Question #137502
Is the ideal generated by x^2+1 in Z2[x] a prime ideal of Z2[x]? give reason.
1
Expert's answer
2020-10-11T17:52:02-0400

An ideal PP of a commutative ring RR is prime if it has the following two properties:

1) If aa and bb are two elements of RR such that their product abab is an element of PP, then aa is in PP or bb is in PP;

2) PP is not the whole ring RR.


Let f(x)=x+1Z2[x]f(x)=x+1\in\mathbb Z_2[x] and g(x)=x+1Z2[x]g(x)=x+1\in\mathbb Z_2[x]. Since 1+1=01+1=0 in the ring Z2\mathbb Z_2, f(x)g(x)=(x+1)(x+1)=x2+x+x+1=x2+(1+1)x+1=x2+1f(x)g(x)=(x+1)(x+1)=x^2+x+x+1=x^2+(1+1)x+1=x^2+1 is in the ideal I=x2+1I=\langle x^2+1\rangle generated by x2+1.x^2+1. Taking into account that f(x)If(x)\notin I and g(x)I,g(x)\notin I, we conclude that the ideal I=x2+1I=\langle x^2+1\rangle generated by x2+1x^2+1 is not prime.


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