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I had this problem where i had the application $\varphi: \mathbb Z[i] \Rightarrow \mathbb Z/(2)$ where $\varphi(a+bi)=\bar{a}+\bar{b}$. I had to find the kernel and prove that is a factor ideal. I proofed that the kernel is formed by all the complex numbers such that $a+b$ is even but I need to find the generator to the ideal.

J. W. Tanner
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bernat tobella
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2 Answers2

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Since $\mathbb{Z}[i]$ is a Euclidean domain and this implies that it is a PID we have that any ideal $I \subset \mathbb{Z}[i]$ is generated by any element in $\mathbb{Z}[i]$ with the smallest norm; where the norm is defined by $a+bi \mapsto a^2 + b^2$. Therefore we need to find a non-zero $a,b$ that satisfies $a+b \equiv 0 \ (\text{mod } 2)$ that has the smallest possible $a^2+b^2$.

Therefore $1+i$ works; $1-i$, $-1+i$, and $1-i$ also work.

Pedro Juan Soto
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    Okay thanks, I guess i need some more theory to understand this proof, but thanks again :) – bernat tobella May 11 '20 at 18:01
  • No worries, let me help you with that. The proof that Euclidean Domain$\subset$PID is a very nice and beautiful algorithmic proof that is easy to understand (and is one of my personal favorites :)). The idea is that "Euclidean Domain" just means that the Euclidean algorithm works, where now the norm takes the place of the remainder in the induction/recursion. To prove that it is a PID you perform a second algorithm on the steps of the Euclidean Algorithm similar to the proof of Bezout's identity https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity. Euc. Dom.$\subset$PID is a generalization. – Pedro Juan Soto May 11 '20 at 18:25
  • @bernattobella In a Euclidean domain a nonzero ideal is generated by any element of least Euclidean value (or least number of prime factors in a PID). See here for the trivial proof. – Bill Dubuque May 11 '20 at 18:34
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A generator of the ideal is $1+i$.

Let $a+bi$ be a multiple of $1+i.$

Then $a+bi= (x+yi)(1+i)=(x-y)+(x+y)i,$ so $a+b=2x$ is even.

On the other hand, if $a+b$ is even, then so is $b-a$,

so $\dfrac{a+bi}{1+i}=\dfrac{(a+bi)(1-i)}2=\dfrac{a+b}2+\dfrac{(b-a)i}2\in \mathbb Z[i]$.

J. W. Tanner
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  • thanks, i guess this proof that all (1+i) generated numbers have that a+b is even and also that if a+b is even, 1+i divides a+bi, I am right? – bernat tobella May 11 '20 at 18:04
  • yes, that's the idea – J. W. Tanner May 11 '20 at 18:05
  • @bernattobella: I thought that's what you were asking for – J. W. Tanner May 11 '20 at 18:13
  • @bernattobella We don't need to pull the generator out of a hat like magic as above. Instead we can derive it from general principles as the smallest element - see my comment on the other answer. – Bill Dubuque May 11 '20 at 18:36
  • it's not really out of a hat, @Gone: we need $n+mi$ to divide $2$, but $1$ and $2$ are clearly not the generator, so we must have $m\ne0$ and $(n+mi)(n-mi)=n^2+m^2=2\implies $ $n^2=m^2=1\implies $ $n+mi=\pm1\pm i$ – J. W. Tanner May 11 '20 at 21:04
  • @J.W.T But it is "magic" since you simply claim the generator is $1+i$ then verify it. You give no motivation for that claim. How is the OP supposed to generalize that? Just pick a random element of an ideal then try to see if it is a generator? I don't follow your argument in your prior comment. What are you trying to show there? – Bill Dubuque May 11 '20 at 21:26
  • Furthermore, the general/theoretical approach finds all 4 solutions instead of this ad hoc guess and verify approach. How would you find the generator of a more general ideal? Just keep guessing? – Pedro Juan Soto May 14 '20 at 16:08