Question about a proof of $\mathbb{Z}[i]/(2+i)\cong \mathbb{Z}/5\mathbb{Z}$

Question

Proof:

Since $\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)$, where $i$ and $x$ correspond,

we have: $\mathbb{Z}[i]/(2+i)\cong \mathbb{Z}[x]/(x^2+1,2+x)$.

In that ring, since $x=-2$, $x^2+1=5$, and

$\mathbb{Z}[x]/(x^2+1,2+x)\cong \mathbb{Z}[x]/(5,2+x)\cong (\mathbb{Z}/5\mathbb{Z})[x]/(2+x)\cong \mathbb{Z}/5\mathbb{Z}$,

since $x=-2$ is in the quotient ring.

Source: problem 1b in https://sites.math.washington.edu/~sullivan/4034s_wi13.pdf

My questions:

-What is meant by $i$ and $x$ correspond? Why is $\mathbb{Z}[i]/(2+i)$ isomorphic to $\mathbb{Z}[x]/(x^2+1,2+x)$. Why is it okay to just replace $i$ with $x$? I understand the $\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)$ part.

-Why is $(\mathbb{Z}/5\mathbb{Z})[x]/(2+x)$ isomorphic to $\mathbb{Z}/5\mathbb{Z}$? T tried but I can't find a homomorphism whose kernel would be $(2+x)$.

Thank you in advance!

Answer 1

• $i$ is the solution to the equation $x^2+1=0$. When you consider $\mathbb{Z}[x]$ and mod out by the ideal $(x^2+1)$, you are creating a new ring in which $x^2+1=0$, that is, you're saying that $x$ satiesfies $x^2+1=0$, so $x$ corresponds to $i$.

• $\mathbb{Z}[i]/(i+2)\cong\frac{\mathbb{Z}[x]}{(x^2+1)}/(x+2)\cong\mathbb{Z}[x]/(x^2+1,x+2)$

• Modding out by $(x+2)$, means that you're forming a new ring in which $x=-2$, but $-2$ is already inside $\mathbb{Z}_5$, so you're not getting anything new.

Answer 2

For your second question: what about

$$\phi:\Bbb Z/5\Bbb Z[x]\to\Bbb Z/5\Bbb Z\;,\;\;\phi(p(x)):=p(-2)\;\;?$$

This is just the evaluation homomorphism (chech it is a ring homomorphism!) ,and its kernel is the set (ideal, in fact) of all polynomials in $\;\Bbb Z/5\Bbb Z[x]\;$ which vanish at $\;x=-2\;$ ...