Chapter 11 (Rings) – Artin – Section 1 (Definition of a Ring)


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“x^2 – 2\sqrt 3 + 8) \times (x^2 + 8 + 2\sqrt 3) = (x^2 + 8)^2 – 12 x^2 \) Share this:Click to share on Twitter (Opens in new window)Click to share on Facebook (Opens in new window)”

1.

Note that a complex number \(\alpha \) is algebraic if it is a root of a(nonzero) polynomial with integer coefficients.
a) Sine we need \(7 + \sqrt [3]{2} \) to be root of an integer polynomial, there fore \((x – (7 + \sqrt [3]{2}) = (x-7 – 2^{\frac{1}{3}}) \) is one of the factors. Clearly \((x-7)^3 – (\sqrt [3]{2})^3 \) \(= (x-7)^3 – 2 \) is the desired polynomial.
b) Again we need \(\sqrt 3 + \sqrt {-5} \) as a root of an integer polynomial. Therefore \((x – (\sqrt 3 + \sqrt {-5})) \) is one of the factors. Take another factor as \((x – (\sqrt 3 – \sqrt {-5})) \) (motivation is to get rid of square roots).
\((x – (\sqrt 3 + \sqrt {-5})) \times (x – (\sqrt 3 – \sqrt {-5})) = x^2 – 2\sqrt 3 + 8 \)
Finally we multiply \(x^2 + 8 + 2\sqrt 3\) to that to get rid of the final square root.
\((x^2 – 2\sqrt 3 + 8) \times (x^2 + 8 + 2\sqrt 3) = (x^2 + 8)^2 – 12 x^2 \)


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