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I found it easier to look at x^4-x^2-2 as x^4+2x^2+1 since x^4-x^2-2 is over Z_3. From here we will be able to factor the equation to a simpler form, (x^2 +1)^2 | I found it easier to look at x^4-x^2-2 as x^4+2x^2+1 since x^4-x^2-2 is over Z_3. From here we will be able to factor the equation to a simpler form, (x^2 +1)^2 | ||
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| + | --- | ||
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| + | Can any explain to me how to find a splitting field? All of this above doesn't seem to be leading to that. | ||
| + | -Herr- | ||
| + | --- | ||
Revision as of 04:20, 4 December 2008
Factoring I found that there are two fields one is in the form F(a) {a + bi| a,b contained in Z_3} the other is (x - B)(x^3+x^2B-x+xB^2-B+B^3) which I got from long division.
Did anyone else get this? --Robertsr 23:51, 2 December 2008 (UTC)
I reduced x^4-x^2-2 to (x^2-2)(x^2+1) so the subfields both have degree 2. I don't know how helpful this is, but I figured I'd post it anyway.
I found it easier to look at x^4-x^2-2 as x^4+2x^2+1 since x^4-x^2-2 is over Z_3. From here we will be able to factor the equation to a simpler form, (x^2 +1)^2
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Can any explain to me how to find a splitting field? All of this above doesn't seem to be leading to that. -Herr- ---
