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Let $\mathbb{Q}[a]=\{r \in \mathbb{R}|r \text{ is algebraic over }\mathbb{Q}(a)\}$, and $f(x)$ be a polynomial with rational coefficients such that $f(p)=q$. Thanks to the answers to my question here is now clear to me that $p$ being algebraic implies $q$ being algebraic too; Now I'm wondering why this implies also $\mathbb{Q}[q]=\mathbb{Q}[p]$. Since we can express $q$ in terms of $p$, it's clear that $\mathbb{Q}[q] \subset \mathbb{Q}[p]$, but I can't see the converse.

EDIT:The statement above is written in the short paper "Versatile Coins" , Szalkai and Velleman, The American Mathematical Monthly 1993, pag.30 (last line).

I give more information about the context: the polynomial $f(x)$ above is of the form $\sum_{i=0}^n a_i p^i(1-p)^i$ with $0\leq a_i \leq {n \choose i}$, and $p,q \in [0,1]$.

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Immanuel Weihnachten
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  • @JyrkiLahtonen yep, I didn't add more details since in that paper it seemed claimed in the general case, thanks – Immanuel Weihnachten Oct 11 '11 at 10:51
  • @JyrkiLahtonen I'm so sorry, was my fault: I've written $\mathbb{Q}[p]={r \in \mathbb{R}|r \text{ is algebraic over }\mathbb{Q}(a)}$ at the beginning of my article instead of $\mathbb{Q}[a]={r \in \mathbb{R}|r \text{ is algebraic over }\mathbb{Q}(a)}$ – Immanuel Weihnachten Oct 11 '11 at 11:04
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    The meaning of your notation ${\mathbb Q}[a]$ is extremely non-standard. Normally that notation means the polynomials in $a$ with rational coefficients. I realize you took the notation from the article mentioned in your edit, but I don't think that's a good reason. :) It's pretty weird that the authors would use the conventional meaning of ${\mathbb Q}(a)$ without even bothering to define it but then write ${\mathbf Q}[a]$ for something so different from its usual meaning. Why not start writing ${\mathbb Q}$ for complex numbers... – KCd Oct 11 '11 at 11:17
  • I replaced my answer with something that is hopefully now an answer to the correct question :-) – Jyrki Lahtonen Oct 11 '11 at 11:18
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    Agree with @KCd. The authors should have introduced another notation to make it clear that they mean the field $\overline{\mathbf{Q}(a)}\cap \mathbf{R}$ instead of the ring of values of polynomials. – Jyrki Lahtonen Oct 11 '11 at 11:29
  • @KCd I'm sorry; as my question show I have no knowledge about the subject, so I didn't know that notation was inappropriate. – Immanuel Weihnachten Oct 11 '11 at 19:14

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Ok. With my misunderstandings about the notation now corrected, the proof goes as follows.

We always have $\mathbf{Q}(q)\subseteq\mathbf{Q}(p)$. If a real number $x$ is algebraic over $\mathbf{Q}(q)$, then it is obviously also algebraic over the bigger field $\mathbf{Q}(p)$. Therefore we have the (easier) inclusion $$ \mathbf{Q}[q]\subseteq\mathbf{Q}[p]. $$ To get the reverse inclusion we need to observe that as a root of the polynomial $f(x)-q$ with coefficients in $\mathbf{Q}(q)$ the number $p$ is algebraic over $\mathbf{Q}(q)$. Therefore (a tower of algebraic extensions is algebraic) any real number algebraic over $\mathbf{Q}(p)$ will also be algebraic over $\mathbf{Q}(q)$. This shows that $$ \mathbf{Q}[p]\subseteq\mathbf{Q}[q] $$ and the claim is proven.

Jyrki Lahtonen
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If $a$ is algebraic, then ${\bf Q}[a]$ (using your definition, rather than the standard one) is just the set of all real algebraic numbers, isn't it? Anything algebraic over $\bf Q$ is trivially algebraic over ${\bf Q}(a)$, and anything that satisfies an algebraic equation $f(x)=0$ with coefficients in ${\bf Q}(a)$ also satisfies the equation you get by multiplying $f$ by all its conjugates, which gives you something with rational coefficients.

So, if $p$ and $q$ are algebraic, then ${\bf Q}[p]={\bf Q}[q]$ since both are equal to the field of real algebraics.

Gerry Myerson
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