Consider the totally real cubic field $K = \mathbb{Q} (\alpha) = \mathbb{Q} [x]/(f)$ defined by $f = x^{3} - x^{2} - 9 x + 10$ (see https://www.lmfdb.org/NumberField/3.3.1957.1). The ring of integers in this particular case is $\mathcal{O}_K = \mathbb{Z} [\alpha] \cong \mathbb{Z}[x]/(f)$.
Now $\mathfrak{p} = (2,\alpha)$ is a prime ideal of norm $2$, and it is not principal, as one can check with computer calculations. (In fact it is a generator of the class group, which is $C_2$.) I was wondering if there's an easy way to see that.
There are algorithms (based on LLL reduction) that verify if any given ideal in $\mathcal{O}_K$ is principal, but I expect some better explanation that may be checked with less computations.
In fact, computer calculations confirm that there are no elements of norm $2$ in $\mathcal{O}_K$, and this would be enough to conclude that $\mathfrak{p}$ is not principal. Curiously, it seems like the equation $N_{K/\mathbb{Q}} (a + b\alpha + c\alpha^2) = 2$, while not having solutions $a,b,c \in \mathbb{Z}$, seems to have solutions in $\mathbb{Z}/p^k\mathbb{Z}$ for all small $p^k$. I checked this experimentally, but the cubic form in question may in fact violate the local-global principle, so there may be no easy way to show that there are no integer solutions.
Is there a good way to see that $\mathfrak{p}$ is not a principal ideal?
In the process of computing the class group, you should have shown that $(\alpha-2) = \mathfrak p^2$. Therefore if $\mathfrak p = (\beta)$ for some $\beta \in \mathcal O_K = \mathbf Z[\alpha]$, we'd have $(\alpha-2) = (\beta)^2 = (\beta^2)$, so $\beta^2 = (\alpha - 2)\varepsilon$ for some unit $\varepsilon \in \mathcal O_K^\times$. We want to show $(\alpha-2)\varepsilon$ is not a square in $\mathcal O_K$ for all $\varepsilon \in \mathcal O_K^\times$. How can we do that?
You might think "Oh no, I have to compute the unit group, which has rank 2. How do you compute a rank 2 unit group by hand?" We don't need the full unit group. What we need is to understand the unit group modulo squares because squared units can be absorbed into the hypothetical $\beta^2$.
The unit theorem tells us that $\mathcal O_K^\times = \pm u^\mathbf Z \times v^\mathbf Z$ for some independent units $u$ and $v$, which would mean $\mathcal O_K^\times/(\mathcal O_K^\times)^2 = \langle -1, \overline{u}, \overline{v}\rangle \cong (\mathbf Z/(2))^3$. A computer algebra package says fundamental units for $K$ are $u := \alpha-1$ and $v := \alpha-3$. It is easy to check $u$ and $v$ have norm $1$, so they are both units. It now suffices to show not that these are fundamental units (that is far more work than we need!), but that they together with $-1$ generate the units modulo squares. Since the structure of the unit group modulo squares is (as a multiplicative group with $\mathbf Z/(2)$ acting as exponents) $3$-dimensional, it suffices to show $-1$, $u$, and $v$ are independent modulo squares: none of the $7$ products (corresponding in linear algebra language to nontrivial "$\mathbf F_2$-linear combinations") $$ -1, u, v, -u, -v, uv, -uv $$ is a square. Obviously $-1$ is not a square in the number field, since cubic fields have a real embedding. To show the remaining six units are not squares, here are two approaches. The second is more robust (viable for testing a number is not a higher power rather than just not a square).
Method 1: If something is a square, then under every real embedding it has to be mapped to a positive number. The polynomial $f(x) = x^3 - x^2 - 9x + 10$ has three real roots, so you have three real embeddings to use. The real roots of $f(x)$ are $\alpha_1 \approx -3.0409$, $\alpha_2 \approx 1.1294$, and $\alpha_3 \approx 2.9115$. I will write $u_j$ and $v_j$ for the images of $u = \alpha-1$ and $v = \alpha -3$ under the real embedding that sends $\alpha$ to $\alpha_j$.
Since $\alpha_1 < 0$ we have $u_1 < 0$, $v_1 < 0$, and $-u_1v_1 < 0$, so $u$ and $v$ and $-uv$ are not squares. Since $\alpha_2$ is between $1$ and $2$, we have $-u_2 < 0$ and $u_2v_2 < 0$, so $-u$ and $uv$ are not squares. That leaves just $-v = 3 - \alpha$, whose image under all real embeddings is positive, so we did not succeed in showing $-v$ is not a square in $\mathcal O_K$ by finding a negative image under some real embedding. It's time to move on to Method 2.
Method 2: To show a number $\gamma$ in $\mathcal O_K$ is not a square, find a prime ideal $\mathfrak q$ such that $\gamma \bmod \mathfrak q$ is not a square in $\mathcal O_K/\mathfrak q$. (This is a fancy version of the fact that 10007 is not a square in $\mathbf Z$ since it's not a square mod $10$, or not a square mod $5$.) I was able to apply this idea to show $u, v, -u, -v, uv, -uv$ are not squares by using the unique prime ideals of norm $5$, $11$, and $13$, and a prime ideal of norm $17$ (there are three of those).
For example, the prime $\mathfrak q$ of norm $5$ is $(5,\alpha)$, so $\alpha \equiv 0 \bmod \mathfrak q$ and thus in the residue field $\mathcal O_K/\mathfrak q \cong \mathbf Z/(5)$ we have $v = \alpha-3 \equiv -3 \equiv 2$ and $2 \not= \Box$ in $\mathbf Z/5\mathbf Z$. So $v$ is not a square in $\mathcal O_K$. Also $-v \equiv 3 \mathfrak q$ and $3 \not= \Box$ in $\mathbf Z/5\mathbf Z$, so $-v$ is not a square in $\mathcal O_K$ either. We handled all the other units by Method 1, so I'll stop here with Method 2, but you should check that Method 2 handles everything by reducing each of the 6 units modulo some prime ideal of norm $5$, $11$, $13$, and $17$ to verify that it is not a square modulo a prime ideal and thus is not a square in $\mathcal O_K$.
Our goal is to show $(\alpha-2)\varepsilon$ is not a square for every unit $\varepsilon$. For this task, $\varepsilon$ only matters modulo unit squares. We know $\mathcal O_K^\times/(\mathcal O_K^\times)^2$ is represented by $\{1,-1,u,v,-u,-v,uv,-uv\}$, so it remains to show the eight products $$ \alpha-2, -(\alpha-2), u(\alpha-2), v(\alpha-2), -u(\alpha-2), -v(\alpha-2), uv(\alpha-2), -uv(\alpha-2) $$ are all not squares in $\mathcal O_K$. You can try this using Method 1 (real embeddings) or Method 2 (reduction modulo well-chosen prime ideals). For example, under the real embedding sending $\alpha$ to $\alpha_1 \approx -3.0409$, we have $\alpha_1 - 2 \approx -5.0409$, $u_1 = \alpha_1 - 1 \approx -4.0409$ and $v_1 = \alpha_1 - 3 \approx -6.0409$, so (i) $\alpha_1-2 < 0$, (ii) $-u_1(\alpha_1-2) < 0$, $-v_1(\alpha_1-2) < 0$, and (iii) $u_1v_1(\alpha_1-2) < 0$. Therefore $\alpha-2$, $-u(\alpha-2)$, $-v(\alpha-2)$, and $uv(\alpha-2)$ are all not squares. It remains for you to check that $-(\alpha-2)$, $u(\alpha-2)$, $v(\alpha-2)$, and $-uv(\alpha-2)$ are all not squares. Have fun!