Answer to Question #105182 in Abstract Algebra for Rahul

Question #105182
Consider the ideal I<x³-1,2x⁴+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
2020-03-12T12:24:14-0400

"p(x)=(x^3-1,2x^4+2x^3+7x^2+5x+5)"  is the greatest common divisor of "x^3-1" and "2x^4+2x^3+7x^2+5x+5" .

Find "p(x)" :

"2x^4+2x^3+7x^2+5x+5:x^3-1"


"\\begin{matrix}\n 2x^4+2x^3+7x^2+5x+5 &&&& |x^3-1 \\\\\n 2x^4-2x &&&& 2x+2\\\\\n&&2x^3+7x^2+5x+5\\\\\n&&2x^3-2&\\\\\n&&&7x^2+7x+7\\\\\n&&&x^2+x+1\n\n\n\\end{matrix}"


"x^3-1:x^2+x+1"


"\\begin{matrix}\n x^3-1 &&& |x^2+x+1 \\\\\n x^3+x^2+x &&&x-1 \\\\\n& -x^2-x-1\\\\\n&-x^2-x-1\\\\\n&&0\n\\end{matrix}"

Then "p(x)=x^2+x+1" .


Then

"Q[x]\/p(x)=Q[x]\n\/<x^2+x+1>=\\\\\n=Q[x]\/Q[x]=\\{0\\}"

It is not a field, because there "x^2+x+1=0"


Answer: "p(x)=x^2+x+1", "Q[x]\/I" is not a field


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