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Let $u=\frac{x^3}{x+1}\in F(x)$, where $F(x)$ is the field of quotients of $F[x]$ ($F$ some field, $x$ an indeterminate over it). Show that $u$ is transcendental over $F$.

This is an exercise in Hungerford.

I'm having some trouble even grasping the concepts involved. For instance, I know that if $v$ is transc. over $F$, then $F[v]\cong F[x]$. Or that if $v$ is transc. over $F$, then $F[v]\subsetneq F(v)$. But I have no idea how to use this to my advantage.

I'm also confused about what it even means for $u$ as above to be transc. over $F$. Am I going to have to consider "polynomials of polynomials"?

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    @DilipSarwate: I think this is implicit, absent a notion of convergence since addition and multiplication are defined for finite sequences of terms (or equivalently, by induction, as binary operations).2016-02-06

2 Answers 2

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Since this is an important basic issue, I'll add a complementary answer to Bill Dubuque's, and the good comments above: the definition of $u$ shows that $x$ satisfies a cubic equation over $F(u)$, so is algebraic over $F(u)$. If $u$ were algebraic over $F$, then, by transitivity of "algebraic extension", $x$ itself would be algebraic over $F$.

This less-explicit but more-qualitative kind of argument can succeed when explicit computations become burdensome.

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The following way of saying it is a bit more concrete, in case anyone prefers it that way.

If the coefficients $c_k$ in the sum $ \sum_{k=0}^n c_k u^k $ are in $F$, then the point is to show that that sum cannot be $0$ unless all of the coefficients are $0$. The sum is $ \sum_{k=0}^n c_k \left( \frac{x^3}{x+1} \right)^k. $ If that $=0$ then multiplying both sides by $(x+1)^n$, one concludes $ \sum_{k=0}^n c_k x^{3k} (x+1)^{n-k} = 0. $ So a polynomial in $x$ evaluates to $0$. Since $x$ is transcendental, that can't happen unless all of the coefficients are $0$.

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    @MichaelHardy How do we prove that there won't be any positive and negative nonzero terms that cancel? E.g. something like $x+x^2-x-x^2$.2016-06-10