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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, $x$ and 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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5  
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 8:57
    
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 9:15
2  
No. That there does not exist a relation of the form $\sum c_i u^i = 0$. –  Qiaochu Yuan Jan 19 '12 at 9:25
    
"... 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

3 Answers 3

up vote 4 down vote accepted

HINT $\rm\ u = x^3/(x+1)\:$ is in lowest terms since $\rm\:(x+1,x^3) = 1.\:$ Therefore, if $\rm\:u\:$ were the root of $\rm\:0\ne f(y)\in F[y]\:$ then the Rational Root Test $\rm\: \Rightarrow\ x^3\ |\ f(0)\in F\backslash0\:,\ $ contradiction. $\:$ Alternatively (equivalently), clear denominators in $\rm\ f(u) = 0\:,\: $ then evaluate at $\rm\ x = 0\ \Rightarrow\ f(0) = 0\ \Rightarrow\Leftarrow$

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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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+1 I was also just about to remark on that viewpoint; my tired hands thank you for saving me the work. –  Bill Dubuque Jan 19 '12 at 19:56

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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