Answer to Question #104949 in Abstract Algebra for Sourav Mondal

Question #104949
Consider the ideal I<x³-1.x⁴+2x³+7x²+5x+5> in Q[x]. Find p(x) ∈Q[x] such that I =<p(x)> .
Is Q[x]/I a field? Give reasons for your answer.
1
Expert's answer
2020-03-10T13:20:16-0400

"p(x)=(x^3-1,x^4+2x^3+7x^2+5x+5)" is the greatest common divisor of "x^3-1" and "x^4+2x^3+7x^2+5x+5", because there are polynomials "P(x)" and "Q(x)" such that "(x^3-1,x^4+2x^3+7x^2+5x+5)=P(x)(x^3-1)+"

"+Q(x)(x^4+2x^3+7x^2+5x+5)", so "\\langle(x^3-1,x^4+2x^3+7x^2+5x+5)\\rangle\\subset" "\\subset\\langle x^3-1,x^4+2x^3+7x^2+5x+5\\rangle" . But "x^3-1" and "x^4+2x^3+7x^2+5x+5" are divisible by "\\langle(x^3-1,x^4+2x^3+7x^2+5x+5)\\rangle", so "\\langle x^3-1,x^4+2x^3+7x^2+5x+5\\rangle\\subset"

"\\subset\\langle(x^3-1,x^4+2x^3+7x^2+5x+5)\\rangle" , so "\\langle x^3-1,x^4+2x^3+7x^2+5x+5\\rangle="

"=\\langle(x^3-1,x^4+2x^3+7x^2+5x+5)\\rangle" .


We have "x^3-1=(x-1)(x^2+x+1)", where "x-1" and "x^2+x+1" are irreducible polynomials in "\\mathbb Q[x]". So we need to check if "x^4+2x^3+7x^2+5x+5" divisible by "x-1" or "x^2+x+1"


"1" is not a solution of "x^4+2x^3+7x^2+5x+5=0", so "x^4+2x^3+7x^2+5x+5" is not divisible by"x-1".


"x^4+2x^3+7x^2+5x+5" also is not divisible by "x^2+x+1", because "x^4+2x^3+7x^2+5x+5=(x^2+x+1)(x^2+x+5)+x"


So "p(x)=(x^3-1,x^4+2x^3+7x^2+5x+5)=1"


Then "\\mathbb Q[x]\/\\langle1\\rangle=\\mathbb Q[x]\/\\mathbb Q[x]=\\{0\\}" . It is not a field, because there "1=0" .


Answer: "p(x)=1" , "\\mathbb Q[x]\/I" is not a field.


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