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Let $R=\mathbb Q[x]/I$ where $I$ is the ideal generated by $1+x^2$. Then is

  1. $y^2 +1$ is irreducible over $R$ ?
  2. $y^2+y+1$ is irreducible over $R$ ?
  3. $y^2-y+1$ is irreducible over $R$ ?
  4. $y^3+y^2+y+1$ is irreducible over $R$ ?

I am completely stuck on it. Please help.

  • 1
    Some hints: If we are working in $R[y]$ then consider $(y+x)(y-x)$, for example. For 4 we have the identity $(y^2+1)(y^2-1)=y^4-1=(y-1)(y^3+y^2+y+1)$. 2 and 3 are standard forms too - where would you look for the roots of those over the rationals?2012-08-31

1 Answers 1

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You can basically think of this quotient ring as "$\mathbb{Q}$ with $\sqrt{-1}=x$ adjoined", or "the complex numbers with rational coefficients". (Fun to say.)

That said, #1 obviously has $x$ as a root.

For 2 and 3, you could check to see what their real roots look like with the quadratic formula... if those roots are in $R$, then they're reducible.

For 4 at the very worst you could actually substitute $ax+b$ and solve the resulting system of equations to see if you can get a solution in your extension. Or you can apply the rational root test and see if it has any rational roots immediately.

I get "no, yes, yes, no"

  • 0
    Since we are just interested in reducibility and not factoring, it's probably easier to apply the rational root test and notice -1 is a root.2012-08-31