# Find elements in quotient ring which satisfies specific condition [on hold]

Let $$\mathbb { R } [ x ]$$ be the polynomial ring in one variable over $$\mathbb { R }$$ . Let $$I$$ be the ideal of $$\mathbb { R } [ x ]$$ generated by the polynomial $$x ^ { 3 } - 8 .$$ Consider the quotient ring $$A = \mathbb { R } [ x ] / I$$ . Find the number of elements $$a$$ of the ring $$A$$ satisfying $$a ^ { 4 } - 1 = 0$$ .

Please give me a method to solve the problem in detail.

## put on hold as off-topic by Servaes, rschwieb, André 3000, Xander Henderson, user26857Feb 11 at 22:33

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First show that every element of $$A$$ can be represented by an element of the form $$ax^2+bx+c,$$ with $$a,b,c\in\Bbb{R}$$. Then compute the representative of $$(ax^2+bx+c)^4-1,$$ keeping in mind that $$x^3-8=0$$ holds in $$A$$. This will yield three polynomial equations in $$a$$, $$b$$ and $$c$$ of total degree $$4$$; solutions to this system of equations yield elements of $$A$$ satisfying the given relation.
The cumbersome work can be simplified by factoring $$x^3-8$$ and applying the Chinese remainder theorem. I'll leave the details to you.