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Let $\pi \in \mathbb{Z}[i]$ irreducible and $\pi$ divides $p$ ($p\in\Bbb Z$ prime). Then show that $N(\pi)=p$.

($N(\alpha)= a^2 +b^2$ where $\alpha = a+ib$.)

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Welcome to Math.StackExchange! Questions that show no effort tend to be frowned upon, and especially so if they're just copy-pasted homework problems. What have you tried? What are your thoughts? – Henry Swanson Apr 4 '14 at 2:04
the reason why I ask is that I have no idea just begin to solve this problem....maybe it tend to be frowned upon but I lose to much time on it...... – jenny Apr 4 '14 at 2:28
Jenny, I took the liberty of typesetting your formulas in TeX. Please check that I didn't distort the meaning. Click the edit-link below your question to see the source code. You can try and use it yourself in the future. – Jyrki Lahtonen Apr 7 '14 at 9:11
up vote 0 down vote accepted

Hint: Definite the norm by $N(\alpha) = N(a + ib) = a^2 + b^2$, where $\alpha \in \mathbb{Z}[i]$. Suppose a prime $p$ factors nontrivially as $p = (a + ib)(c + id)$. Then $N(p) = p^2 = (a^2 + b^2)(c^2 + d^2)$ because $N$ is multiplicative. Since $p$ factors nontrivially, neither $a^2 + b^2$ or $c^2 + d^2$ is a unit. Hence $p = a^2 + b^2 = c^2 + d^2$.

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Let us write your element as $\alpha=a+ib$ with $a,b\in\mathbb{Z}$ and $N(a)=q=(a+ib)(a-ib)=\alpha\overline{\alpha}$.

You assume that $\alpha$ divides $p$, hence $p=\alpha\beta$ for some $\beta\in\mathbb{Z}[i].$ Observing that the norm is multiplicative (follows from its definition and straight forward calculation), you get $$p^2=N(p)=N(\alpha\beta)=N(\alpha)N(\beta)= qN(\beta).$$ Hence $q$ divides $p^2$. Since $q$ is a positive integer and $p$ is a prime number you have either $q=1$, $q=p$ or $q=p^2$.

$a)$ $q=1$ implies that $(a,b)\in \{(0,\pm 1),(1,\pm 1)\}$ and thus that $\alpha$ is unitary, which is not true by hypothesis (you assume that $\alpha$ is irreducible).

$b)$ $q=p^2$ implies that $N(\beta)=1$, so $\beta$ is unitary. Since $\alpha$ is irreducible, then so is $p=\alpha\beta$. But you assumed that $p=1\pmod{4}$, so by Fermat's theorem on sum of squares you can write $p=c^2+d^2=(c+id)(c-id)$, which implies that $p$ is not irreducible.

The only remaining case is $q=p$, which gives $N(\alpha)=p$ as you wanted.

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The element a is not the same in a^2 +b^2 the element is any....I should write c maybe – jenny Apr 7 '14 at 19:07
Yes, I wrote $\alpha=a+ib$ instead. – Jérémy Blanc Apr 7 '14 at 20:10

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