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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.

  • 2
    Do you understand how to manipulate ring quotients? Can you write down what $y^2 + 1$ is, for example?2012-08-31
  • 2
    Is this homework? Hint: Look for factors of $y^2+1$ because mod $1+x^2$ these are $0$.2012-08-31
  • 9
    The question does not make any sense to me. If $y$ is, indeed, the coset $x+I$, then e.g. $y^2+y+1$ is just an element of $R$. But $R$ is a field. A field does not have irreducible elements. A field has only units. If OTOH $y$ would be another variable, then we could discuss, whether these polynomials in $y$ are irreducible elements in the polynomial ring $R[y]$!!2012-08-31
  • 2
    $y$ should not be $x+I$ but a new indeterminate.2012-08-31
  • 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"

  • 1
    Did you read the question? It specifies that $y$ is the coset $x+I$. In other words, in the natural interpretation $R=\mathbb{Q}[i]$ we have $y=i$, so $y^2+1=0$, $y^2+y+1=i$ et cetera. I think that your interpretation is the one that turns the question into a meaningful one, but as it is written, this does not yet fit. +1 for the most sensible guess of what the question really was :-)2012-08-31
  • 0
    @JyrkiLahtonen You're absolutely right, but this seemed like the most plausible intended question so I went with it.2012-08-31
  • 1
    for the last polynomial multiply by $(y-1)$ to get $y^4-1$ whihc factors as $(y^2+1)(y+1)(y-1)$2012-08-31
  • 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