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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?

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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 $v$. – Kevin Carlson Oct 18 '12 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. – user1306 Oct 18 '12 at 12:38
I think it always works when $a$ and $b$ have degree two over $\mathbb{Q}$. – Conrad Oct 18 '12 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. – Conrad Oct 18 '12 at 12:48
@KevinCarlson: Any idea how to construct a counter-example to always? – Conrad Oct 18 '12 at 13:09
up vote 6 down vote accepted

Yes, the answer is "always".

(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


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

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Beautiful! I filled in just about all the justifications-hope you don't mind my cluttering up your aesthetic. – Kevin Carlson Oct 18 '12 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!). – Conrad Oct 19 '12 at 9:01
Things got messed up at that point. I corrected it - should be OK now. – Hagen Knaf Oct 19 '12 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 $[F:\mathbb{Q}(v)]=1$. – Conrad Oct 19 '12 at 11:01

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