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I bought the third edition of "Abstract Algebra" by Dummit and Foote. In my opinion this is the best "algebra book" that has been written.

I found several solution manual but none has solutions for Chapters 13 and 14 (Field extensions and Galois theory respectively)

Is there a solution manual for these chapters?

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  • $\begingroup$ As far as I've seen, there isn't any, though scattered problems have solutions online. Nevertheless, if you have specific problems you are interested in, I'm sure that the folks here will be more than happy to help! $\endgroup$ – Alex Wertheim Aug 3 '13 at 2:25
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    $\begingroup$ @Gaston Burrull maybe you should write one now that this opportunity has presented itself. All you need is a few hundred spare hours. $\endgroup$ – James S. Cook Aug 3 '13 at 2:34
  • $\begingroup$ @JamesS.Cook In that case I need someone who validate my solutions (which is almost the same as having a solution manual) Sometimes, I feel very insecure if I answered rightly, I sometimes skip details. $\endgroup$ – Gaston Burrull Aug 3 '13 at 2:38
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    $\begingroup$ @GastónBurrull I hear you. I have a student who was working through those chapters, some problems are really simple, others are pages of calculation. That said, you can always post questions of the form: I think blah blah blah then ask is this correct? And to be greedy, follow that up by, is there a better way to see it? You could learn a lot from this. $\endgroup$ – James S. Cook Aug 3 '13 at 3:05
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    $\begingroup$ try solving each and every Question and if you have any doubt or if You think Your Justification is not clear Post it here. I am doing the same for the same part, 13th and 14th Chapter of Dummit Foote for the last 10 days and i am receiving reasonably good responses from other Users. All the best :) $\endgroup$ – user87543 Aug 3 '13 at 10:54
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If anyone is interested, I made a full solution manual for Chapter 13 - Field Theory.

You can find it here https://positron0802.wordpress.com/algebra-dummit-chapter-13-solution/.

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    $\begingroup$ I checked your solutions and all are correct and very straightforward. But some solutions can be shortened a lot. For example in 13.6 exercise 6 you can say the following: Since the two polynomials are irreducible polynomials, you only need to find one common root for both, but $-\zeta_n$ is obviously a root of both, by problem 1, since 2 and $n$ are coprime and $-1=\zeta_2$. 13.2 exercise 5 can be shortened by the observation that 2 and 3 are coprime numbers and degree considerations. The solution of 13.2 ex 16 is just noticing that F[r]=F(r) by the main theorem of that section. $\endgroup$ – Gaston Burrull Jan 28 '17 at 6:21
  • $\begingroup$ 13.4 exercise 3 just note that $(x^4+x^2+1)(x^2-1)=x^6-1$ so the splitting field is the ciclotomic field and $\phi(6)=2$ $\endgroup$ – Gaston Burrull Jan 28 '17 at 6:25
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    $\begingroup$ Thanks and congratulations for your hard work on this! $\endgroup$ – Gaston Burrull Jan 28 '17 at 6:27
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    $\begingroup$ Thanks! I modified Exercise 13.6.6 according to your recommendation. I didn't change the others exercises as I considered they were short enough. I think that in Exercise 13.4.3 one can't use $\varphi(6)=2$ as it has not been proved that the $n$th cyclotomic polynomial has degree $\varphi(n)$. I appreciate any more comments and suggestions on the solutions. I might write solutions to Chaper 14 when I finish my semester in the university in June. $\endgroup$ – positrón0802 Jan 28 '17 at 19:38
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    $\begingroup$ You're right, you still can't use cyclotomic degree for $n$ composite, only for primes. I'm impatient for your future solutions. You have pretty nice solutions to be an undergraduate student! $\endgroup$ – Gaston Burrull Jan 29 '17 at 8:18
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With my friends we make a partial solution manual for chapter 13 in spanish (since is our native language) as @James S. Cook proposed.

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There's a mistake in the solution to problem 17. The problem asks to prove that the degree n of an irreducible polynomial f(x) divides the degree of an irreducible factor of the composite f(g(x)), but the proof purports to show the opposite, that m divides n. The problem is that \alpha and \g(\alpha) have gotten switched around in the final line. Otherwise, everything is okay

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  • $\begingroup$ Already corrected. $\endgroup$ – positrón0802 Jul 23 '17 at 20:23

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