Generators of the Relations of a Galois Extension Let $K$ be a Galois extension of $\mathbb{Q}$ of degree $n$. Pick some primitive element and take the roots $a_1, ..., a_n$ of its minimal polynomial. Then the evaluation map $\mathbb{Q}[x_1, ..., x_n] \to K$ with $x_i \mapsto a_i$ gives a surjection to $K$. The kernel is a maximal ideal $M$ of algebraic relations between $a_1, ..., a_n$. Is it true that $M$ is generated as an ideal by the polynomials of $\mathbb{Q}[x_1, ..., x_n]$ that are fixed by the Galois group of $K$ acting on $\mathbb{Q}[x_1, ..., x_n]$ and are contained within $M$?
 A: A local version of the assertion seems to be true: the maximal ideal $M\mathbb{Q}[x_1,\ldots x_n]_M$ of the local ring $\mathbb{Q}[x_1,\ldots x_n]_M$ is generated by the set of Galois-invariant polynomials contained in $M$, or of course even a finite subset of them.
Let $H$ be the group of automorphisms of $B:=\mathbb{Q}[x_1,\ldots x_n]$ obtained through lifting the elements $g\in G$. That is the lift $h$ of an element $g$ permutes the $x_i$ in the same way, in which $g$ permutes the roots $a_i$.
Let $L\subset\mathbb{Q}(x_1,\ldots ,x_n)$ be the fixed field of $H$. Let $A:=B\cap L$. Then $A$ consists of all $H$-invariant polynomials.
The ring extension $A\subset B$ is integral: all the conjugates of an element $b\in B$ lie in $B$. Hence the coefficients of the minimal polynomial of $b$ over $L$ lie in $A$, that is $b$ is integral over $A$.
As a polynomial ring $B$ is integrally closed and thus $A$ is integrally closed in $L$. Moreover $B$ is the integral closure of $A$ in $\mathbb{Q}(x_1,\ldots ,x_n)$. Altogether we have a situation, in which the ramification theory of prime ideals can be applied - see for example Abhyankar's book "Ramification theoretic Methods in Algebraic Geometry".
Let $P:=M\cap A$. Then $P$ is maximal and $A/P=\mathbb{Q}$.
For every $h\in H$ we have $h(M)=M$, because by definition of $H$ we have
$(h(p(x_1,\ldots ,x_n)))(a_1,\ldots,a_n)=g(p(a_1,\ldots ,a_n))=0$
for every $p(x_1,\ldots ,x_n)\in M$. (Here $h\in H$ is the lift of $g\in G$.)
Since the prime ideals of $B$ lying over $P$ are $H$-conjugates, we conclude that $M$ is the only prime of $B$ lying over $P$. Or in other words: the decomposition group $D$ of $M$ equals $H$.
The map $H\rightarrow G$, that exists by construction, can also be described as follows: an automorphism $h$ is mapped to the automorphism
$\overline{h}: B/M\rightarrow B/M,\quad b+M\mapsto h(b)+M$.
In ramification theory the kernel $I$ of the map $h\mapsto\overline{h}$ is called the inertia group of $M$. In our case by construction $I$ is the trivial group.
On general on the other hand ramification theory yields the following facts: let $B(D)$ respectively $B(I)$ be the integral closures of $A$ in the fixed fields $\mathbb{Q}(x_1,\ldots ,x_n)^D$ and $\mathbb{Q}(x_1,\ldots ,x_n)^I$. Then
$(M\cap B(D))\, B(I)_{M\cap B(I)}$ (*)
is the maximal ideal of the local ring $B(I)_{M\cap B(I)}$.
In the present case $I$ is the trivial group, so that $B(I)=B$. Moreover $D=H$, so that $B(D)=A$. Consequently we get
$PB_M = MB_M$.
Since $P$ consists of all $H$-invariant polynomials contained in $M$, this proves the assertion.
Note that the proof works for an arbitrary field instead of $\mathbb{Q}$.
Remark: in the case of Dedekind domains the prime factorization allows to prove the equation (*) without localization, thus giving the assertion in the form originally posted. However in the present situation $B$ is never Dedekind.
A: $\def\QQ{\mathbb{Q}}$ Yes. Moreover, this is true more generally. Let $f$ be any separable polynomial over $\QQ$. Let $K$ be the splitting field of $f$, let $\theta_1$, ..., $\theta_n$ be the roots of $f$ in $K$ and let $G = \mathrm{Gal}(K/\QQ)$. Map the polynomial ring $\QQ[x_1, \ldots, x_n]$ to $K$ by $x_i \mapsto \theta_i$ and let $M$ be the kernel of this map. Then I claim that $M$ is generated by $G$-invariant polynomials. This is stronger than what you ask for, in that I don't assume that the $\theta_i$ are primitive in $K$, or even that $f$ is irreducible. Also, I use $\QQ$ for notational convenience; this argument works over any field.
As in Hagen's answer, let $B = \QQ[x_1, \ldots, x_n]$ and let $A$ be the ring of invariant functions, $A=B^G$. Let $P = A \cap M$ and set $Q = BP$. We want to prove that $M=Q$, the obvious statement is $M \subseteq Q$. Let $X$ denote the set $\{ \theta_1, \theta_2, \ldots, \theta_n \}$.
Lemma For each $r$, the polynomial $f(x_r)$ is in $Q$.
Proof Let $f(x) = x^n - f_1 x^{n-1} + f_2 x^{n-2} - \cdots \pm f_n$, so the $f_k$ are in $\QQ$. Let $e_k$ denote the $k$-th elementary symmetric polynomial in the $x_i$. So $e_k - f_k$ is $G$-invariant (in fact, $S_n$ invariant) and is in $\mathrm{Ker}(B \to L) = M$. So $e_k - f_k \in P$.
Now, we have $0 = \prod_i (x_r - x_i)$ because one of the factors is $x_r-x_r$. Expanding, 
$$0 = x_r^n - e_1 x_r^{n-1} + e_2 x_r^{n-2} - \cdots = f(x_r) - (e_1-f_1) x_r^{n-1} + (e_2 - f_2) x_r^{n-2} - \cdots.$$
So $f(x_r) = (e_1-f_1) x_r^{n-1} - (e_2 - f_2) x_r^{n-2} + \cdots$ as an valid identity in the polynomial ring $B$, and we deduce that $f(x_r)$ is in the $B$-ideal generated by $e_k-f_k$. We already showed $e_k - f_k \in P$, so this shows that $f(x_r) \in BP = Q$. $\square$.
Unfortunately, it is not true that $\QQ[x_1, \ldots, x_n]/\langle f(x_1), \ldots, f(x_r) \rangle = K$. Note that $K[x_1, \ldots, x_n]/\langle f(x_1), \ldots, f(x_r) \rangle$ is a direct sum of copies of $K$, indexed by $X^n$, so $\QQ[x_1, \ldots, x_n]/\langle f(x_1), \ldots, f(x_r) \rangle$ is a subring of this. Functions in $\QQ[x_1, \ldots, x_n]$ which vanish on the $G$-orbit of $(\theta_1, \ldots, \theta_n)$, but not on the rest of $X^n$, will be in $M$ but not in $\langle f(x_1), \ldots, f(x_r) \rangle$. By thinking about elements of $K[x_1, \ldots, x_n]/\langle f(x_1), \ldots, f(x_r) \rangle$ as functions on $X^n$, we will eventually find an element $h$ of $P$ so that $M = \langle f(x_1), \ldots, f(x_n), h \rangle$. We go through several intermediate steps:
Choose a polynomial $\lambda(x_1, \ldots, x_n)$ in $L[x_1, \ldots, x_n]$ which vanishes at $(\theta_1, \ldots, \theta_n)$ but not at any other point of $X^n$. (If our ground field is infinite, just choose random constants $c_i$ in the ground field and the linear equation $\sum c_i (x_i - \theta_i)$ will not pass through any of the other points.
Let $\mu(x_1,\ldots, x_n) = \prod_{g \in G} \lambda^g(x_1, \ldots, x_n)$, where the group action is on the coefficients of $\lambda$ and not on the variables. Then the coefficients of $\mu$ are $G$-invariant, so $\mu \in \QQ[x_1,\ldots, x_n]$, and $\mu$ vanishes on the $G$-orbit of $(\theta_1, \ldots, \theta_n)$, while not vanishing on any other point of $X^n$.
Set $\nu(x) = \prod_{g \in G} \mu(x_{g(1)}, \ldots, x_{g(n)})$. Then $\nu$ is a $G$-invariant element of $k[x_1, \ldots, x_n]$. In other words, $\nu \in P$. Moreover, $\nu$ vanishes on the $G$-orbit of $(\theta_1, \ldots, \theta_n)$ and nowhere else on $X^n$. 
It is in fact already true that $M = \langle f(x_1), f(x_2), \ldots, f(x_r), \nu \rangle$. However, to make the proof easier, we go one step further. $Y$ be the set of values of $\nu$ on  $X^n \setminus G \cdot (\theta_1, \ldots, \theta_n)$. So $Y$ is a Galois invariant set not containing $0$. Set $h = 1-\prod_{y \in Y} (1-t/y)$. Since $Y$ is Galois invariant, $h$ is a polynomial in $\nu$ with $k$ coefficients, and hence in $A$. Moreover, $h$ vanishes at $(\theta_1, \ldots, \theta_n)$, so $h \in P$. Finally, $h$ is $1$ at all points of $X^n$ other than the $G$-orbit of $(\theta_1, \ldots, \theta_n)$.
I will show that $M = \langle f(x_1), \ldots, f(x_n), h \rangle$. Let $s \in M$. Then $s(1-h)$ vanishes on $X^n$. This shows that $s(1-h) = \sum g_i f(x_i)$ for some $g_i \in B$. But then $s = sh + \sum g_i f(x_i)$, proving the claim. 
