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 $I=\left<a^3 - 3, b^2 - 7, c^4-2, \tilde{\alpha}-(a+bc)\right> \subset \mathbb{Q}[a,b,c,\tilde{\alpha}]$ 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.


(N.b. I've adited this for accuracy: I was assuming before that this gives minimal polynomials, rather than just ones that vanish.)

  • $\begingroup$ Is it really a method that is supposed to find a minimal polynomial or just some (nonzero) polynomial that has the sum and product of algebraic numbers as a root? It seems like the "choose to be irreducible" assumption is the only reason you're supposed to get a minimal polynomial out of this. $\endgroup$ – KCd Apr 19 '12 at 0:23
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    $\begingroup$ See Theorem 2.3 and Example 2.4 at math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod2.pdf for a method using tensor products that produces a monic polynomial over a given field having a sum or product of numbers as a root when you already know monic polynomials over that field with the individual terms as roots. There's no assurance that the result is going to be minimal, but "random" examples will likely have that feature (in the same sense that a random irreducible has Galois group $S_n$). $\endgroup$ – KCd Apr 19 '12 at 0:26
  • $\begingroup$ $I$ is certainly contained in ${\bf Q}[a,b,c,\tilde\alpha]$, so if you have proved $P^N$ is in $I$, then you have proved that $P^N$ is in ${\bf Q}[a,b,c,\tilde\alpha]$, no? Or have I misunderstood your "but I couldn't get there"? $\endgroup$ – Gerry Myerson Apr 19 '12 at 0:31
  • $\begingroup$ @GerryMyerson, the later $I$ mentioned is indeed the $IV(I)=I\mathbb{C}[a,b,c,\tilde{\alpha}]$ in the OP's meaning. $\endgroup$ – wxu Apr 19 '12 at 0:52
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    $\begingroup$ The convention is, if you find that the problem as stated is not the problem you wanted to state, then edit, but do it transparently (e.g., write enough stuff so people won't wonder why there are comments that don't jibe with the question). $\endgroup$ – Gerry Myerson Apr 19 '12 at 2:12

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 $P(y)\in \mathbb{Q}[y]$.

  • $\begingroup$ That's great. Thanks very much! $\endgroup$ – Tom H Apr 20 '12 at 17:31

*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 $\mathbb{Q}$.

  • $\begingroup$ I'm afraid you lose me slightly in the third paragraph: why does $f(y) = P(y)G(y)$? Thanks again! $\endgroup$ – Tom H Apr 18 '12 at 20:50
  • $\begingroup$ I donot know how to edit a wrong answer, maybe I should delete the origin words? $\endgroup$ – wxu Apr 19 '12 at 0:49
  • $\begingroup$ I can just about follow this, but there was no Galois theory to be seen in this course, and 'fiber' is a rather alien word to me. I can't help but wonder if there's a slightly more basic proof to be found...? :/ $\endgroup$ – Tom H Apr 19 '12 at 2:29
  • $\begingroup$ @TomH, Ok, let me think a think. $\endgroup$ – wxu Apr 19 '12 at 4:47

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