# Method for determining irreducibles and factorising in $\mathbb Z[\sqrt{d}]$

I know that $\mathbb Z[\sqrt{7}]$ is a UFD, and I can write the equation $(2 + \sqrt{7})(3 - 2\sqrt{7}) = (5 - 2\sqrt{7})(18 + 7\sqrt{7})$. So clearly these are not all irreducibles. How do I determine which of these aren't irreducible? How do I factor those into irreducibles?

It seems difficult when dealing with $\mathbb Z [\sqrt{d}]$ with $d$ positive, since the norm map isn't as nice (it's not always positive).

Thanks

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I added the algebraic-number-theory tag. In your case, $\mathbb Z[\sqrt 7]$ happens to be the ring of integers in the quadratic number field $\mathbf Q(\sqrt 7)$. This is an infinitely interesting class of objects. –  Dylan Moreland Feb 2 '12 at 18:42
@DylanMoreland. Thanks. I actually placed that tag there myself, but someone removed it. –  Matt Feb 2 '12 at 20:49
The page link in this answer may be useful, since it gives a complete description of what happens to the rational primes in the quadratic extensions of $\mathbb{Q}$ (in terms of ideals, of course). –  Arturo Magidin Feb 2 '12 at 22:08

An irreducible of $\mathbb{Z}[\sqrt{7}]$ must be a prime of $\mathbb{Z}$, or must lie above a prime of $\mathbb{Z}$.

Since $\mathbb{Z}[\sqrt{7}]$ is the splitting field of $x^2-7$, you want to determine how $x^2-7$ factors modulo $p$, for $p$ prime. In other words, for which primes is $7$ a square modulo $p$?

If $p=2$ or $p=7$, $x^2-7$ factors as a perfect square; so $2$ and $7$ ramify; that is, there is a unique irreducible (up to units) that divides $2$ and that divides $7$. It is not hard to find $3+\sqrt{7}$ as a divisor of $2$: $(3+\sqrt{7})(3-\sqrt{7}) = 2$. Note that $N(3+\sqrt{7}) = 2$ is a prime, so that means that $3+\sqrt{7}$ is necessarily irreducible. And of course, $\sqrt{7}$ divides $7$, and is irreducible.

For $p\neq2, 7$, we have: for $p\equiv 1\pmod{4}$, $$\left(\frac{7}{p}\right) = \left(\frac{p}{7}\right) = \left\{\begin{array}{ll} 1 & \text{if }p\equiv 1,2,4\pmod{7}\\ -1 & \text{if }p\equiv 3,5,6\pmod{7}. \end{array}\right.$$ For $p\equiv 3\pmod{4}$, $$\left(\frac{7}{p}\right) = -\left(\frac{p}{7}\right) = \left\{\begin{array}{ll} 1 & \text{if }p \equiv 3,5,6\pmod{7}\\ -1 &\text{if }p\equiv 1,2,4\pmod{7} \end{array}\right.$$ So a rational prime $p$ splits into a square (times a unit) if $p=2$ or $p=7$; factors into two distinct irreducibles in $\mathbb{Z}[\sqrt{7}]$ if and only if $p\equiv 1, 3, 9, 19, 25, 27\pmod{28}$; and remains irreducible if $p\equiv 5, 11, 13, 15, 17, 23$.

The units are precisely the elements of norm $\pm 1$; in fact, it cannot be $-1$, since that would require $a^2 -7b^2 =-1$, which has no solutions modulo $7$. You can find, through the "usual" methods, that every unit is of the form $\pm(8+3\sqrt{7})^k.$

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The norm is good enough for getting a good idea about the factors: $2 + \sqrt{7}$ has prime norm, hence is irreducible, and the same goes for the other three factors. Thus the factors with equal norm must be associates, i.e. their quotients are units. You can find these by writing down the corresponding fractions and making the denominators rational.

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What if I chose an example where the norm isn't prime? Take $2 \cdot 11 = (5 + \sqrt{3})(5 - \sqrt{3})$ in $\mathbb Z[\sqrt{3}]$. I know $2 = (1 + \sqrt{3})(-1+\sqrt{3})$ isn't irreducible, but what about the other terms? –  Matt Feb 2 '12 at 17:45
@Matt: Since $x^2-3=(x-5)(x+5)\pmod{11}$, $(11)=(11,5+\sqrt{3})(11,5-\sqrt{3})$, so $11$ is not prime in $\mathbb{Z}[\sqrt{3}]$. (If $d$ is a square modulo $p$, then $p$ is not prime in $\mathbb{Z}[\sqrt{d}]$). –  Arturo Magidin Feb 2 '12 at 17:55
@ArturoMagidin. But using the norm $N(a+b\sqrt{3}) = a^2 - 3b^2$, we see that $N(11) = 11^2$. So if $11$ factorises into irreducibles, they must both have norm $11$. But $a^2 - 3b^2 = 11$ implies $a^2 = 2$ mod $3$, which has no solutions. So 11 must be irreducible? –  Matt Feb 2 '12 at 18:11
And then using the same logic on $5+\sqrt{3}$ and $5 - \sqrt{3}$, we get that $(1 + \sqrt{3})(-1 + \sqrt{3})\cdot 11 = (5+\sqrt{3})(5 - \sqrt{3})$, where all terms are irreducible and none are units. So this breaks unique factorisation. What am I doing wrong? –  Matt Feb 2 '12 at 18:19
@Matt: You are forgetting that there are numbers with negative norm. You could have $a^2-3b^2=-11$. And indeed, $(1-2\sqrt{3})(-1-2\sqrt{3}) = 11$, with $N(1-2\sqrt{3})=N(-1-2\sqrt{3}) = -11$. And similarly, $N(5+\sqrt{3}) = 22$, which can be expressed as a product of an element of norm $-11$ and an element of norm $-2$. –  Arturo Magidin Feb 2 '12 at 18:21