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I have been going through the exercises from Contemporary Abstract Algebra, and there are a few questions that I just couldn't solve.

  1. Let $F=Q(\pi^3)$. Find a basis for $F(\pi)$ over $F$. I know the structure of $Q(\pi)$ (kinda like the quotient field on $Q$), but I don't know how to proceed from there.

  2. Let $F$ be a field with characteristic $p\neq 0$. Show that the polynomial $x^{p^n}-x$ has distinct zeros. I know that a polynomial $f$ has a multiple zero if and only if $f$ and $f'$ have a common factor of positive degree. However, I'm kinda stuck on verifying that $x^{p^n}-x$ and $p^nx^{p^n-1}-1$ have no common factor.

  3. If $F$ is a field and the multiplicative group of nonzero elements of $F$ is cyclic, show that $F$ is finite. I honestly don't even know how to start thinking about this question. A hint would be much appreciated.

  4. Show that $\sqrt{2} \notin Q(\pi)$. Here is my train of thought: assume otherwise. Then \begin{equation} \sqrt{2} = \frac{a_m\pi^m+\cdots+a_1\pi+a_0}{b_n\pi^n+\cdots+b_1\pi+b_0} \end{equation} so that \begin{equation} 2 = \frac{(a_m\pi^m+\cdots+a_1\pi+a_0)^2}{(b_n\pi^n+\cdots+b_1\pi+b_0)^2} \text{.} \end{equation} But I'm not sure how to continue from here.

  5. Suppose $a$ is algebraic over a field $F$. Show that $a$ and $1+a^{-1}$ have the same degree over $F$. The only observation that I can make is that $F(a)=F(1+a^{-1})$, which may or may not help me solve this question. I feel like there must be a simple trick that I'm simply overlooking.

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This should probably be split into multiple questions. I'm certain as well that 3 has been answered multiple times on this site. Your observation in 5 is incorrect. Consider $\omega$ as a primitive third root of unity, then $\omega^{-1}=\omega^{2}$ and furthermore $\omega+\omega^2$ is real. – JSchlather Dec 17 '12 at 4:24
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Too many questions, too little self effort shown. – DonAntonio Dec 17 '12 at 4:24
    
@JacobSchlather I meant $1+a^{-1}$ of course. – Aden Dong Dec 17 '12 at 4:40
    
I'm sure 1. was asked here not too long ago. – Gerry Myerson Dec 17 '12 at 4:52

1-Observe that $\pi$ is algebraic over $F$ and therefore $F(\pi)=F[\pi]$. The element $\pi$ is a root of the polynomial $x^3-\pi^3\in F[x]$ and therefore $F(\pi)$ is generated as a $F$-vector space by the set $\{1,\pi,\pi^2\}$. It remains to prove that this set is linearly independent. If not, there exist polynomials $f,g,h\in Q[X]$ not all zero such that $f(\pi^3)+\pi g(\pi^3)+\pi^2h(\pi^3)=0$. The degrees in $\pi$ of each term in $f(\pi^3)$ are $\equiv 0\mod 3$, in $\pi g(\pi^3)$ they are $\equiv 1\mod 3$ and in $\pi^2 h(\pi^3)$ they are $\equiv 2\mod 3$. Then we have a non trivial polynomial equation in $\pi$ with rational coefficients which is impossible since $\pi$ is trascendental.

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Is $p(x)=x^3-\pi^3$ irreducible in $F[x]$? If $p(x)$ is irreducible, then $F[x]/<p(x)>\cong F(\pi)$ by the isomorphism $\phi(f(x)) = f(\pi)$, and $\{1,\pi,\pi^2\}$ is a basis of $F(\pi)$. But I don't know if $p(x)$ is irreducible or not. Could you explain please? – Myath Feb 3 at 7:09
    
It follows from my proof that $x^3-\pi^3$ is irreducible and this is because we've seen that: 1) $[F(\pi):F]=3$, which implies that the irreducible polynomial of $\pi$ has degree 3, and 2) the polynomial $\x^3-\pi^3$ annihilates $\pi$ so it must be the irreducible polynomial of $\pi$. – Diego Feb 4 at 14:23

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