The Grothendieck-Serre Correspondence : Obstructions to the construction of the cyclic group of order 8 as a Galois group? I am reading this note called as The Grothendieck-Serre Correspondence by Leila Schneps where this quote occurs:

the author still recalls Serre’s unexpected reaction of spontaneous delight upon being shown a very modest lemma on obstructions to the construction of the cyclic group of order 8 as a Galois group, simply because he had never spotted it himself.

I am quite confused. I read the statement as:

obstruction to the [construction of cyclic group of order 8] as a Galois group

We can create a cyclic group of order 8: $(\mathbb Z/8\mathbb Z, +, 0)$. So I'm not sure what the "galois group obstruction is". Is the above quote to be read as:

obstruction to the [construction of cyclic group of order 8 as a Galois group] ?

I am still confused, because I thought the field $\mathbb Q(\zeta_8)/\mathbb Q$ has galois group $\mathbb Z/8\mathbb Z$ (where $\zeta_8$  is the $8$-th root of unity), from Kummer theory?
So what obstruction is the quote referring to which delighted Serre?
 A: This is more of a long comment than an answer (I believe I found an exact reference to the obstruction, but my summary of it might not be correct).
One guess is the obstruction discussed in "On Cyclic Field Extensions of Degree 8" (a paper written by the author of the article you cite, and which also mentions Serre).
The correct statement of the problem is: there is no "versal" polynomial for $\mathbb Z/8\mathbb Z$ over $\mathbb Q$, a polynomial with generic coefficients whose Galois group over its coefficient field is that group, and whose specializations give rise to all the polynomials with that galois group (see the article for more details).
The references therein explain the obstruction. I only skimmed things quickly and might be badly misunderstanding what is going on, but I think the idea is that if there was such a polynomial existed then you could find a specialization unramified and inert over $2$, but that contradicts the fact that every extension of $\mathbb Q$ with Galois group $\mathbb Z/8\mathbb Z$ such that 2 doesn’t split is ramified over $2$. Hopefully someone more knowledgeable can jump in.
