Show that (2) is prime ideal in $\mathbb{Z}[i]$

Question

I want to show that (2) is a prime ideal in $\mathbb{Z}[i]$. I know that I need to show that the quotient ring $\mathbb{Z}[i]/2$ is an integral domain. I'm not sure how this can be done or what $\mathbb{Z}[i]/2$ looks like. I also know that $\mathbb Z[i]\simeq \mathbb Z[x]/(x^2+1)$, but I'm not sure why.

Answer 1

I want to show that (2) is a prime ideal in $\mathbb{Z}[i]$.

Did you consider $2=(1+i)(1-i)$?

I'm not sure how this can be done or what $\mathbb{Z}[i]/2$ looks like.

Maybe you should read a couple other solutions on the site about understanding quotient rings, like this one

I also know that $\mathbb Z[i]\simeq \mathbb Z[x]/(x^2+1)$, but I'm not sure why.

You can prove that using the first homomorphism theorem for rings with a map from $\mathbb Z[x]\to \mathbb Z[i]$ given by substituting $i$ into polynomials in $\mathbb Z[x]$.

Answer 2

This claim is false. Observe that $0 = (x+1)(x+1) \in \mathbb{Z}[x]/(x^2+1,2)$. Conclude that the quotient is not a domain, hence that (2) is not prime.

Answer 3

I don't think $(2)$ is a prime ideal of $\mathbb{Z}[i]$. I hope that this can answer your second question: The map $\varphi: \mathbb{Z}[X] \to \mathbb{Z}[i], f \mapsto f(i)$ is a well defined ring homomorphism. One can now check that $\text{ker}(\varphi)=(x^2+1)$ and that $\varphi$ is indeed surjective, hence $\mathbb{Z}[X]/(x^2+1) \cong \mathbb{Z}[i]$.

Answer 4

Yes, $\,R = \Bbb Z[i]\cong \Bbb Z[x]/f,\ f =x^2\!+\!1,\,$ follows immediately from the First Isomorphism Theorem, as in adh's answer. You seek to prove that $2$ is not prime in $\,R\,$ by showing $\,R/2\,$ is not a domain. This is easy by quotient reciprocity, which we do for $\,f = x^2-d,\,$ and $\,2\to p\,$ below

$$\begin{align} \Bbb Z[\sqrt d]/(p) &\,\cong\, (\Bbb Z[x]/f)\,/\,(p,f)/f\\[.2em] &\,\cong\, \ \Bbb Z[x]/(p,f)\\[.2em] &\,\cong\, (\Bbb Z[x]/p)\,/\,(f,p)/p\\[.2em] &\,\cong\ \ \Bbb F_p[x]/(f) \\[.2em] &\,\cong \ \ \Bbb F_2[x]/(x\!+\!1)^2\ \ {\rm in\ OP\!:}\,\ p\!=\!2,\,d\!=\!-1 \end{align}\qquad$$

The final quotient is not a domain since there $\,x\!+\!1 \neq 0$ but $\,(x\!+\!1)^2 = 0.\,$ This is called algebra of dual numbers over $\Bbb F_2.\,$ Dual numbers are useful as algebraic models of tangent and jet spaces.

Generally, using $\,\Bbb Z[\sqrt d]/(p)\,\cong\,\Bbb F_p[x]/(x^2-d)\,$ from the quotient reciprocity above, then employing $\,I\,$ prime $\iff R/I$ domain, $ $yields the following

$\!\begin{align}{\bf Prime\ Reciprocity}\qquad\quad\ \ &\ \ \ \ \ \ \ \ \ p\, \text{ is prime in } \Bbb Z\sqrt d\\[,2em] \iff\ & x^2-d\,\ \text{is prime in}\ \Bbb F_p[x]\\[.2em] \iff\ &\ \ d\,\ \text {is nonsquare in } \Bbb F_p\end{align}$

Similarly, quotient reciprocity allows us to deduce the ideal factorization of rational primes in higher degree number rings (a fundamental result of Dedkeind, e.g. see here).