# Minimal Polynomials for Algebraic Dependence?

Hello all!

I recently had a question concerning algebraic dependence that has thus far gone unanswered from my professors and texts, that I hope I can phrase properly here. When answering, please reference any papers or texts that you may happen to be citing so that I can look them up later! The statement I would like to be true is the following:

Let {$\alpha$,$\beta$} be a set of real numbers that are algebraically dependent over $\mathbb{Q}$. Then $\exists! f \in \mathbb{Z}[x,y]$ of lowest degree such that $f$ is primitive (in the sense of ring theory) and $f(\alpha,\beta) = 0$.

Does anyone know if this statement, or a close cousin perhaps, is true? If it is false, can we add/remove some hypotheses to make it true?

Thank you all in advance for your time and help!

-Richard

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Do you mean that $f$ is of lowest degreee in each variable? If not, we can take $\alpha = 1$, $\beta$ an irrational algebraic integer of minimal polynomial $p$, and then both $f(X,Y) = p(Y)$ and $f(X,Y) = p(Y)-p(0)X$ are solutions. –  AFK Jan 25 '11 at 2:12
Sorry, what does "primitive (in the sense of ring theory)" mean? –  Gerry Myerson Jan 25 '11 at 2:30
I'm confused. Are you asking whether "Q[X,Y]/I has transcendence degree 1 over Q" implies "I is principal"? –  Yaakov Baruch Jan 25 '11 at 3:40
@Gerry: I presume primitive means that there is no nonunit simultaneously dividing all the coefficients. –  Pete L. Clark Jan 25 '11 at 4:48
Perhaps to meet YBL's objection one wants those two reals to be transcendental? –  Gerry Myerson Jan 25 '11 at 6:36

It is true if both your numbers are transcendental. It follows from the fact that height one prime ideals of $\mathbb{Q}[x,y]$ are principal. This can be found in standard algebra textbooks. Otherwise, I believe it's false. There are already counterexamples in the comments.