# Transcendental extensions of $\mathbb{Q}$ containing algebraic elements.

Suppose that $v$ is transcendental over $\mathbb{Q}$ and $a,b\notin\mathbb{Q}$ are algebraic over $\mathbb{Q}$. When does $\mathbb{Q}(v,av+b)$ contain an element algebraic over $\mathbb{Q}$ but not in $\mathbb{Q}$?

Is it possible that the answer is always?

• I think it's more like never. Any element of the extension in $av+b$ will have some term in $av$, and you have no way to cancel that with a rational function of $v$. Oct 18, 2012 at 10:46
• @KevinCarlson There is a trivial case that $a=b$, and then $a=\frac{av+b}{v+1}$. For a less trivial example, let $a=\sqrt{2}$ and $b=\sqrt{6}$. Then $(4v)^{-1}((av+b)^2-2v^2-6)=\sqrt{3}$. So it does happen. Oct 18, 2012 at 12:38
• I think it always works when $a$ and $b$ have degree two over $\mathbb{Q}$. Oct 18, 2012 at 12:44
• i.e. if $a$ is degree two then it has the form $c+d\sqrt{e}$ where $c,d,e\in\mathbb{Q}$. So with out loss of generality we may assume that $a=\sqrt{\alpha}$ and $b=\sqrt{\beta}$ for some $\alpha,\beta\in\mathbb{Q}$. Then if we square $av+b$ we can easily get an algebraic number not in $\mathbb{Q}$ out of it. Oct 18, 2012 at 12:48
• @KevinCarlson: Any idea how to construct a counter-example to always? Oct 18, 2012 at 13:09

(1) Let $f\in\mathbb{Q}[x]$ be an irreducible polynomial, let $F$ be some extension field of $\mathbb{Q}$ and assume that $f$ becomes reducible over $F$. Then $F$ contains an element $a\not\in\mathbb{Q}$ that is algebraic over $\mathbb{Q}$.

Proof: the coefficients of the irreducible factors of $f$ over $F$ are algebraic over $\mathbb{Q}$ and at least one of them is not in $\mathbb{Q}$.

(2) The rational function field $\mathbb{Q}(v)$ contains no elements $a\not\in\mathbb{Q}$ algebraic over $\mathbb{Q}$.

This is well-known and has already been discussed on this site several times.

(3) Let $a,b\not\in\mathbb{Q}$ be algebraic over $\mathbb{Q}$ and consider the field $F:=\mathbb{Q}(v,av+b)$, where $v$ is transcendental over $\mathbb{Q}$. Then $F$ contains elements $c\not\in\mathbb{Q}$ algebraic over $\mathbb{Q}$.

Proof: let $f\in\mathbb{Q}[x]$ be the minimal polynomial of a primitive element of the algebraic extension $\mathbb{Q}\subseteq\mathbb{Q}(a,b)=:K$. Then by (2) $f$ remains irreducible over $\mathbb{Q}(v)$. Hence $[K:\mathbb{Q}]=[K(v):\mathbb{Q}(v)]$.

By construction we have $FK=K(v)$, since $a=v^{-1}(av), b=v^{-1}(bv)$ in $FK$. Moreover $[FK:F]=[K(v):F]\leq [K(v):\mathbb{Q}(v)]$.

Assume now that $F$ contains no $c\not\in\mathbb{Q}$ algebraic over $\mathbb{Q}$. Then by (1) $[FK:F]=[K(v):\mathbb{Q}(v)]$.

Since degree of field extensions is multiplicative, we have the equation

$[FK:\mathbb{Q}(v)]=[FK:F][F:\mathbb{Q}(v)]=[FK:K(v)][K(v):\mathbb{Q}(v)]$

we then get $[F:\mathbb{Q}(v)]=[FK:K(v)]=[K(v):K(v)]=1$, hence the contradiction $a,b\in\mathbb{Q}$.

• Beautiful! I filled in just about all the justifications-hope you don't mind my cluttering up your aesthetic. Oct 18, 2012 at 20:25
• I don't understand the line "Then by (1) $$F:\mathbb{Q}(v)$=1$" i.e. $F$ might contain no elements algebraic over $\mathbb{Q}$ but still contain elements algebraic over $\mathbb{Q}(v)$. Could you explain this? Thanks for the answer. I think I can construct my own proof (which is heavily inspired by your proof) but I'd like to understand this step in your proof (and then accept your answer!). Oct 19, 2012 at 9:01
• Things got messed up at that point. I corrected it - should be OK now. Oct 19, 2012 at 10:33
• I think maybe I get it now. If we take $f\in\mathbb{Q}[x]$ to be the minimal polynomial of the primitive element of $\mathbb{Q}(a,b)$ then $f$ does not become reducible over $F$ but does become reducible over $K(v)$. Thus, $[K(v):F]$ is the degree of $f$ and so is $[K(v):\mathbb{Q}(v)]$. Thus $[F:\mathbb{Q}(v)]=1$. Oct 19, 2012 at 11:01
• @KevinArlin why $av+b\in \mathbb Q(v)$ implies that $a,b \in \mathbb Q$? What i get is that $av+b \notin \mathbb Q$, hence transcendental over $\mathbb Q$. Jul 6, 2021 at 6:53