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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"?

user26857
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FPP
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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? – Qiaochu Yuan Jan 19 '12 at 08:57
  • 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 $c_i\in F$? – FPP Jan 19 '12 at 09:15
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    No. That there does not exist a relation of the form $\sum c_i u^i = 0$. – Qiaochu Yuan Jan 19 '12 at 09:25
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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 $F$.) – Dilip Sarwate Jan 19 '12 at 21:00
  • @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). – Vandermonde Feb 06 '16 at 01:22

2 Answers2

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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.

paul garrett
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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 $0$.

Michael Hardy
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  • Thank you, I actually do like it better this way. Can you also comment on how one would prove that x is transcendental? – Maryam Feb 05 '16 at 22:47
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    @Maryam : It's transcendental since it's an "indeterminate". Indeterminates are things from which one builds up polynomials by writing $$\sum_{k=0}^n a_k x^k. \tag 1$$ The polynomial $(1)$ is the zero polynomial only if all the coefficients $a_0,\ldots, a_n$ are $0$. If $(1)$ could be the zero polynomial for some nonzero coefficients $a_0,\ldots,a_n$, then $x$ would be algebraic since it would be a zero of that polynomial. But only if there is such a polynomial is $x$ algebraic. Since there is none, $x$ is not algebraic, but transcendental. $\qquad$ – Michael Hardy Feb 06 '16 at 00:09
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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 $x+x^2-x-x^2$. – yoyostein Jun 10 '16 at 13:50