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.
You actually already proved the difficult part.
If $J$ is maximal, then by the correspondence of ideals you mention $J'$ has to be maximal, too. If it weren't, there would be a proper ideal $T'\supset J'$, which would give a proper ideal $T\supset J$.
Now assume $J$ prime and let $a'b'\in J'$, where $a',b'$ are the classes in $R/I$ of some $a,b\in R$. This means that there are $i_a,i_b\in I$ such that $(a+i_a)(b+i_b)=ab+ai_b+bi_a+i_ai_b\in J+I=J$, as $I\subseteq J$. Thus $ab\in J$ and, say, $a\in J$. Hence $a'\in J'$, so $J'$ is prime.
Finally, suppose $J$ radical and let $(a')^r\in J'$ for some $a\in R$, $r\in\Bbb N$. Then there are some $i_a,i\in I$ such that, by binomial expansion and since $I$ is an ideal, $(a+i_a)^r=a^r+i\in J+I=J$. Hence $a^r\in J$, so $a\in J$. Therefore $a'\in J'$ and $J'$ is radical.
Best Answer
The radical ideals are precisely those which can be written as intersections of prime ideals.
In a PID, the prime ideals are $\{0\}$ and $(p)$ for prime elements $p$. Hence, the radical ideals are $(0)$ and $\cap_i (p_i)$ for prime elements $p_i$. If there are infinitely many of them, the intersection is $(0)$. If not, we get $(\prod_i p_i)$. Hence, the radical ideals are those ideals $(n)$ for which $n$ is square-free.