Prime number $p=4n+1$ in $\mathbb{Z}$ is not prime in $\mathbb{Z}[i]$
Prime number $p=4n+1$ is not prime in $\mathbb{Z}[i]$
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abstract-algebra
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0any prime number $p\in \mathbb Z$ with $p\equiv 3(mod 4)$ is prime in $\mathbb Z[i]$ , for more details in the book *Algebraic number theory, Mollin* – 2017-02-05
2 Answers
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Hint: Use the famous theorem of Fermat:
If $p > 2$ is prime then $p = a^2 + b^2$, for some natural numbers $a$ and $b$, if and only if $p = 4n+1$ for some natural number $n$.
Now once you have written $p = a^2 + b^2$ then you can factorize it in the Gaussian Integers $\mathbb{Z}[i]$ and hence it is not prime!
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0I want to prove that $p$ is not prime in $\mathbb{Z}[i]$ and then prove the Fermat's theorem. – 2017-02-05
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0but your questions does not seems to ask so!! – 2017-02-05
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0There are many other ways to prove the fermat's theorem! – 2017-02-05
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See $\mathbb{Z}[i]$ is practically $ \frac{\mathbb{Z}[X]}{X^2 + 1}$. Now p is prime if $ \frac{\mathbb{Z}[X]}{(X^2 + 1,p)} \cong \frac{\mathbb{Z}_p[X]}{\overline{X}^2 + 1}$ is an integral domain. where $ \overline{X}$ is the image of X in modulo p . So this is integral domain whether $X^2 + 1$ is reducible in modulo p or not. And by an elementary number theorem ,we get $X^2 \equiv -1$ modulo p has a solution iff $ p = 4n + 1$.