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Let $P(X)$ be an irreducible polynomial in $\mathbb Q[X]$. The field $L = \mathbb Q[X]/(P(X))$ will have at least one root $\alpha$ of $P(X)$ but it may have more.

Can all the other roots $L$ has be expressed as polynomials in $\alpha$ and how are those polynomials computed?

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    All the other roots in $L$ or all the other roots in a splitting field of $P$? – lhf Apr 28 '20 at 12:41
  • @lhf, $L$, thank you for the question. –  Apr 28 '20 at 12:51
  • By the very definition of $L$, any element $a\in L$ is the class modulo $P$ of some polynomial $R\in \mathbb{Q}[X]$. Then of course $a=R(\alpha)$, so any element of $L$ can be expressed as a polynomial in $\alpha$. – Captain Lama Apr 28 '20 at 15:03
  • See also https://math.stackexchange.com/questions/3647084/the-galois-group-of-polynomial-px-in-mathbbkx-is-cyclic-and-is-generated – lhf Apr 28 '20 at 22:11

1 Answers1

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Let $P(X)$ be an irreducible polynomial over $\mathbb Q$ with degree $d$ and pick a root $\alpha$. Let us have the fields $A = \mathbb Q(\alpha)$ and $B$ the splitting field of $P(X)$. $A$ has degree $d$ and is contained in $B$.

If $A = B$ then $P(X)$ splits into linear factors, this is equivalent to the roots being expressible as polynomials in $\alpha$.

$A = B$ happens iff $[B : \mathbb Q] = d$ in other words the Galois group has size d.

Example 1: degree 5 polynomial with group $C_5$

polynomial from https://www.lmfdb.org/NumberField/?hst=List&galois_group=C5&search_type=List

? p(x) = x^5 - 10*x^3 - 5*x^2 + 10*x - 1
%23 = (x)->x^5-10*x^3-5*x^2+10*x-1
? lift(factor(Mod(p(x),p(a))))
%24 = 
[                                              x - a 1]

[  x + (-3/7*a^4 + 2/7*a^3 + 24/7*a^2 + 6/7*a - 6/7) 1]

[x + (-2/7*a^4 - 1/7*a^3 + 23/7*a^2 + 18/7*a - 25/7) 1]

[   x + (1/7*a^4 - 3/7*a^3 - 8/7*a^2 + 19/7*a + 9/7) 1]

[ x + (4/7*a^4 + 2/7*a^3 - 39/7*a^2 - 36/7*a + 22/7) 1]

Example 2: degree 4 with group $V_4$

polynomial from https://www.lmfdb.org/NumberField/?hst=List&galois_group=4T2&search_type=List

? p(x) = x^4 + 9
%25 = (x)->x^4+9
? lift(factor(Mod(p(x),p(a))))
%26 = 
[      x - a 1]

[      x + a 1]

[x - 1/3*a^3 1]

[x + 1/3*a^3 1]

Non-Example. Degree 6 with group $S_3 \times C_3$.

? p(x) = x^6 - x^3 + 7
%37 = (x)->x^6-x^3+7
? lift(factor(Mod(p(x),p(a))))
%38 = 
[                 x - a 1]

[x + (-1/3*a^4 + 2/3*a) 1]

[ x + (1/3*a^4 + 1/3*a) 1]

[       x^3 + (a^3 - 1) 1]

In this case we get 3 linear factors and then another non-linear factor. I don't know how to characterize this sort of situation.