I've just asked this question in another post, but I'm reposting it here since it wasn't the only question there and it was suggested to me to ask just one question per post.
I would like to know if there is a generic method to prove or disprove that an ideal is principal. Before closing the question, someone commented that there isn't a single way to do it, so if that's the case I would like some review on two exercises I was given, where the ideals are in $\Bbb Z[X]$.
$(5, X^2 +3)$: I believe this ideal to not be principal, as if it were principal then the generator would need to be of degree $0$ to generate $5$, but $5$ is irreducible so it can only be generated by $1$ or $5$. The number $5$ cannot be a generator since $X^2 +3$ is in the ideal, but it also cannot be $1$ as the ideal is different from $\Bbb Z[X]$, for example $X$ is not in it,
$(X^2 +1, X+2)$: I believe this ideal to also not be principal. Those polynomials are irreducible and not associate, so if the ideal were to be generated by one element it would need to be a unit. But the remainder of division of $X^2+1$ by $X+2$ is $5$, so every constant in this ideal is multiple of $5$ and so the ideal is different from $\Bbb Z[X]$.
Are there some mistakes?
Best Answer
There is no general method, but here are suggestions in two cases: $\mathbf Z[x]$ and $\mathbf Z[\sqrt{d}]$.
Let $I$ be an ideal in $\mathbf Z[x]$. Every ideal in $\mathbf Z[x]$ is finitely generated (and your two examples are even explicitly defined with two generators each), so we can write $I = (f_1,\ldots,f_r)$. Let $d$ be the gcd of $f_1,\ldots,f_r$ in $\mathbf Z[x]$, which could be computed using irreducible factorizations of the $f_i$'s since $\mathbf Z[x]$ has unique factorization. This gcd is determined up to sign.
Claim: if $I$ is principal then $d$ must be a generator of $I$.
To prove the claim; suppose $I = (g)$. Then from $f_i \in I$ we have $f_i \in (g)$, so $g \mid f_i$ for all $i$. Thus $g \mid d$ by properties of unique factorization domains.
Since $g \in I = (f_1,\ldots,f_r)$, we have $g = f_1h_1+\cdots+f_rh_r$ for some $h_i$ in $\mathbf Z[x]$. Then $d$ dividing each $f_i$ implies $d \mid g$. That together with $g \mid d$ tells us $g = \pm d$, so $I = (g) = (d)$. This concludes the proof of the claim.
Your two examples are $(f_1(x),f_2(x))$ in $\mathbf Z[x]$ where $f_1(x)$ and $f_2(x)$ are relatively prime in $\mathbf Z[x]$. Therefore $\gcd(f_1(x),f_2(x)) = 1$ in $\mathbf Z[x]$. So if the ideal $(f_1(x),f_2(x))$ in $\mathbf Z[x]$ is principal then it must be $(1)$. Now it remains to check that $1 \not\in (f_1(x),f_2(x))$ in $\mathbf Z[x]$ in your two examples.
While ideals in $\mathbf Z[x]$ are not always principal, ideals in $\mathbf Q[x]$ are all principal and we can extend ideals in $\mathbf Z[x]$ to ideals in $\mathbf Q[x]$. When $A$ is a ring contained in a ring $B$ (like $\mathbf Z$ contained in $\mathbf Q$) each ideal $I$ in $A$ has an extension $IB$ to an ideal in $B$ that is defined to be the smallest ideal in $B$ containing $I$. Check
(i) $IB$ consists of all sums $\sum_{k=1}^n i_kb_k$ where $n \geq 1$, $i_k \in I$, and $b_k \in B$,
(ii) when $I = aA$ we have $IB = aB$,
(iii) when $I = a_1A + \cdots + a_mA$, we have $IB = a_1B + \cdots + a_mB$.
We can abbreviate (ii) and (iii) to $I = (a) \Rightarrow IB = (a)$ and $I = (a_1,\ldots,a_m) \Rightarrow IB = (a_1,\ldots,a_m)$, where you're expected to realize that the parentheses in the descriptions of $IB$ mean ideals generated in $B$ rather than in $A$.
Example. If $I = (2,x)$ in $\mathbf Z[x]$, then $I \not= \mathbf Z[x]$ since $1 \not\in I$, as all constants in $I$ are even, but $I\mathbf Q[x] = (2,x) = \mathbf Q[x]$ since $2$ is a unit in $\mathbf Q[x]$.
Starting with an ideal $I = (f_1(x),\ldots,f_r(x))$ in $\mathbf Z[x]$, its extended ideal $I\mathbf Q[x]$ must be principal because $\mathbf Q[x]$ is a PID. Then $I = (f(x))$ for some $f(x)$ in $\mathbf Q[x]$. By (iii) above, $I\mathbf Q[x] = (f_1,\ldots,f_r)\mathbf Q[x]$, so by the same argument used in $\mathbf Z[x]$ we know $f$ has to be the gcd of the $f_i$'s in $\mathbf Q[x]$, which is determined up to a nonzero constant multiple. Every nonzero polynomial in $\mathbf Q[x]$ has a constant multiple that is a primitive polynomial in $\mathbf Z[x]$, so we can take $f$ to be the gcd of $f_1,\ldots,f_r$ in $\mathbf Q[x]$ that's primitive in $\mathbf Z[x]$.
Suppose $I = (f_1,\ldots,f_r)$ is principal in $\mathbf Z[x]$, so $I = d\mathbf Z[x]$ where $d$ is the $\mathbf Z[x]$-gcd of $f_1,\ldots,f_r$. Then $(d) = (f)$ as ideals in $\mathbf Q[x]$, so $d = sf$ where $s \in \mathbf Q^\times$. Write $s$ in reduced form as $m/n$. Then $nd = mf$, an equation in $\mathbf Z[x]$. Compute the content (gcd of coefficients) of both sides: $n\,{\rm cont}(d) = m\,{\rm cont}(f) = m$ in $\mathbf Z$ since $f$ is primitive, so ${\rm cont}(d) = m/n = s$. Thus $d = sf = {\rm cont}(d)f$. If some $f_i$ is primitive then $d$, a factor of $f_i$ in $\mathbf Z[x]$, is also primitive, so ${\rm cont}(d) = 1$ and thus $d = f$: if $I$ is principal and some $f_i$ is primitive then $I$ must be generated as an ideal in $\mathbf Z[x]$ by the $\mathbf Q[x]$-gcd of the $f_i$'s that is scaled to be primitive in $\mathbf Z[x]$.
Everything I wrote about $\mathbf Z[x]$ carries over to $R[x]$ when $R$ has unique factorization.
Now let's look at a different family of rings: $\mathbf Z[\sqrt{d}]$ where $d$ is an integer that is not a perfect square.
Two key facts: (i) every nonzero ideal $I$ in $\mathbf Z[\sqrt{d}]$ has finite index (meaning the quotient ring $\mathbf Z[x]/I$ is finite) and (ii) each nonzero principal ideal $(a+b\sqrt{d})$, where $a$ and $b$ are integers, has index $|a^2 - db^2|$ in $\mathbf Z[x]$.
Here is how you can prove (i): pick a nonzero element $a+b\sqrt{d}$ in $I$. Then its norm $n := (a+b\sqrt{d})(a-b\sqrt{d})$ is a nonzero integer that's also in $I$ and $(n) \subset I \subset \mathbf Z[\sqrt{d}]$, so it suffices to prove principal ideals $(n)$ with nonzero $n$ in $\mathbf Z$ have finite index in $\mathbf Z[\sqrt{d}]$ and that follows from writing $(n) = \mathbf Z{n} + \mathbf Z{n}\sqrt{d}$, since it implies $\mathbf Z[\sqrt{d}]/(n) \cong (\mathbf Z/n\mathbf Z)^2$ as additive groups (not as rings!) and thus $(n)$ has index $n^2$ in $\mathbf Z[\sqrt{d}]$.
Here is how you can prove (ii): when $I = (a+b\sqrt{d})$, set $n = (a+b\sqrt{d})(a-b\sqrt{d})$. Then $(n) \subset I \subset \mathbf Z[\sqrt{d}]$, so $$ n^2 = [\mathbf Z[\sqrt{d}]:(n)] = [\mathbf Z[\sqrt{d}]:I][I:(n)], $$ where $[I:(n)]$ is the cardinality of the quotient group $I/(n) = (a+b\sqrt{d})/((a+b\sqrt{d})(a-b\sqrt{d}))$. In any integral domain $R$, with nonzero elements $\alpha$ and $\beta$, $\alpha R/(\alpha\beta)R \cong R/\beta R$ as additive groups, so $I/(n) \cong \mathbf Z[\sqrt{d}]/(a-b\sqrt{d})$. The conjugation mapping $x+y\sqrt{d} \mapsto x-y\sqrt{d}$ is a ring automorphism of $\mathbf Z[\sqrt{d}]$ that sends the ideal $(a+b\sqrt{d})$ to the ideal $(a-b\sqrt{d})$, so conjugation induces a ring isomorphism from $\mathbf Z[\sqrt{d}]/(a+b\sqrt{d})$ to $\mathbf Z[\sqrt{d}]/(a-b\sqrt{d})$. Thus $[\mathbf Z[\sqrt{d}]:(a-b\sqrt{d})] = [\mathbf Z[\sqrt{d}]:(a+b\sqrt{d})]$, so $[I:(n)] = [\mathbf Z[\sqrt{d}]:(a-b\sqrt{d})] = [\mathbf Z[\sqrt{d}]:(a+b\sqrt{d})]$. Hence $$ n^2 = [\mathbf Z[\sqrt{d}]:(n)] = [\mathbf Z[\sqrt{d}]:I][I:(n)] = [\mathbf Z[\sqrt{d}]:I]^2. $$ Now take square roots to get $[\mathbf Z[\sqrt{d}]:I] = |n|$.
Back to the general setting, when $I$ is a nonzero ideal in $\mathbf Z[\sqrt{d}]$ with index $n$, in order that $I$ be principal we have to be able to solve the equation $x^2 - dy^2 = n$ or $x^2 - dy^2 = -n$ in integers $x$ and $y$. So if those two equations have no integral solutions, then $I$ is not principal.
Example. In $\mathbf Z[\sqrt{5}]$ the ideal $I = (2,1+\sqrt{5})$ turns out to have index $2$, but the equations $x^2 - 5y^2 = \pm 2$ have no integral solutions because they have no solutions mod $5$. Thus the ideal $(2,1+\sqrt{5})$ in $\mathbf Z[\sqrt{5}]$ is nonprincipal.
Theorem 3.3 and examples in Section 4 here show you how to determine whether $x^2 - dy^2 = \pm n$ has an integral solution without having to rely on the modular arithmetic trick above.