The setup is that $a \in \mathbb C$ is a root of $g(x)=x^3 – x + 1$. I'm trying to argue that the quotient $\mathbb Z[a] / (a+2)$ is isomorphic to $\mathbb Z_5$. I know that given our hypothesis, $\mathbb Z[a]$ must be a finitely generated $\mathbb Z$-module, with basis $\{1,a,a^2\}$. The strategy is to show that the map $f : \mathbb Z[a] \to \mathbb Z_5$ given by $f(r_1 + r_2a + r_3a^2) = r_1 -2r_2 + 4r_3$ is an onto homomorphism with kernel $(a+2)$. I'm having trouble showing that the kernel is contained in $(a+2)$. Any hints would be appreciated.
Abstract Algebra – Proving Z[a]/(a+2) ? Z_5
abstract-algebra
Best Answer
Note that $x^3-x+1$ has a root in $\mathbb{Z_5}$, which is $3$. So this gives an idea to define $\varphi:\mathbb{Z}[a]\to\mathbb{Z_5}$ by $f(a)\to\bar{f}(3)$, where $\bar{f}\in\mathbb{Z_5}[x]$ is the reduction of $f$ mod $5$, i.e we take all the coefficients mod $5$. The map $\varphi$ is well defined (check this!), and is clearly a surjective ring homomorphism. We want to show that its kernel is the ideal $(a+2)$ of $\mathbb{Z}[a]$.
It is obvious that $(a+2)$ is contained in the kernel. Conversely, assume $f\in\mathbb{Z}[x]$ is a polynomial such that $f(a)\in\text{Ker}(\varphi)$. We can divide $f$ by $x+2$ with remainder in $\mathbb{Z}[x]$. (possible because $x+2$ has an invertible leading coefficient) So there are $q\in\mathbb{Z}[x]$ and $r\in\mathbb{Z}$ such that $f(x)=(x+2)q(x)+r$. Then $f(a)=(a+2)q(a)+r$, and hence:
$0=\varphi(f(a))=\varphi(a+2)\varphi(q(a))+\varphi(r)=0+\varphi(r)=r \pmod{5},$
which means that $5|r$. So it remains to show that $5\in (a+2)$, and then it will follow that $f(a)=(a+2)q(a)+5s$ (where $r=5s$) belongs to $(a+2)$ as well.
To show that $5\in (a+2)$, note first that $a^3+3=[a+2]+[a^3-a+1]=a+2\in (a+2)$. Also, $a^3+8$ is divisible by $a+2$ ($-2$ is a root of the polynomial $x^3+8$), and so belongs to $(a+2)$ as well. Thus, $5=[a^3+8]-[a^3+3]\in (a+2)$, as required.