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

WARNING: unbalanced footnote start tag short code found.

If this warning is irrelevant, please disable the syntax validation feature in the dashboard under General settings > Footnote start and end short codes > Check for balanced shortcodes.

Unbalanced start tag short code found before:

“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.
$$(x^2 – 2\sqrt 3 + 8) \times (x^2 + 8 + 2\sqrt 3) = (x^2 + 8)^2 – 12 x^2$$