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This question is quit a straightforward use of the isomorphism theorem. I think I just have to show that $\ker(\varphi) = (X^4+X^3+X^2+X+1)$ while $\varphi$ is the evaluation homomorphism. That $\ker(\varphi) \supseteq (X^4+X^3+X^2+X+1)$ can be shown trivially. However, I'm struggling with the other direction. Does anyone have a hint? I think I have to show that for all $p \in \ker(\varphi)$ we can write something like $p(x) = q(x)(x^4+x^3+x^2+x+1)$. Is this the right approach?

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Use Gauss' Lemma and the fact that $x^4+\dotsc+1$ is the minimal polynomial of $\zeta_5$. –  Martin Brandenburg Feb 2 '13 at 22:08
    
Please edit your LaTeX. I think you want to use $\mathbb{Z}$ and you certainly do not want $\matchcal$ and you do want $\backslash supseteq$ to get $\supseteq$ instead of \superseteq. That is a pain, but it is so easy to make LaTeX errors. –  Barbara Osofsky Feb 2 '13 at 22:16

2 Answers 2

Your idea is correct. Namely, we have

$$\mathbb{Z}[e^{2\pi i/ 5}] \cong \mathbb{Z}[X]/ (f) $$

where $(f) = \ker(\phi)$ and $\phi: \mathbb{Z}[X] \rightarrow \mathbb{C}$ evaluates a polynomial at $X = e^{2\pi i/ 5}$. (Note that the kernel does not need to be an principal ideal in general, but here it is..)

Know what is your problem to continue? You are basically done, once you realize that $f = X^4 + X^3 +X^2 +X + 1$ is irreducible and vanishes on $e^{2\pi i/ 5}$.

In particular, why does $\ker(\phi) \subset (f)$ hold? Assume $h \in ker(\phi)$, and decompose $h = h_1\cdots h_n$ into primfactors ($\mathbb{Z}[X]$ is a unique factorization domain). Then some factor $h_i$ must vanish on $\zeta_5$, i.e. $h \in (h_i)$ and we must show that $h_i \in (f)$. Now write $h_i = a \cdot \tilde{h}$ with $a \in \mathbb{Z}$ and $\tilde{h} \in \mathbb{Z}[X]$ primitiv. Then by the Gauss-Lemma $\tilde{h} \in \mathbb{Q}[X]$ is irreducible. To finish consider the map $\tilde{\phi}: \mathbb{Q}[X] \rightarrow \mathbb{C}$ which is also given by evaluation on $\zeta_5$ and note that the kernel of this map is uniquely generated by a irreducible polynomial. Since $f$ is irreducible and $f \in ker(\tilde{\phi})$, it follows $\ker (\tilde{\phi}) = (f)$ and hence $\tilde{h} \in g f$ for some $g \in \mathbb{Q}[X]$. But since $\tilde{h}$ is primitiv, $g$ already has integral coefficients.

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Take $p\in\ker\varphi$. Write $p(X)=q(X)(X^4+X^3+X^2+X+1)+r(X)$ with $\deg r<4$ and $q,r\in\mathbb Q[X]$. Since $p(e^{2\pi i/5})=0$ it follows that $r(e^{2\pi i/5})=0$. Set $f=X^4+X^3+X^2+X+1$. Assume that $r\neq 0$ and let $d=(f,r)$ (here $(a,b)$ denotes the greatest common divisor of the polynomials $a$ and $b$). Since $d\mid f$ and $f$ is irreducible over $\mathbb Q$ (why?) it follows $d=f$ or $d=1$. The case $d=f$ is not possible because $d\mid r$ and $\deg r<\deg f$. It remains $d=1$. Then there exist $g,h\in\mathbb Q[X]$ such that $f(X)g(X)+r(X)h(X)=1$. Now, by sending $X$ to $e^{2\pi i/5}$ and taking into account that $f(e^{2\pi i/5})=r(e^{2\pi i/5})=0$, we get $0=1$, a contradiction. In conclusion, $r=0$ and this shows that $p\in(X^4+X^3+X^2+X+1)$.

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