I am reading ring theory from Abstract Algebra by Dummit & Foote and currently in the chapter Unique Factorization Domain. In this book on Page no: 290 authors state the following (in the 3rd paragraph):

"It follows by what we just saw that $~p$ factors in$~\mathbb{Z}[i]$ into precisely two irreducibles if and only if $p=a^2+b^2$ is the sum of two integer squares (otherwise $p$ remains irreducible in $\mathbb{Z}[i])$ "

I know that this statement is basically drawn on the basis of the previous two paragraphs. But actually those paragraphs are not totally clear to me.

What I understand is:

  • In the 1st paragraph (Page:290) authors proved that "The prime elements in the ring $\mathscr{O}=\mathbb{Z}[\sqrt D]$ (for any integer $D$ which is not perfect square) are factors of some prime number of $\mathbb{Z}$ (this mean the factors of the prime numbers of $\mathbb{Z}$ in the larger ring $\mathscr{O})$."
  • For this, they FIRST proved (in 1st paragraph P-290) If we take any prime $\pi \in \mathscr{O}$ then $<\pi>$ is a prime ideal and it is clear that $~<\pi> \cap \mathbb{Z}$ is a prime ideal in $\mathbb{Z}$, and hence $~<\pi> \cap \mathbb{Z}=<p>$, for some prime number $p\in \mathbb{Z}$. Consequently, $\pi$ is a divisor of $p$ in $\mathscr{O}$.
  • and SECONDly they proved (in 1st paragraph P-290), that if $\pi \in \mathscr{O}$ is an irreducible element and $\pi$ is a factor of some prime number $p$ of $\mathbb{Z}$. i.e., $p=\pi \pi'$ in $\mathscr{O}$. Then it can be proved (as they have shown) either $p$ remains irreducible in $\mathbb{Z}[i]$ (when $\pi'$ becomes an unit ) or $p$ is precisely product of two irreducible in $\mathbb{Z}[i]$.

But I cannot understand how these result the statement about $\mathbb{Z}[i]$ as a special case...(which I mentioned at first).

Please help. Thank you.


If $p$ is the product of an irreducible $a+ib$ and another number $u+iv$, then the imaginary part $av+bu$ of this product must be zero, i.e., $u-iv$ is a rational multiple of $a+ib$. As we clearly need that $a,b$ are coprime, $u-iv$ is an integer multiple of $a+ib$. So far we have $p=(a+ib)\cdot n(a-ib)=n\cdot (a^2+b^2)$, hence $a^2+b^2=1$ (which makes $a+ib$ a unit) or, as desired, $a^2+b^2=p$. Conversely, if $p=a^2+b^2$, then $p=(a+ib)(a-ib)$ and $a+ib$ is irreducible because the norm of any divisor would have to divide its norm, a prime.

  • $\begingroup$ ..Why should $av+bu=0$ implies $u-iv$ to be a rational multiple of $a+ib$? $\endgroup$ – indrajit Apr 25 '18 at 2:57
  • $\begingroup$ Can I proceed in this way: Let $p$ is product of two irreducibles in $\mathbb{Z}[i]$, say, $p=(a+bi)(u+vi)$ taking norms we get $p^2=(a^2+b^2)(u^2+v^2)$ then $a^2+b^2=p ~\text{or}~ p^2$ but if it is $p^2$ then $u+iv$ would be a unit. so $p=a^2+b^2$......Does this suffice the necessity part.. $\endgroup$ – indrajit Apr 25 '18 at 6:02

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