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The Shafarevich conjecture states that the Galois group $\mathrm{Gal}({\overline{\mathbf{Q}}/\mathbf{Q}^{ab}})$ is a free profinite group, where $\mathbf{Q}^{ab}$ is the maximal abelian extension of $\mathbf{Q}$.

As far as I know, there is some cohomological evidences (Bloch-Kato conjecture/theorem) supporting the Shafarevich conjecture but it seems that we are still far to answer the conjecture in a clear way.

What will be the consequences in arithmetic and arithmetical algebraic geometry if the Shafarevich conjecture was true?

Would it be the last step in our "understanding" of the absolute Galois group?

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The Shafarevich conjecture belongs to the broader program of Inverse Galois theory, and in that context it is just another step in that particular approach to understanding $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$.

So the answer to the second question is definitely not. For example we could prove the Shafarevich conjecture and still don't know all the finite quotients of the absolute Galois group of the rationals.

Even our understanding of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}^{ab})$ could still be far from complete, since a lot the results are not constructive. For example Iwasawa solved the solvable part of the conjecture in the 50s, but I doubt that we know how to explicitly generate most finite solvable groups over $\mathbb{Q}^{ab}$.

On a side note, you might get a better idea of the impact of solving the conjecture by seeing what we have learned in the one case we have managed to solve, $\mathbb{Q}^{tr}(\sqrt{-1})$, where $tr$ indicates generated by all totally real algebraic numbers ($\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}^{tr}(\sqrt{-1}))$ is a free profinite group of countable rank by results of Pop and others).

The answer to the first question might be no as well, see there other two MO questions in the same spirit (1, 2). Trivially, the Shafarevich conjecture implies the inverse Galois problem over $\mathbb{Q}^{ab}$.

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  • $\begingroup$ Thank you for your viewpoint. Am I right if I say that your interpretation of "Understanding absolute Galois group" is quasi-exclusively guided by solving the inverse Galois problem ? $\endgroup$ – Muhammed Ali Dec 10 '16 at 14:26
  • $\begingroup$ No, I don't think so. There's serveral approaches to understanding it, which give different information about $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. For example the study of Galois representations (Artin, l-adic, mod p), the theory of dessin d'enfant, the Grothendieck-Teichmüller conjectures. There's plenty of conjectures and research programs (Langlands, anabelian geometry...) which try to understand the absolute Galois group in one way or another. $\endgroup$ – Myshkin Dec 10 '16 at 16:56
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Note that the Shafarevich conjecture was proved by Florian Pop for function fields in "Étale Galois covers of affine smooth curves. The geometric case of a conjecture of Shafarevich. On Abhyankar’s conjecture." Invent. Math. 120, No. 3, 555-578 (1995), and Iwasawa proved that the maximal prosolvable quotient of $G_{\mathbf{Q}^\mathrm{ab}}$ is a free prosolvable group, see [Neukirch-Schmidt-Wingberg, Cohomology of number fields], IX.5.

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One additional aspect may be worth mentioning: When trying to understand the structure of an absolute Galois group ${\rm Gal}(\overline{K}/K)$ of a field $K$, it is often helpful to keep in mind its action on the roots of unity, that is, to study the pair $({\rm Gal}(\overline{K}/K),\chi_K)$, where $\chi_K\colon {\rm Gal}(\overline{K}/K)\to\widehat{\mathbb{Z}}^\times$ is the cyclotomic character. Now by the Kronecker-Weber theorem, $\mathbb{Q}^{\rm ab}$ is obtained from $\mathbb{Q}$ by adjoining all roots of unity. Therefore the group ${\rm Gal}(\overline{\mathbb{Q}}/\mathbb{Q}^{\rm ab})$, which the Shafarevich conjecture attempts to describe, is just the kernel of cyclotomic charcter $\chi_{\mathbb{Q}}$, and this makes it particularly important.

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