First some simple facts:
The ring $\mathbb{Z}[X]$ has dimension $2$ and it is a unique factorization domain with unit group $\{\pm 1\}$. We immediately conclude:
- Every ideal has height $0$, $1$, or $2$.
- The only ideal of height $0$ is $(0)$.
- If the primary decomposition of an ideal contains only height $1$ ideals, then:
- The ideal is principal
- These ideals are in one-to-one correspondence with nonzero polynomials with positive leading coefficient
Now let $I$ be a nonprincipal ideal. Suppose $I \cap \mathbb{Z} = (0)$. Then $I \otimes \mathbb{Q}$ is a proper, nonzero ideal of $\mathbb{Q}[X]$, generated by a primitive polynomial $g(x)$ (i.e. integer coefficients of gcd 1). It follows that $I = g(x) J$ where $J \cap \mathbb{Z} \neq (0)$.
Alternatively, $g(x)$ can be obtained a the gcd of all elements of $I$. (the $g(x)$ so constructed would have content $n$ rather than $1$ if $I$ is divisible by $(n)$). But the above is the method of proof I thought of to show the cofactor contains an integer.
Taking a different tactic, essentially by the Euclidean algorithm any ideal has a basis of the form
$$ I = \langle g_i(x) \mid 1 \leq i \leq k \rangle $$
where
- $g_i(x) = c_i x^{n_i} + f_i(x)$
- where $c_i \in \mathbb{Z}$, $c_i > 0$, $\deg f_i(x) < n_i$
- $n_i < n_{i+1}$
- $c_{i+1} | c_i$ and $c_i \neq c_{i+1}$
For example, the ideal
$$ \langle 8, 4x + 4, 2x^2 + 2 \rangle$$
In fact, I'm pretty sure this works out to a normal form by the algorithm:
- For each $d$, pick (if any) the smallest polynomial of degree $d$ in $I$
- Throw out any polynomial whose leading term is divisible by a picked polynomial of lesser degree
"Smallest", here, is determined by first comparing the coefficients on $x^d$, and if those are equal comparing the coefficients on $x^{d-1}$, and so forth. Integers are compared by absolute value first, and if equal, $n$ is considered smaller than $-n$.
e.g. the polynomial $4x + 4$ is considered smaller than $8x$ and than $4x-4$.
Unfortunately, not every choice of $g_i$'s is admissible. For example, the ideal
$$ \langle 4, 2x+1 \rangle $$
actually has normal form
$$ \langle 1 \rangle $$
I conjecture a sequence of reduced $g_i$'s (meaning we reject $\langle 8, 4x-4 \rangle$ because we can reduce $4x-4$ to $4x+4$ by a monomial multiple of $8$) satisfying the above properties is the normal form of an ideal if and only if $c_i \mid f_i(x)$.
Alas, I don't know enough about the theory of Groebner bases over Euclidean rings to say for sure.
In fact, the inclusion $IJ \subseteq I \cap J$ is always true. Given ideals $I$ and $J$, an element of $IJ$ is of the form $x = a_1 b_1 + \cdots + a_t b_t$ for some $t \in \mathbb{Z}_{\geq 0}$ and $a_k \in I$ and $b_k \in J$. Since ideals are closed under arbitrary multiplication, then each term of the above sum is in both $I$ and $J$, and since ideals are closed under addition, then the sum is in $I$ and $J$, so $x \in I \cap J$.
Now, let's consider the reverse inclusion. For ease of notation, let's just consider two primary ideals $Q_1$ and $Q_2$. By part (b) we have $Q_1 = (p_1^m)$ and $Q_2 = (p_2^n)$ for some primes $p_1, p_2$ and, as per my comment, assume $p_1$ and $p_2$ are non-associate. Given $x \in Q_1 \cap Q_2$, then $x = p_1^m a$ and $x = p_2^n b$ for some $a,b \in R$. Then $p_1 \mid p_2^nb$, so $p_1 \mid p_2^n$ or $p_1 \mid b$. If $p_1 \mid p_2^n$, then by repeated using the fact that $p_1$ is prime we find that $p_1 \mid p_2$. Then $p_2 = p_1 c$ for some $c \in R$. But since primes are irreducible, then $c$ must be a unit, which contradicts that $p_1$ and $p_2$ are non-associate. Thus $p_1 \mid b$, so $b = p_1 b_1$ for some $b_1 \in R$.
Now we have
$$
p_1^m a = x = p_2^n b = p_2^n p_1 b_1 \, .
$$
Since $R$ is a domain, then we cancel the factor of $p_1$, which yields $p_1^{m-1} a = p_2^n b_1$. Repeating the same argument as above (or by induction, if you want to be rigorous), then $p_1 \mid b_1$ so $p_1^2 \mid b$, and so on and so forth until we get $p_1^m \mid b$. Then $b = p_1^m b_m$ for some $b_m \in R$, so
$$
x = p_2^n b = p_2^n p_1^m b_m \in (p_1^m)(p_2^n) = Q_1 Q_2 \, .
$$
We can use the above in the induction step to prove this result for any finite number of primary ideals. I'll leave the details to you.
Suppose we know the result for $k$ primary ideals. Then \begin{align*} Q_1 \cap \cdots \cap Q_k \cap Q_{k+1} &= (Q_1 \cap \cdots \cap Q_k) \cap Q_{k+1} = (Q_1 \cdots Q_k) \cap Q_{k+1} \, . \end{align*}
Given $x \in \bigcap_{i=1}^{k+1} Q_i = (Q_1 \cdots Q_k) \cap Q_{k+1}$, then $$x = p_1^{n_1} \cdots p_k^{n_k} a \qquad \text{and } \qquad x = p_{k+1}^{n_{k+1}} b $$ for some $a,b \in R$. You can use the same proof as above to show that $p_{k+1}^{n_{k+1}} \mid a$ which will finish the induction.
Best Answer
Hint: in a Dedekind (or Prüfer) domain $\:(I+J)\: (I\cap J)\: =\: IJ\ \ $ (gcd $*$ lcm law)