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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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    What you are being asked to show is that $u$ does not satisfy a polynomial with coefficients in $F$. Is this clearer?2012-01-19
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    Sort of? Would long division help? I'm not sure what I'm headed towards, though... That $u$ cannot be written as $\sum c_ix^i$, for some $c_i\in F$?2012-01-19
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    No. That there does not exist a relation of the form $\sum c_i u^i = 0$.2012-01-19
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    "... there does not exist a relation of the form $\sum c_iu^i=0$" in which a finite positive number of the $c_i$ are nonzero (to relate back to Qiaochu Yuan's previous comment that $u$ does not satisfy a _polynomial_ with coefficients in $F$.)2012-01-19
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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

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