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Is there a name for an irreducible (separable, if you want) polynomial over a field such that adjoining one root of the polynomial splits the polynomial?

Such polynomials are discussed here for example.

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    I don't understand what you mean, do you mean adjoining a root to the ground field (that in which the coefficients are taken). In that case you mean that the field generated by one root contains all the other roots. THen the degree of the splitting field equals the degree as does the order of the Galois group which is minimal so the extension is cyclic. – 2017-02-12
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    @MarcBogaerts Yes, I mean adjoining a root to the ground field. As for your point, suppose I have a finite Galois extension $E/F$, where the Galois group isn't cyclic. Because the extension is finite and separable, by the primitive element theorem, it is simple, so $E = F(\alpha)$. So the minimal polynomial $m_{\alpha, F}$ of $\alpha$ is such that adjoining one root of it gives $F(\alpha)$, a normal extension. So $m_{\alpha, F}$ splits completely in $F(\alpha)= E$, and yet the Galois group of $m_{\alpha, F}$ (which equals $\operatorname{Gal}(E/F)$) is not cyclic.. – 2017-02-12
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    I'm a bit confused because I think of the case of a cubic with one real root and two complex (conjugate) roots. Then adjoining the real root to the rationals will only yield a field with real elements. So this field is not the splitting field and the Galois group is not cyclic since it contains the transposition of the two comples roots. – 2017-02-13
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    @MarcBogaerts If $f$ is an irreducible polynomial in $F[x]$, and $\alpha$ is a root of $f$. the proposition "$F(\alpha)$ splits $f \implies \operatorname{Gal}_F(f)$ is cyclic" is true when $f$ is of prime degree, but false in general. The reason its true in prime degree is that then the Galois group is a subgroup of $S_p$ of order $p$ acting transitively on $\{1, \dotsc, p\}$. Such a thing is necessarily a $p$ cycle when $p$ is prime. – 2017-02-13
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    @MarcBogaerts But there can be subgroups of $S_n$ of order $n$ acting transitively on $\{1, \dotsc, n\}$ for nonprime $n$. For instance, the subgroup $\{e, (12)(34), (13)(24), (14)(23) \} \triangleleft S_4$ is of order $4$ and acts transitively on $\{1, 2, 3, 4\}$. But it is not cyclic. – 2017-02-13
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    Indeed, this is another example where the Galois group has order equal to the degree of the polynomial, but this is not always the case, consider an example of a 4-th degree polynomial with Galois group $S_4$, so the splitting field has degree $24$ but the field generated by one root has only degree $4$. I agree that by the primitive element theorem the splitting field is generated by an element, but its minimal polynomial will have degree $24$. – 2017-02-13
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    The polynomial $x^4-4x^2+1$ has the property and its Galois group is non-cyclic of order 4. – 2017-02-15

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Your question is whether there’s an established name for these polynomials. As far as I know, there is no agreed-upon name, and it even seems to me that people haven’t bothered giving nonce-names to the concept either.

But now that I’m on my soapbox, why don’t I point out that if $L\supset K$ is a Galois extension and $\alpha$ is a primitive element (of which there are many many, of course), then the minimal $K$-polynomial for $\alpha$ has your property. If you want a ready-made infinite family of polynomials of this kind, you have the cyclotomic polynomials $\Phi_n$.