of ideals in $$\mathbb{Z}$$ that does not mean that any ideal of $$\mathbb{Z}$$ is an intersection of prime ideals. Indeed $$12\mathbb{Z}=3\mathbb{Z}\cap4\mathbb{Z}$$ is not an intersection of prime ideals (the ideal $$4\mathbb{Z}$$ is primary but not prime).

Trivia fact: the first important results in the general study of primary decompositions in commutative rings were found by Emanuel Lasker in his 1905 Ph.D. thesis. At that time Lasker was the reigning world chess champion.