# Show that $a + b\sqrt{n}$ is an irreducible element of $\mathbb{Z}[\sqrt{n}]$

Let $n(\neq 0,1)$ be a square-free integer. Suppose that $|a^2 - nb^2|$ is a prime integer for $a,b \in \mathbb{Z}$. Show that $a + b\sqrt{n}$ is an irreducible element of $\mathbb{Z}[\sqrt{n}]$. Then, show that $p = |a^2-nb^2|$ is not irreducible in $\mathbb{Z}[\sqrt{n}]$.

I can't prove the first part. I think the question is a bit un clear too. Are all $a + b\sqrt{n}$ is an irreducible element of $\mathbb{Z}[\sqrt{n}]$? Or just those with norm being a prime? For the second, I find it weird because i thought all primes in an integral domain is irreducible. Is $\mathbb{Z}[\sqrt{n}]$ not an integral domain? If so, can someone give me an example of a zero divisor in $\mathbb{Z}[\sqrt{n}]$ or a proof of $\mathbb{Z}[\sqrt{n}]$ not being an integral domain?

Thanks.

My attempt:

Claim 1 $N((a+b\sqrt{n})(c+d\sqrt{n}) = N(a+b \sqrt{n})N(c+d\sqrt{n})$

Proof \begin{align*} & N((a+b\sqrt{n})(c+d\sqrt{n}) \\ &= N((ac + bdn) + (ac + bd)\sqrt{n}) \\ &= |(ac+bdn)^2 - n(ac+bd)^2| \\ &= |(a^2c^2 + 2abcdn + b^2d^2n^2) - n(a^2c^2 + 2abcd + b^2d^2)| \\ &= |a^2c^2 + 2abcdn + b^2d^2n^2 - a^2c^2n - 2abcdn - b^2d^2n| \\ &= |a^2c^2 + b^2d^2n^2 - a^2c^2n - b^2d^2n| \\ &= |(a^2 - nb^2)(c^2 - nd^2)| \\ &= |a^2 - nb^2||c^2 - nd^2)| \\ &= N(a+b\sqrt{n})N(c+ d \sqrt{n}) \end{align*}

Claim 2. $a+b\sqrt{n}$ is a unit in $\mathbb{Z}[\sqrt{n}]$ if and only if $N(a+b\sqrt{n}) = 1$.

Proof

Since $a+b\sqrt{n}$ is a unit in $\mathbb{Z}[\sqrt{n}]$, we can find $(c+d \sqrt{n}) \in \mathbb{Z}[\sqrt{n}]$ such that $(a+b\sqrt{n})(c+d \sqrt{n}) = 1$. This gives us $N(a+b\sqrt{n})N(c+ d \sqrt{n}) = N((a+b\sqrt{n})(c+d \sqrt{n})) = N(1) = 1$. From the way the norm is defined, $N(a+b\sqrt{n}) \geq 0$ and $N(c+ d \sqrt{n}) \geq 0$. Therefore, we must have $N(a+b\sqrt{n}) = 1$ and $N(c+ d \sqrt{n}) = 1$.

I am having trouble proving the other direction

Proof for the question

First part Suppose $a + b\sqrt{n}$ is reducible in $\mathbb{Z}[\sqrt{n}]$, then $a + b\sqrt{n} = (\alpha_1 + \beta_1 \sqrt{n})(\alpha_2 + \beta_2 \sqrt{n})$ for some non-units $\alpha_1 + \beta_1 \sqrt{n}, \alpha_2 + \beta_2 \sqrt{n} \in \mathbb{Z}[\sqrt{n}]$. This gives us $|a^2 - nb^2| = N(a + b\sqrt{n}) = N(\alpha_1 + \beta_1 \sqrt{n})(\alpha_2 + \beta_2 \sqrt{n}) = N(\alpha_1 + \beta_1 \sqrt{n})N(\alpha_2 + \beta_2 \sqrt{n})$. By claim 2, $N(\alpha_1 + \beta_1 \sqrt{n}) \neq 1$ and $N(\alpha_2 + \beta_2 \sqrt{n} \neq 1$ as $\alpha_1 + \beta_1 \sqrt{n}, \alpha_2 + \beta_2 \sqrt{n}$ are non-units. This implies that $N(\alpha_1 + \beta_1 \sqrt{n}) = p$ and $N(\alpha_2 + \beta_2 \sqrt{n}) = p$ as $p$ is prime. This presents us with a contradiction as $p = N(a + b\sqrt{n}) = N(\alpha_1 + \beta_1 \sqrt{n})N(\alpha_2 + \beta_2 \sqrt{n}) = p^2$. Hence, $a + b\sqrt{n}$ is irreducible in $\mathbb{Z}[\sqrt{n}]$.

Second part If $p$ is irreducible in $\mathbb{Z}[\sqrt{n}]$, then if $p = ab$ for some $a,b \in \mathbb{Z}[\sqrt{n}]$, $a$ is a unit or $b$ is a unit . This gives us $p^2 = N(p) = N(a)N(b)$. If $a$ is a unit, then $N(b) = p^2$.

I don't know how to continue

If $|ab| \neq |a||b|$, then my proof probably fail, but i think this is true.

Note that $(a^2-nb^2)(c^2-nd^2)=(ac+bdn)^2-n(ad+bc)^2$, which is the multiplicativity of norms, related to $(a-b\sqrt n)(c-d\sqrt n)=(ac+bdn)-(ad+bc)\sqrt n$.

Suppose you have a factorisation $a-b\sqrt n=(c-d\sqrt n)(e-f \sqrt n)$ then you also have $$\pm p=a^2-nb^2=(c^2-nd^2)(e^2-nf^2)$$ and this is a factorisation in $\mathbb Z$, so one of the factors, say the first, must be $\pm 1$.

This then gives $(c-d\sqrt n)(c+d\sqrt n)=c^2-nd^2=\pm 1$ so that $c-d\sqrt n$ is a unit (we have found an explicit multiplicative inverse).

For the second part $\pm p=(a+b \sqrt n)(a-b\sqrt n)$

Suppose $1=(a\pm b\sqrt n)(c\pm d\sqrt n)=(ac+bdn)\pm (ad+bc)\sqrt n$ (taking the same sign both times) so that one factor is a unit. We know that $n$ is square free, so $\sqrt n$ is irrational. But this equation expresses $\sqrt n$ as a rational number unless $ac+bdn=1$ and $ad+bc=0$. Let $q=c^2-nd^2$

Now we have $\pm pq=(a^2-nb^2)(c^2-nd^2)=(ac+bdn)^2-n(ad+bc)^2=1$ which is a contradiction in $\mathbb Z$ because $p$ is not a unit (and we cannot, from this equation, have $q=0$).

The language of norms codifies such calculations. The fact that norms are multiplicative allows us to take questions about factorisation from a quadratic ring (or ring of integers in any number field) into $\mathbb Z$. It is important at this stage to get an accurate understanding of what is happening here. It is particularly useful to note that the formula for the norm gives an explicit inverse for an element of norm $\pm 1$ and therefore a direct proof that an element of norm $\pm 1$ is a unit.

Note that it is not necessarily true that a prime in $\mathbb Z$ is a prime in the quadratic field. If we set $n=-1$, for example, we find $2=i(1-i)^2$ where $i$ is a unit, and $5=(1+2i)(1-2i)$. The prime $5$ splits into two distinct factors, while the double factor for $2$ is an example of "ramification". It turns out that this is a significant difference.

• In the first part of your answer you've used $(a^{2}-nb^{2})$ and related this to $(a-b \sqrt{n})$ but shouldn't it be $(a + b \sqrt{n})$? Secondly, by finding a multiplicative inverse, is this proving that the element is irreducible? Is the proof they've given not complete? – MathsIsFun Apr 7 '20 at 11:02

The first part of the question is asking you to show that if $|a^2-nb^2|=p$ is prime as an integer, then $a+b\sqrt{n}$ is irreducible in $\mathbb Z[\sqrt{n}]$. The second part asks you that this $p$ is not irreducible (and therefore not prime) viewed as an element of $\mathbb Z[\sqrt{n}]$; that is, $p$ has a factorization as a product of non-units in $\mathbb Z[\sqrt{n}]$. For example, $3$ is not prime as an element of $\mathbb Z[\sqrt{3}]$, since $3=\sqrt{3}\sqrt{3}$ and $\frac{1}{\sqrt{3}}=\frac{1}{3}\sqrt{3}\not\in\mathbb Z[\sqrt{3}]$.

(It isn't the case that $\mathbb Z[\sqrt{n}]$ is not an integral domain; it clearly is a domain, since it is a subring of the real numbers.)

Both parts follow from the following lemmas:

Lemma 1: Let $||a+b\sqrt{n}||=|a^2-nb^2|$. Then for any $\alpha=a+b\sqrt{n}$ and $\beta=c+d\sqrt{n}$, $||\alpha\beta||=||\alpha||\cdot||\beta||$.

Idea of proof: Note that $||a+b\sqrt{n}||=|a+b\sqrt{n}|\cdot|a-b\sqrt{n}|$ and compute $(a\pm b\sqrt{n})(c\pm d\sqrt{n})$.

Lemma 2: Suppose $\alpha\in \mathbb Z[\sqrt{n}]$. Then $\alpha^{-1}\in\mathbb Z[\sqrt{n}]$ if and only if $||\alpha||=1$.

Idea of proof: If $\beta\in \mathbb Z[\sqrt{n}]$ then $||\beta||$ is a non-negative integer so Lemma 1 implies one direction. The other direction just requires a little high school algebra: rationalize a fraction with denominator $a+b\sqrt{n}$ by multiplying the numerator and denominator by the conjugate $a-b\sqrt{n}$.

• If it is an integral domain, shouldn't prime implies irreducible? What am i misunderstanding here? math.stackexchange.com/questions/405759/… – user10024395 Apr 24 '15 at 4:20
• Yes, you are correct that prime implies irreducible in an integral domain, but the property of being prime depends on the ambient set. You are given that $|a^2-nb^2|$ is prime in $\mathbb Z$, but it might not be prime viewed as an element of another ring. As a silly example, note that while $6$ is not prime in $\mathbb Z$, it IS prime in $2\mathbb Z$ (the ring of even integers), since there is no way to write $6$ as a product of two EVEN integers. In the opposite direction, no integer is prime as a real number (since there are no primes in a field). – Sean Clark Apr 24 '15 at 4:25
• hi is |ab| = |a||b|. i can't seem to prove the lemma without using this. – user10024395 Apr 24 '15 at 6:18