2
$\begingroup$

Let $I_1$ be the ideal generated by $x^2 +1$ and $ I_2$ be the ideal generated by $x^3 – x^2 + x -1$ in $\mathbb{Q}[x]$. If $R_1= \mathbb{Q}[x]/ I_1$ and $R_2= \mathbb{Q}[x]/ I_2$ then which of the followings are true?
1. $R_1$ and $ R_2$ are fields
2. $R_1$ is a field and $ R_2$ is not a field
3. $R_1$ is an integral domain, but $ R_2$ is not an integral domain
4. $R_1$ and $ R_2$ are not integral domains.

Here $x^2+1$ is irreducible but $x^3 – x^2 + x -1$ is not irreducible. So I think $R_1$ is field but $R_2$ is not. So 2 and 3 are true. Am I right?

  • 0
    Your argument shows that $R_2$ is not a field. Why then $R_2$ is not an integral domain?2012-12-18

2 Answers 2

1

Yes, you're right, yet as YAPC asks, you haven't yet explained why $\,R_2\,$ is not an integral domain and why $\,R_1\,$ is, as trivial as the explanation might be.

Maybe a little explanation about "a pol. here is irreducible iff the principal ideal it generates is prime iff this ideal is maximal iff the quotient ifn..." etc. will round up the answer, though this may heavily depend on what you can rely on.

Please note that the above chain of reasoning between quote marks may not apply if we work on general rings, or even on polynomial rings over rings that are not fields...

1

$R_2$ is isomorphic to $\mathbb Q[x]\left/\left<(x^2+1)( x-1) \right>\right.$ which, since $\mathbb Q[x]$ is a PID and the prime factors in the denominator are relatively prime, is isomorphic to $\mathbb Q[i] \times \mathbb Q $ by the Chinese Remainder Theorem, and this is clearly not a domain.

I might suggest an exercise if this question arose in an introductory abstract algebra course. Use the Euclidean Algorithm to compute what the identities of the factors in the direct product are.

  • 0
    Thanks for catching the typo which was produced by a cut and paste operation not followed by the necessary edit.2013-01-26