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Let $R = Q[x] / (x^4 - 3x^2+ 6x)$. How can we prove that $x^2 + 1$ is invertible in $R$? How can we prove that $R$ is isomorphic to a direct sum of two fields?

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We would appreciate it to see some thoughts of your own on this. Also, it is a bit rude to command us to prove it. – Gregor Bruns Dec 13 '12 at 8:10
@SugataAdhya: Please do not use $\Bbb R$ ("blackboard" R) for an abstract ring; $\Bbb R$ denotes the real numbers, and obviously we are not working with the real numbers here. – anon Dec 13 '12 at 8:11
I'm sorry. I didn't notice. – Sugata Adhya Dec 13 '12 at 8:23

Hint: You can factor $x^4-3x^2+6x$ as $x(x^3-3x+6)$. I'm sure you can prove that these two factors are coprime, and so by the CRT $R\cong \mathbb{Q}[x]/(x)\times\mathbb{Q}[x]/(x^3-3x+6)$. Now, you want to prove that these two factors are fields. Recall that if you have a field $F$ then $F[x]/(p(x))$, by definition, will be a field if and only if $(p(x))$ is a (BLANK) ideal, but since $F[x]$ is a PID, it suffices to prove that $(p(x))$ is (BLANK). Now, verifying that $(p(x))$ is (BLANK) is simple because you only have to verify that $p(x)$ is a (BLANK) polynomial. Once you have filled in these blanks, I think that this should be easy for you since $x$ is clearly (BLANK) and $x^3-3x+6$ is (BLANK) by (BLANK)'S (BLANK).

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(BLANK) you :-) – robjohn Dec 14 '12 at 16:11
@robjohn ;) you're welcome – Alex Youcis Dec 14 '12 at 16:13

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