Polynomial relation between two algebraic numbers $p$ and $q$ implies $\mathbb{Q}[q]=\mathbb{Q}[p]$ 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 $p,q \in [0,1]$.
 A: 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.
A: 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.  
