page1882 Rings Matthias Lorentzen...mattegrisenforlag.com
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Question
Let `alpha = x + langle x^2 + 1 rangle in RR(alpha)`, so
`alpha ^2 = x^2 + langle x^2 + 1 rangle`.
We have that
`x^2 -(x^2 + 1 ) = x^2 - x^2 - 1 = -1`
`(mod langle x^2 + 1 rangle)`.
Does this mean that `alpha^2 + 1 = 0`
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