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	<title>Comments on: Galois Theory for Dummies &#8211; Part I</title>
	<atom:link href="http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/feed/" rel="self" type="application/rss+xml" />
	<link>http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/</link>
	<description>Strain your Brain</description>
	<lastBuildDate>Tue, 31 Aug 2010 17:39:33 -0700</lastBuildDate>
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		<title>By: Alfred Neunzoller</title>
		<link>http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/comment-page-1/#comment-27683</link>
		<dc:creator>Alfred Neunzoller</dc:creator>
		<pubDate>Fri, 19 Mar 2010 01:15:57 +0000</pubDate>
		<guid isPermaLink="false">http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/#comment-27683</guid>
		<description>The article was awesome! Please, please, write a part II!</description>
		<content:encoded><![CDATA[<p>The article was awesome! Please, please, write a part II!</p>
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		<title>By: César Bravo</title>
		<link>http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/comment-page-1/#comment-16681</link>
		<dc:creator>César Bravo</dc:creator>
		<pubDate>Thu, 04 Jun 2009 14:24:07 +0000</pubDate>
		<guid isPermaLink="false">http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/#comment-16681</guid>
		<description>Thanks for your answer...

At+
César</description>
		<content:encoded><![CDATA[<p>Thanks for your answer&#8230;</p>
<p>At+<br />
César</p>
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		<title>By: yaniv</title>
		<link>http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/comment-page-1/#comment-16679</link>
		<dc:creator>yaniv</dc:creator>
		<pubDate>Thu, 04 Jun 2009 09:51:31 +0000</pubDate>
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		<description>Cesar,

1. The original phrasing is correct - g(x) is not irreducible. We are using it in order to create the polynomial f(x) which will be irreducible.
2. You are right, thanks!
3. Right.
4. v*u = sig-1(sig(u)*sig(v)) = sig-1((p+f(x))*(s+f(x))) = sig-1(p*s+f(x)) = sig-1(1+f(x)) = 1.
5. As the article says - the Galois group is the automorphism group of E/F. Part 2 of the article will explain the importance of the Galois group.</description>
		<content:encoded><![CDATA[<p>Cesar,</p>
<p>1. The original phrasing is correct &#8211; g(x) is not irreducible. We are using it in order to create the polynomial f(x) which will be irreducible.<br />
2. You are right, thanks!<br />
3. Right.<br />
4. v*u = sig-1(sig(u)*sig(v)) = sig-1((p+f(x))*(s+f(x))) = sig-1(p*s+f(x)) = sig-1(1+f(x)) = 1.<br />
5. As the article says &#8211; the Galois group is the automorphism group of E/F. Part 2 of the article will explain the importance of the Galois group.</p>
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		<title>By: César Bravo</title>
		<link>http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/comment-page-1/#comment-16667</link>
		<dc:creator>César Bravo</dc:creator>
		<pubDate>Tue, 02 Jun 2009 20:16:55 +0000</pubDate>
		<guid isPermaLink="false">http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/#comment-16667</guid>
		<description>Hi 
Some observations:
1.-
Page 2, line 2 reads:
&quot;The problem is that g(x) is not irreducible.&quot;
I think the text must be
&quot;The problem is that g(x) is irreducible over Q.&quot;
or perhaps
&quot;The problem is that g(x) is not reducible over Q.&quot;

2.- I think that the symbols &quot;&quot; are swapped on the
first line of Gauss´s Lemma proof
&quot;()&quot;
That is:
(=&gt;) If f(x) splits on Z, obviously it splits on Q, since Z is contained on Q.

3.- Whitout loss of generality I must assume that
&quot;WLOG&quot;
stands for
&quot;Whitout Loss Of Generality&quot;,
right?

4.- I can cope with the arguments up to page 3 where is claimed that
&quot;Let v = \sigma^{-1}(s). It is clear that v \times u = 1&quot;
justifying the claim with
&quot;(if this is not clear to you at this point try to prove it)&quot;
well I recognize that I tried to prove it and was unable to do it, so:
can you sketch the proof giving working hints to a standard proof?

5.- About the &quot;Fields Automorphisms&quot; sections: elsewhere I found a
figure relating what seemed to me as extension fields (perhaps splitting fields)
and Galois Groups, but there was not any explication and I can not 
reformulate the underline relation myself. So, it would be great if you
can explain that relation.

Regards
César</description>
		<content:encoded><![CDATA[<p>Hi<br />
Some observations:<br />
1.-<br />
Page 2, line 2 reads:<br />
&#8220;The problem is that g(x) is not irreducible.&#8221;<br />
I think the text must be<br />
&#8220;The problem is that g(x) is irreducible over Q.&#8221;<br />
or perhaps<br />
&#8220;The problem is that g(x) is not reducible over Q.&#8221;</p>
<p>2.- I think that the symbols &#8220;&#8221; are swapped on the<br />
first line of Gauss´s Lemma proof<br />
&#8220;()&#8221;<br />
That is:<br />
(=&gt;) If f(x) splits on Z, obviously it splits on Q, since Z is contained on Q.</p>
<p>3.- Whitout loss of generality I must assume that<br />
&#8220;WLOG&#8221;<br />
stands for<br />
&#8220;Whitout Loss Of Generality&#8221;,<br />
right?</p>
<p>4.- I can cope with the arguments up to page 3 where is claimed that<br />
&#8220;Let v = \sigma^{-1}(s). It is clear that v \times u = 1&#8243;<br />
justifying the claim with<br />
&#8220;(if this is not clear to you at this point try to prove it)&#8221;<br />
well I recognize that I tried to prove it and was unable to do it, so:<br />
can you sketch the proof giving working hints to a standard proof?</p>
<p>5.- About the &#8220;Fields Automorphisms&#8221; sections: elsewhere I found a<br />
figure relating what seemed to me as extension fields (perhaps splitting fields)<br />
and Galois Groups, but there was not any explication and I can not<br />
reformulate the underline relation myself. So, it would be great if you<br />
can explain that relation.</p>
<p>Regards<br />
César</p>
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	<item>
		<title>By: Dani</title>
		<link>http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/comment-page-1/#comment-10275</link>
		<dc:creator>Dani</dc:creator>
		<pubDate>Mon, 11 Aug 2008 07:56:03 +0000</pubDate>
		<guid isPermaLink="false">http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/#comment-10275</guid>
		<description>There is also a really nice proof for Eisenstein’s Criterion (with coefficients in Z) in Wikipedia.
http://en.wikipedia.org/wiki/Eisenstein%27s_criterion#Basic_proof</description>
		<content:encoded><![CDATA[<p>There is also a really nice proof for Eisenstein’s Criterion (with coefficients in Z) in Wikipedia.<br />
<a href="http://en.wikipedia.org/wiki/Eisenstein%27s_criterion#Basic_proof" rel="nofollow">http://en.wikipedia.org/wiki/Eisenstein%27s_criterion#Basic_proof</a></p>
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		<title>By: Nissan</title>
		<link>http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/comment-page-1/#comment-373</link>
		<dc:creator>Nissan</dc:creator>
		<pubDate>Sun, 18 Nov 2007 22:47:08 +0000</pubDate>
		<guid isPermaLink="false">http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/#comment-373</guid>
		<description>Thank you very much for the great pleasure I had reading the article. I try to imagine the level of excitement I could have gotten should I understood it.</description>
		<content:encoded><![CDATA[<p>Thank you very much for the great pleasure I had reading the article. I try to imagine the level of excitement I could have gotten should I understood it.</p>
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		<title>By: yaniv</title>
		<link>http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/comment-page-1/#comment-177</link>
		<dc:creator>yaniv</dc:creator>
		<pubDate>Tue, 09 Oct 2007 09:50:21 +0000</pubDate>
		<guid isPermaLink="false">http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/#comment-177</guid>
		<description>Dan,

The topic is covered in the (excellent) Algebra B 2 course.

For the second part of your comment: most of the algebra books I read did postpone the proof of the existance of the said polynomial until the end. Actually, as is mentioned in the introduction of the article I chose to present many thing in the reverse order. Now, I completely agree with you that the existance of such polynomials is not as important to the subject as the fundamental theorem, for example, but I really liked the proof!</description>
		<content:encoded><![CDATA[<p>Dan,</p>
<p>The topic is covered in the (excellent) Algebra B 2 course.</p>
<p>For the second part of your comment: most of the algebra books I read did postpone the proof of the existance of the said polynomial until the end. Actually, as is mentioned in the introduction of the article I chose to present many thing in the reverse order. Now, I completely agree with you that the existance of such polynomials is not as important to the subject as the fundamental theorem, for example, but I really liked the proof!</p>
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	<item>
		<title>By: Dan</title>
		<link>http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/comment-page-1/#comment-175</link>
		<dc:creator>Dan</dc:creator>
		<pubDate>Mon, 08 Oct 2007 21:19:28 +0000</pubDate>
		<guid isPermaLink="false">http://yaniv.leviathanonline.com/blog/math/galois-theory-for-dummies-part-i/#comment-175</guid>
		<description>I liked the article. Quite informative, and offers a nice introduction. On a related note, do you perchance know which of the &#039;Algebra B&#039;s touches on Galois theory?

A little criticism: The proof that you provided in part 3, as well as the related lemmas, are quite elegant. However, I feel that it is much easier for the reader to &#039;believe&#039; in the existence of such a polynomial without proof than in the theorem about it. Therefore, unless it was necessary for developing the rest of the theory, I think that part 3 might have been better fit in a later installment. Nonetheless, it made for an enjoyable read by itself.</description>
		<content:encoded><![CDATA[<p>I liked the article. Quite informative, and offers a nice introduction. On a related note, do you perchance know which of the &#8216;Algebra B&#8217;s touches on Galois theory?</p>
<p>A little criticism: The proof that you provided in part 3, as well as the related lemmas, are quite elegant. However, I feel that it is much easier for the reader to &#8216;believe&#8217; in the existence of such a polynomial without proof than in the theorem about it. Therefore, unless it was necessary for developing the rest of the theory, I think that part 3 might have been better fit in a later installment. Nonetheless, it made for an enjoyable read by itself.</p>
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