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I wish to find out if $t^{2}$ is a prime element of $\mathbb{F}_{2}(t^{2},s^{2})$ so I can justify the use of Eisenstein on the polynomial $x^{2}-t^{2}\in\mathbb{F}_{2}(t^{2},s^{2})[x]$

I believe that it does since $\mathbb{F}_{2}(t^{2},s^{2})/\langle t^{2}\rangle\cong\mathbb{F}_{2}(s^{2})$ is an integral domain (since it is a field)

Is my argument correct and I may use Eisenstein ? (I already noted that $t^{4}$ does not divide $t^{2}$)

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$\mathbb{F}_2(t^2, s^2)$ is a field. It has no prime elements, and the ideal generated by $t^2$ is the whole thing, so the quotient is zero. –  Qiaochu Yuan Oct 11 '12 at 5:24
    
@QiaochuYuan - I see, thank you. It seems that if I replace the brackets with $[ ]$ then it is a prime elemnt. Can I justify using Eisenstein now ? –  Belgi Oct 11 '12 at 5:26
    
Yes, but you don't need to. You can give a direct proof. –  Qiaochu Yuan Oct 11 '12 at 5:38
    
@QiaochuYuan If you can think of such an argument I would love to hear it, please take the time to write it. Its about the principle (I 'feel' Eisenstein and I can't argue the conditions). The question itself is not as important. if I can't seem to think of how to justify the use of Eisenstein (and you say it can be justified) then I either missing something important or not understanding some concept and this troubles me –  Belgi Oct 11 '12 at 5:43

2 Answers 2

up vote 1 down vote accepted

Two comments.

  1. A general form of Eisenstein's criterion shows that $x^2 - t^2$ is irreducible in $\mathbb{F}_2[t^2, s^2][x]$. Since $\mathbb{F}_2[t^2, s^2]$ is a UFD, a general form of Gauss's lemma shows that $x^2 - t^2$ is irreducible in $\mathbb{F}_2(t^2, s^2)[x]$, and primeness is equivalent to irreducibility in a UFD.

  2. Neither of the above general facts are necessary in this case. If you only want to show primeness in $\mathbb{F}_2[t^2, s^2][x]$, it suffices to observe that the quotient by $x^2 - t^2$ is $\mathbb{F}_2[s^2, x]$, which is an integral domain. If you want to show primeness in $\mathbb{F}_2(t^2, s^2)[x]$, you can observe directly that the only form a nontrivial factorization can take is $(x - a)(x + a)$ for some $a \in \mathbb{F}_2(t^2, s^2)$, and by degree considerations no such $a$ satisfies $a^2 = t^2$. This shows irreducibility and, again, irreducible is equivalent to prime in a UFD.

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Can you please add explanation about the use of Eisenstein ? Where exactly do I need $t^2$ to be a prime element, if I say that $\mathbb{F}_{2}[t^{2},s^{2}][x]/\langle t^{2}\rangle\cong\mathbb{F}_{2}[s^{2}]$ then I have what I need ? I really appriciate your help! –  Belgi Oct 11 '12 at 5:58
    
@Belgi: I don't really understand the question. –  Qiaochu Yuan Oct 11 '12 at 6:00
    
I will try to be more clear: I wish to use Eisenstein in its general form to show that $x^{2}-t^{2}\in\mathbb{F}_{2}(t^{2},s^{2})[x]$ is irreducible . I believe we have shown $x^{2}-t^{2}\in\mathbb{F}_{2}[t^{2},s^{2}][x]$ using Eisenstein with the integral domain $R=\mathbb{F}_{2}[t^{2},s^{2}]$ and the prime ideal $\langle t\rangle$ of $R$. But I want $x^{2}-t^{2}$ to be irreducible in $\mathbb{F}_{2}(t^{2},s^{2})[x]$ not just in $\mathbb{F}_{2}[t^{2},s^{2}]$. –  Belgi Oct 11 '12 at 6:08
    
@Belgi: yes, that follows from Gauss's lemma (en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial)). –  Qiaochu Yuan Oct 11 '12 at 6:13
    
Oh, So I have to use both Eisnstien and Gauss's lemma and not only Eisnstien , right ? thank you for the help (+1) –  Belgi Oct 11 '12 at 6:15

Hint $\ $ prime $\rm t\in R[t]\iff$ domain $\rm\:R[t]/(t)\cong R$

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