My first step in computing the class group of $\mathbb Q(\sqrt{-55})$ was to compute the Minkowski bound. Initially, I said $\lambda(-55)=2\sqrt{-55}/\pi<2(8)/3<6$ and I went the normal way of finding an ideal of norm 5, $(5, \sqrt{-55})$ and this has a contribution to the class group since it is non-principal. But then I realised that I could get a better bound on $\lambda(-55)=2\sqrt{-55}/\pi<2(7.5)/3=5$. So in fact, I should have got no contribution from the ideal $(5, \sqrt{-55})$ and should have got that this ideal is in the same class as an ideal of norm $<5$.
I don't see how to get an ideal of norm $<5$ from $(5, \sqrt{-55})$ and I'm not sure why my argument above was wrong.
In more complicated situations, I'm worried that by making a mistake similar to the above, I would get much larger class groups than is correct. How do I avoid this mistake.
Is there a way of spotting if an ideal is in the same ideal class as an integral ideal of smaller norm, in general?
The first thing to note is that ideal classes are different from ideals, we say $I$ is in the class of $J$ if $\exists (a),(b)$ with $(a)I=(b)J$. With this in mind, consider $(4\sqrt{-55})(5,\sqrt{-55})=(20\sqrt{-55},-220)=(\sqrt{-55})$. Now consider the ideal $(2,\frac{1+\sqrt{-55}}{2})$. I claim this is in the same class as $(5,\sqrt{-55})$.
Indeed, consider $$(10\sqrt{-55})(2,\frac{1+\sqrt{-55}}{2})=(20\sqrt{-55},5\sqrt{-55}+275)=(\sqrt{-55})$$
This answers your original question.
To get an ideal of norm $<5$, you want to consider ideals with norm $2$ or $3$. The way we do this is to determine whether $(2) , (3)$ are ramified in $K=\mathbb{Q}(\sqrt{-55})$.
Since $-55\equiv 1\pmod{4}$ we know that Disc($K)=-55$, and since neither $2$ nor $3$ divides 55, we see that neither prime is ramified. Thus, it remains to decide whether they are split completely or inert.
$(3)$ remains a prime in $K$, since $3$ is not a quadratic residue modulo $-55$, so we can discount ideals of norm $3$.
$(2)\equiv1\pmod{8}$, so we know $(2)$ splits completely in $K$. The ideal I=$(2,\frac{1+\sqrt{-55}}{2})$ is a prime ideal of norm $2$, since $(2,\frac{1+\sqrt{-55}}{2})(2,\frac{1-\sqrt{-55}}{2})=(2)$. A computation shows that $I^2$ is not principal, and indeed, that $I^2\neq \bar I$. Hence, $I$ is not of order $3$. We compute either $I^3$, to give us $\bar I $, or $(I^2)^2$, to give us a principal ideal. Either way, we see that $I$ has order 4, and the class group is henceforth $\mathbb{Z}/4\mathbb{Z}$.