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Denote $F=\mathbb{C}(t)$ and consider $p(x)=x^{3}-t\in F[x]$

Is it true that $p$ is irreducible over $F$ ?

My thoughts:

I think that since it is not true that $t^{2}\mid t$ (I don't know how to type not divide) and since $\mathbb{C}(t)/\langle t\rangle\cong\mathbb{C}$ is an integral domain then by Eisenstein the claim follows

Am I correct ? I am not sure that indeed $t$ is prime and that I can apply Eisenstein in my case

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What is $\langle t\rangle\subseteq\Bbb C(t)$? If it is an ideal, it must be the whole field. –  Andrew Oct 11 '12 at 5:11
    
@Andrew - so I am wrong and I did not give a correct argument to say why I may use Eisenstein ? –  Belgi Oct 11 '12 at 5:13
    
It seems you are using a general version of Eisenstein? It looks like you need $\langle t\rangle$ to be prime, which it is not. –  Andrew Oct 11 '12 at 5:16
1  
In order to use Eisenstein, you have to view this as a polynomial over $\mathbb{C}[t]$, then $\langle t \rangle$ is indeed a prime ideal, and you can conclude that the polynomial is irreducible over $\mathbb{C}[t]$. You would need an additional argument that it is irreducible over $\mathbb{C}(t)$, I think. –  Lukas Geyer Oct 11 '12 at 5:16
    
@LukasGeyer - Can you please suggest a way to claim that if we show it is irreducible over $\mathbb{C}[t]$ then it is also irreducible over $\mathbb{C}(t)$ ? –  Belgi Oct 11 '12 at 5:20

2 Answers 2

up vote 1 down vote accepted

Isn't it much easier? If this polynomial was reducible, then it would have a linear factor $x-a$ with $a \in \mathbb{C}(t)$ and $a^3 = t$. However, if $d$ denotes the degree of $a$ (as a rational function in $t$), then $a^3$ has degree $3d$. Since $d$ is an integer, $3d \ne 1$, so there can not be such an $a$.

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I agree, but can you please tell me if my argument is also correct ? –  Belgi Oct 11 '12 at 5:11

If $p(x)$ is reducible, write $p(x)=f(x)g(x)$. Then the degree of either $f$ or $g$ must be $1,$ and we will have found a root of $p(x)$, meaning $t$ has a cube root in $\Bbb C(t).$

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To say this can not be I wanted to use Eisenstein, and so I don't want to argue this. I appriciate your answer –  Belgi Oct 11 '12 at 5:08
    
As it stends it is not clear why such a cube root does not exist... –  Belgi Oct 11 '12 at 5:08
    
I agree that you should justify it doesn't exist. –  Andrew Oct 11 '12 at 5:13
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If $t=a^3$ with $a\in\mathbb C(t)$, then $a=\frac{f(t)}{g(t)}$ with $f,g\in\mathbb C[X]$. Then $tg^3(t)=f^3(t)$. The degree on the right is $\equiv 0\pmod 3$, on the left it is $\equiv 1\pmod 3$. –  Hagen von Eitzen Oct 11 '12 at 5:52

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