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Related to this question here Adjoining elements to $\mathbb{Z}$ given a set of generators:

I want to determine the structure of $R'$ obtained by adjoining $\alpha$ to $\mathbb{Z}$ with generators $\alpha^3+\alpha^2+\alpha+1 = 0, \alpha^2 +\alpha = 0$.

I was able to reduce $(\alpha^3+\alpha^2+\alpha+1, \alpha^2 +\alpha)$ to $(\alpha^2 +\alpha,\alpha+1)$. Here, we can see that $\alpha^2 +\alpha = \alpha(\alpha+1)$, so this is equivalent to the principal ideal $(\alpha+1)$.

My question is, what more can we say about $\mathbb{Z}[x]/(x+1)$? I'm still struggling with this concept, so I'm not sure where I can go from here.

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Lost
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    $\alpha^3 + \alpha^2 + 1 - \alpha(\alpha^2 + \alpha) = 1$... – bzc Dec 12 '13 at 04:20
  • @BrandonCarter I apologize, but I made a mistake: it should've been $\alpha^3 + \alpha^2 + \alpha + 1$. The computations are based on that. – Lost Dec 12 '13 at 04:27

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By quotienting $\mathbb{Z}[x]$ by the ideal generated by $x+1$, you are effectively dictating that $x+1=0$, or that $x=-1$. Therefore, $\mathbb{Z}[x]/(x+1) \cong \mathbb{Z}[-1] \cong \mathbb{Z}$. An isomorphism $\phi$is given by $\phi([f]) = f(-1)$, where $[f]$ is the coset of the polynomial $f$, or, more simply, evaluation of a polynomial at $-1$. This map is well-defined because, if $x+1$ divides a polynomial $f$, then $f(-1) = 0$.

William Ballinger
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  • I was wondering if we were allowed to simply solve for $x$, so I was hesitant about proceeding with that. I assume it's because the ideal is principal.

    I do see why $\mathbb{Z}[-1] \cong \mathbb{Z}$. Thanks for the help.

     – Lost Dec 12 '13 at 04:24