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i need to prove that: $\mathbb{Q}[X]/\left(X^{2}-1\right)$ isomorphic to $\mathbb{Q} \times \mathbb{Q}$, but i don't know where to start. I first wanted to use the Chinese value theorem, but i can't see how that should fit in. Hints/tricks/other theorems would be much apreciated!

Thanks!

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  • $\begingroup$ Hint for using CRT: $X^2 - 1 = (X-1)(X+1)$. $\endgroup$ – Tobias Kildetoft Oct 17 '13 at 9:41
  • $\begingroup$ @tobias this is very useful indeed, now i can use the chinese theorem because $X-1 + X+1 = R$ $\endgroup$ – Kees Til Oct 17 '13 at 10:06
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Hint: Recall that $X^2 - 1 = (X+1)(X-1)$, so let us consider the map $T \colon \mathbb Q [X] \ni q \mapsto \bigl(q(1), q(-1)\bigr)\in \mathbb Q^2$. What can you say about $T$? Is it onto? What is its kernel?

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