(Minimal?) Polynomials using the Nullstellensatz I'm struggling with an exercise that was asked in class:

Let $\alpha = \sqrt[3]{3} + \sqrt{7}\sqrt[4]{2}.$ Show that there is a polynomial $p$ in
  the ideal $I=\left&lta^3 - 3, b^2 - 7, c^4-2, \tilde{\alpha}-(a+bc)\right> \subset \mathbb{Q}[a,b,c,\tilde{\alpha}]$ depending only on $\tilde{\alpha}$ with $p(\alpha) = 0$ (ie substituting the real number $\alpha$ for the variable $\tilde{\alpha}$ gives zero). If we can
  choose this polynomial to be irreducible, it is the minimal polynomial of $\alpha$. You may assume the Nullstellensatz.

The approach I tried was to say that over $\mathbb{C}$, $V(I)$ is the points $(\omega^d\sqrt[3]{3},\pm\sqrt{7},\pm i\sqrt[4]{2}, \omega^d\sqrt[3]{3}+ \pm\sqrt{7}\pm i\sqrt[4]{2})$, where $\omega$ is a cube root of 1.
Hence the polynomial $ P = \Pi (\tilde{\alpha} -  (\omega^d\sqrt[3]{3}+ \pm\sqrt{7}\pm i\sqrt[4]{2}))$ is in $I(V(I))$, so there is some $N > 0$ with $P^N \in I \subset \mathbb{C}[a,b,c,\tilde{\alpha}]$ (and obviously $\alpha$ is a root of this).
I was then trying to show that this must be in $\mathbb{Q}[a,b,c,\tilde{\alpha}]$; but I couldn't get there (also don't think this is true in general).
Can anyone offer any guidance?
I know this is really representative of a general method of finding the polynomials that vanish on the sums and products of algebraic numbers given their minimal polynomials. Eg if $\alpha, \beta$ solve $f, g$, this gives polynomials that vanishe on $\alpha + \beta$ or $\alpha\beta$... Any insight on why this works in the general case would be great too.
Thanks!
(N.b. I've adited this for accuracy: I was assuming before that this gives minimal polynomials, rather than just ones that vanish.)
 A: I should rewirte this answer.
Lemma
Let $\mathfrak{p}$ be a prime ideal of $\mathbb{Q}[x_1,\ldots,x_n]$. Let $F/\mathbb{Q}$ be a finite Galois extension. Let $Gal(F/\mathbb{Q})=\{\sigma_1,\ldots,\sigma_f\}$. Let $S=\{\mathfrak{p}_i\}$ be the set of prime ideals of $F[x_1,\ldots,x_n]$ which are lying over $\mathfrak{p}$. Then $G=Gal(F/\mathbb{Q})$ may act transitively on $S$. In particular, $S$ is finite. 
Proof 
First, $G$ acts on $S$. Suppose $\sigma\in G$. Then $\sigma(\mathfrak{p}_i)\cap \mathbb{Q}[x_1,\ldots,x_n]\supset \mathfrak{p}(=\sigma(\mathfrak{p}))$ and $\sigma^{-1}(\sigma(\mathfrak{p}_i)\cap \mathbb{Q}[x_1,\ldots,x_n])\subset \mathfrak{p}_i\cap \mathbb{Q}[x_1,\ldots,x_n]=\mathfrak{p}$, hence $G$ acts on $S$. 
Second, $G$ acts transitively on $S$. Let $\mathfrak{p}_1,\mathfrak{p}_2\in S$ and $x\in \mathfrak{p}_1$. Then $\sigma_1(x)\sigma_2(x)\cdots \sigma_f(x)\in \mathfrak{p}\subset \mathfrak{p}_2$. Hence there exists some $\sigma$ such that $\sigma(x)\in \mathfrak{p}_2$, in other word, $x\in \sigma^{-1}(\mathfrak{p}_2)$. So $\mathfrak{p}_1\subset \bigcup_{i}\sigma_i(\mathfrak{p}_2)$. By prime avoidance theorem, $\mathfrak{p}_1=\sigma(\mathfrak{p}_2)$ for some $\sigma\in G$.
Now, come back to our problem: 
Let $I\subset \mathbb{Q}[x_1,\ldots,x_n,y]:=\mathbb{Q}[X,Y]$ be an ideal such that $\mathbb{Q}[X,Y]/I$ is of dimension zero. Suppose $L$ is a field finite over $\mathbb{Q}$ such that the maximal ideals $\mathfrak{m}_i,i=1,2,\ldots,l$ containing $I$ are $L$-coefficients. Write $\mathfrak{m}_i=(X-A_i,Y-b_i)$, and $P(Y)=(Y-b_1)(Y-b_2)\cdots(Y-b_l)$. We ask if $P(Y)\in \mathbb{Q}[Y]$?
The answer is Yes.
Let $F$ be a finite Galois extension of $\mathbb{Q}$, which contains $L$. Then it is clear to see that $(X-A_i,Y-b_i)$ are already maximal in $F[X,Y]$. So all maximal ideals in $F[X,Y]$ containing $I$ are exactly $\mathfrak{m}_iF[X,Y]$. Thus we may assume $L$ is a finite Galois extension of $\mathbb{Q}$.
Let $\sqrt{I}=\bigcap \mathfrak{n}_{i}$ in $\mathbb{Q}[X,Y]$. Let $S_i$ be the set of maximal ideals which lying over $\mathfrak{n}_i$ in $L[X,Y]$. Denoted $S$ the set of the maximal ideals $\mathfrak{m}_i$. Then $S$ is a disjoint union of $S_i$. And $G$ may act on $S$ having orbits $S_i$, $P(Y)=\prod_i P_i(Y)$ each $P_i(Y)\in \mathbb{Q}[Y]$. 
Example Let us consider a concrete example. Consider $(x^2-2,y)$ in $\mathbb{Q}[x,y]$ is maximal, and $(x-\sqrt{2},y),(x+\sqrt{2},y)$ in $\mathbb{Q}[\sqrt{2}][x,y]$. Now our $P(y)=y^2$ which is not irreducible. However, this example suggests that $P(y)$ is a power of an irreducible polyonomial. 
This is true!
lemma
Let $\mathfrak{n}$ be a maximal ideal of $\mathbb{Q}[X,y]$ and $L$ a finite Galois extension of $\mathbb{Q}$ such that the fiber of $\mathfrak{n}$ in $L[X,y]$ are all $L$-points. We construct $P(Y)$ as in the way above. Let $S$ be the set of maximal ideals $\mathfrak{m}_i=(X-a_i,Y-b_i)$ in $L[X,Y]$ which lying over $\mathfrak{n}$. Let $\mathfrak{n}\cap \mathbb{Q}[y]=(f(y))$ where $f(y)$ is a monic polynomial thus a monic irreducible polynomial. Then $P(Y)$ is a power of $f(y)$.
Proof
Observing that $G$ acts transitively on $S$, thus the group $H_i=\{\sigma\mid \sigma(b_i)=b_i\}$ are all conjugate. It follows that $P(y)=f(y)^{|H_1|}$. In particular, if all $b_i$ are different, then $P(y)=f(y)$.
Practice In practice, let $\alpha_1,\ldots,\alpha_s$ be algebraic numbers and $\beta=g(\alpha_1,\ldots,\alpha_s)$ for $g\in \mathbb{Q}[x_1,\ldots,x_s]$. We let $I=(f_1(x_1),\ldots,f_s(x_s),y-g(x_1,\ldots,x_s))$ where $f_i$ are minimal polynomials of $\alpha_i$ over $\mathbb{Q}$. If certain good conditions holds, for example, the extension $\mathbb{Q}[\alpha_1,\ldots,\alpha_i]/\mathbb{Q}[\alpha_1,\ldots,\alpha_{i-1}]$ has degree $deg(\alpha_i)$ for $i=1,\ldots,s$, then our $I$ is maximal. More, if we require $g(x_1,\ldots,x_s)$ are all distinct for substituting the roots of $f_i$. Then our $P(y)$ is irreducible polynomial of $\beta$ over $\mathbb{Q}$.

No Galois theory, no Nullstellensatz, but use ``The fundamental theorem of symmetric polynomials''.
Let $\alpha_1,\alpha_2,\ldots, \alpha_s$ be algebraic numbers. Let $\beta=g(\alpha_1,\ldots,\alpha_s)$ for some $g\in \mathbb{Q}[x]$. Let $f_i$ be the minimal polynomial of $\alpha_i$. Let $P(y)=\prod (y-g(\theta_1,\ldots,\theta_s))$ where $\theta_i$ runs through the roots of $f_i$. Then $P(y)$ is a polynomial of $\mathbb{Q}[y]$.
Proof
Let $\gamma_{i}$ be the roots of $f_1$. Then $\prod_i(y-g(\gamma_i,x_2,\ldots,x_s))$ is a symmetric polynoimal of $\gamma_i$ in $\mathbb{Q}[x_2,\ldots,x_s,y][\gamma_1,\ldots,\gamma_t]$ (View $\gamma_i$ being indeterminate). Since the element symmetric polynomials of $\gamma_1,\ldots,\gamma_t$ have vaule in $\mathbb{Q}$. Hence  $\prod_i(y-g(\gamma_i,x_2,\ldots,x_s))$ is a polyonomial in  $\mathbb{Q}[x_2,\ldots,x_s,y]$. Finally $P(y)\in \mathbb{Q}[y]$.
A: *Update: there is something wrong in the proof as pointed out. We should use Galois extension. And take Galois actions to $P(y)$. *
Let $\mathfrak{n}$ be a maximal ideal of $\mathbb{Q}[x_1,x_2,\ldots,x_n,y]:=K$.  We may find a finite field extension $L/\mathbb{Q}$, such that $V(\mathfrak{n})=\cup\{(x_1-a_1,\ldots,x_n-a_n,y-b)\}=\{\mathfrak{m}_1,\ldots,\mathfrak{m}_k\}$(i.e. the closed points containing $I$ are $\mathbb{Q}[\alpha_1,\ldots,\alpha_m]$-points) in $\operatorname{Spec}L[x_1,\ldots,x_n,y]$
Now we get a polynomial $P$ by multiplying all $(y-b)$, and we ask that if $P(y)\in\mathbb{Q}[y]$?
This answer is yes! 
update: Now we may assume $L/\mathbb{Q}$ is Galois, suppose $Gal(L/\mathbb{Q})=\{\sigma_1,\ldots,\sigma_l\}$, then for $\sigma_j$ act on $\mathfrak{m}_i$ will be a maximal ideal containing $\sigma_j(\mathfrak{n})=\mathfrak{n}$, that is to say $Gal(L/\mathbb{Q})$ act on the set $\{\mathfrak{m}_1,\ldots,\mathfrak{m}_k\}$(the fiber of $\mathfrak{n}$), so we will obtain $P(y)\in \mathbb{Q}[y]$, but we have not said that $P(y)$ is irreducible.
The map $\mathbb{Q}[y]\to\mathbb{Q}[x_1,\ldots,x_n,y]\to L[x_1,\ldots,x_n,y]$ sends closed points to closed points. And $\mathfrak{m}_1\cdots\mathfrak{m}_k\cap \mathbb{Q}[y]=(f(y))=\mathfrak{n}\cap \mathbb{Q}[y]$. Hence $f(y)=P(y)G(y)$ for some $G(y)\in L[y]$(Here we just let $x_i=a_i$ to get this equation). But all the closed points in $f(y)$ are appeared, we obtain that $G$ is a constant. Now $P(y)$ is monic, we can see $P(y)\in \mathbb{Q}[y]$.
Notice: here we use the separablity of $L/\mathbb{Q}$!. 

For general, we only suppose $I$ is an ideal of $\mathbb{Q}[x_1,x_2,\ldots,x_n,y]$ such that $\mathbb{Q}[x_1,x_2,\ldots,x_n,y]/I$ is of dimension zero. We find a finite field extension $L/\mathbb{Q}$ such that the closed points over $I$ in $L[x_1,\ldots,x_n,y]$ are $L$-points. Then $\mathfrak{m}_1\cdots\mathfrak{m}_k\cap\mathbb{Q}[y]=(f(y))$, now $f(y)$ is a monic nonzero square-free polynomial. So our $P(y)$ equals $f(y)$.

In the OP's case, the ideal $I$ is maximal so the polynomial $P(y)$ is indeed an irreducible polynomial over $\mathbb{Q}$.
