The standard method (I think) for proving that decomposition fields of polynomials exist is by considering an irreducible polynomial $P$ (over the field $K$), and then noticing that $X + (P)$ is a root of $\tilde{P}$ in $K[X]/(P)$ (where $\tilde{P} \in (K[X]/(P))[Y]$ is the canonical image of $P$).

My question is about how many roots of $P$ one has added ?

Indeed, in cases like $K= \mathbb{R}$, since irreducible polynomials are either of degree $1$ (and then the root is in $\mathbb{R}$) or of degree $2$, but then $\tilde{P} = (Y-(X+(P))(Y+(X+(P)))$, so both roots were added.

However in other cases such as non perfect fields, a root can appear multiple times in an irreducible poynomial, and so $n$ roots weren't added ($n$ being the degree of $P$). And it seems to me that even in some perfect cases the question is non trivial, for instance for $K= \mathbb{Q}$ and $P= \Phi_n$, how many of the $n$th primitive roots of unity were added ?

So my question concerns the general case, but particular instances would be helpful as well (for instance how it works for perfect fields): how many roots does $\tilde{P}$ have in $K[X]/(P)$, when $P$ is an irreducible polynomial over $K$ ?

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    $\begingroup$ By construction, $\tilde P$ is guaranteed only to have at least one root in $K[X]/(P)$. For example, this is the case with $K=\Bbb Q$ and $P=X^3-2$. The full splitting field of $P$ would have degree 6 rather than 3. We will have several roots only if there exists a non-trivial polynomial $Q$ of degree $<\deg P$ such that $P(Q(X))$ is divisible by $P$. $\endgroup$ – Hagen von Eitzen Apr 9 '17 at 10:31
  • $\begingroup$ Ok I see. Isn't this last condition sufficient ? Indeed if $degQ < degP$, and if $Q\neq X$, then $Q +(P) \neq X +(P)$, so that $Q+(P)$ and $X+(P)$ are two distinct roots of $\tilde{P}$. And is this condition equivalent to anything "nicer" ? $\endgroup$ – Max Apr 9 '17 at 16:23

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