Answer to Question #180453 in Abstract Algebra for Anuj

Question #180453

a) Find all the units of Z[ − 7]. b) Check whether or not < 8x + 6x − 9x + 24 > [x] 3 2 Q is a field. c) Construct a field with 125 element


1
Expert's answer
2021-05-09T15:51:16-0400

(a) I know some vague facts about this, but nothing too concrete. I know that if "(a+b\\sqrt{7})" is invertible, there exists some "(c+d\\sqrt{7})" such that "(ac+7bd)+(ad+bc)\\sqrt{7}=1" , so "ac+7bd=1" and "ad+bc=0" . I'm also aware that there's a natural norm on "Z[\\sqrt{7}]" , namely "N(a+b\\sqrt{7})=a^2\u22127b^2" . As the norm is multiplicative, we know that for "x\u2208Z[\\sqrt{7}]\\times N(x)=\u00b11" .


(b).With p = 3, we have "a_0 = a_1 = a_2 = 0" (mod p) since 24 = −9 = 6 = 0

(mod 3),"a_n = a_3 \\neq 0 (mod 3)" since "a_3 = 8 = 2 (mod 3)" , and "a_0 \\neq 0 (\\text{ mod }p^2)\\text{ since } a_0 = 24 = 6 (mod 9)." So, by the Eisenstein Criterion, "8x^3 + 6x^2 \u2212 9x + 24" is irreducible over Q.

"< 8x^3 + 6x^2 \u2212 9x + 24 > Q[x]" is not a field,.


(c)


If a cubic polynomial of "F_5[x]" is reducible, then it splits into a linear factor and a quadratic factor or into the product of three linear factors. Linear factors are very easy to test for, as (x−a) is a factor of f if and only if f(a)=0.


So you might choose a random degree 3 polynomial and test for the five possible roots.


For instance, I'm tempted to try "f(x)=x^3+2x+1" . Then

"f(0)=1,\\\\f(1)=4,\\\\f(2)=8+4+1\u22613,\\\\f(3)=27+6+1\u22614,\\\\f(4)=64+8+1\u22613" .

As none of these are zero, f(x) is an irreducible cubic.



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