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Is there a non-advanced theorem (or some other simple way) to show that $\langle 1+i \rangle$ is prime in $\mathbb{Z}[i]$? I'm somewhat stuck. I've so far argued that any even integer and any number of the form $a+ib$, where $a,b$ are either both even or both odd, is in this ideal. But any number of this form that is divisible into some number $\ne 1+i$ is also of the same form, so it must still be in this ideal. Hence, this ideal is prime.

But I don't feel this is quite rigorous.

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    Can you see how many elements the quotient $Z[i]/(1+i)$ has? – 2017-02-13
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    Looks like 3 elements? $\langle 1+i \rangle$, $1+ \langle 1+i \rangle$, $i+ \langle 1+i \rangle$. – 2017-02-13
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    Not so fast! $1-i=-i(1+i)\in\langle 1+i\rangle$, so two of those cosets are equal. Draw a picture of $\Bbb{Z}[i]$ on the $xy$-plane, and circle the elements of $\langle 1+i\rangle$. You should get a picture that reminds you of a chessboard. – 2017-02-13
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    So according to Jyrki's observation there are at most two elements in the quotient. Are there really two elements or just one? – 2017-02-13
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    I don't understand. How about the cosets $1+ \langle 1+i \rangle$ and $i+ \langle 1+i \rangle$? They are not in $\mathbb{Z}[i]/ \langle 1+i \rangle$? – 2017-02-13
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    @sequence: Those are the same coset. – 2017-02-13
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    How can one see that they are the same coset? – 2017-02-13
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    The solution goes the same way as in [this duplicate question](http://math.stackexchange.com/questions/361099/show-that-2i-is-a-prime-ideal). – 2017-02-13

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He asked for the simplest way, so probably he does not want to use quotients. Let $a+bi, c+di \notin \langle 1+i \rangle$, i.e. by the definition of a prime ideal we have to show that the product is also not contained in the ideal.

You have already figured out that an element is contained in the ideal if and only if the parity of the real and imaginary part are the same.

This translates into $a-b = c-d = 1 \mod 2$. Note that there are no signs mod 2, i.e. also $c+d=a+b=1 \mod 2$.

We have $$(a+bi)(c+di)=ac-bd + (ad+bc)i$$ and compute $$ac-bd-(ad+bc) = a(c-d)-b(c+d)=a-b=1 \pmod 2,$$ hence $$(a+bi)(c+di)=ac-bd + (ad+bc)i \notin \langle 1+i \rangle.$$

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    If $a$ and $b$ are both even or both odd then $a-b = 0 \mod 2$, I think. – 2017-02-13
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    Thats true, but i started with elements that are NOT in the ideal. – 2017-02-13
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HINT: Consider the map $\phi : \Bbb Z[i] \to \Bbb Z_2$ by $a+ib \to a-b$. See that the Kernel of the map $\phi$ is $(1+i)$.

Now $\Bbb Z[i]/(1+i) \simeq Z_2$.

Since $\Bbb Z_2$ is a field we conclude that $(1+i)$ is a maximal ideal and hence a prime ideal.

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    I'm not sure how I can use this fact. Seems like I don't know something. – 2017-02-13
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    now you understood – 2017-02-13
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    @user8795 Maybe use the thoughts the OP already gave in his question? – 2017-02-13
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    @user8795 Sorry, I'm not sure what I can conclude about $\langle 1+i \rangle$ from this. Perhaps I'm missing some theory. – 2017-02-13
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    got it now/..?? – 2017-02-13
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A proper ideal $I$ of a ring $R$ is prime if and only if $R/I$ is an integral domain (i.e. it has no zero divisors).

One of the most useful calculation devices for computing rings is that the "evaluation at $a$" map is an isomorphism

$$ R[x] / (x-a) \to R $$

We can use this if we convert the problem to one of polynomial rings:

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

and consequently,

$$ \begin{align*} \mathbb{Z}[i] / \langle i+1 \rangle &\cong \left( \mathbb{Z}[x] / (x^2 + 1) \right) / \langle x+1 \rangle \\&\cong \mathbb{Z}[x] / (x^2 + 1, x+1) \\&\cong \left( \mathbb{Z}[x] / (x + 1) \right) / \langle x^2 + 1 \rangle \\&\cong\mathbb{Z} / \langle (-1)^2 + 1 \rangle \\&\cong \mathbb{Z} / \langle 2 \rangle \\&\cong \mathbb{F}_2 \end{align*}$$

Since $\mathbb{F}_2$ is an integral domain, conclude $\langle i+1 \rangle$ is prime. In fact, we can say more: because the result is a field, we know $\langle i+1 \rangle$ is a maximal ideal.

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    What does $\mathbb{F}_2$ look like? – 2017-02-13
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For any positive integer $n$ we consider the homomorphism $\chi: \mathbb{Z} \rightarrow \mathbb{Z}[i]/ I, z\mapsto z+ I$, where $I$ is the principal ideal of $\mathbb{Z}[i]$ generated by the element $n+i$ of $\mathbb{Z}[i]$. A simple calculation shows that $\ker\chi= J$, where $J$ is the principal ideal of $\mathbb{Z}$ generated by the element $n^2+1$. Moreover, $\forall a, b\in \mathbb{Z}$ it holds $(a+bi)+ I=(a-nb)+I$. Hence, $\chi$ is an epimorphism. Now the first isomorphism theorem tell us that $\mathbb{Z}/J\cong \mathbb{Z}[i]/ I$.

For $n=1$, we have $J=(1+i)$, $I=(2)$ and $\mathbb{Z}/(2)\cong \mathbb{Z}[i]/ (1+i)$ Since $\mathbb{Z}/(2)$ is a field $(1+i)$ is maximal ideal and hence prime.

Generally, $(n+i)$ is a maximal ideal of $\mathbb{Z}[i]$ iff $n^2+1$ is a prime integer.