# Are Hilbert primes also Hilbert irreducible ? Furthermore, are Hilbert primes also primes in $\mathbb{ Z}$?

Consider the set $\mathcal H$ of Hilbert numbers (numbers of the form $4n + 1$, for $n \ge 0$). Define a Hilbert prime as any number $h$ in the Hilbert set satisfying $h \neq 1$ and if $h \mid ab$ where $a \in \mathcal H$, $b\in \mathcal H$ then $h \mid a$ or $h \mid b$. Define a member of the Hilbert set $q$ as Hilbert irreducible as if and only if $q \neq 1$ and $q$ cannot be expressed as the product of two smaller Hilbert numbers.

I am trying to determine if Hilbert prime implies Hilbert irreducible, however I am not particularly strong at number theory. I have been unsuccessful in finding a counterexample, and I am starting to believe the implication holds. The textbook I pulled this example from for self study (Rings, Fields, and Groups: An Introduction to Abstract Algebra by Reg Allenby) has answers in the back of the text. However, the solution only says that Hilbert primes are also primes in $\mathbb{Z}$. Any help would be greatly appreciated, I can't see to find any proof regarding this concept anywhere!

Suppose $h$ is not Hilbert irreducible. Then there are $m,n\in\mathbb{N}$ such that $m,n\neq 0$ and $h=(4m+1)(4n+1)$. Then $h|(4m+1)(4n+1)$, but certainly $h\nmid (4m+1)$ and $h\nmid (4n+1)$, so $h$ is not Hilbert prime. Therefore, by the contrapositive, any Hilbert prime is Hilbert irreducible.

• The interesting thing is actually the other direction: Not every Hilbert irreducible is a Hilbert prime! Sep 22 '15 at 20:30
• With $h=9$ and $a=b=15$ then $h| ab$ but $h$ does not divide $a$ nor $b.$ So $9 \in H$ and $9$ is not a Hilbert prime. But $9$ is Hilbert irreducible. Jun 7 '17 at 5:44
• @DanielWainfleet Thank you, that's a good example illustrating my comment above. Jun 7 '17 at 21:23

Since Hilbert primes are also primes in $\mathbf{Z}$, and since primes in $\mathbf{Z}$ are irreducible in $\mathbf{Z}$, then Hilbert primes must be Hilbert irreducible.

More explicitly, suppose $h$ is a Hilbert prime which is not Hilbert irreducible. Then we can write $h=pq$ where $p,q$ are smaller Hilbert numbers. But Hilbert numbers are integers, so we've just found a non-trivial factorization of the $\mathbf{Z}$-prime $h$, which is a contradiction.

• but how do I prove that a Hilbert prime is a prime in $\mathbb{Z}$ ? Sep 22 '15 at 20:35

It's easier to prove that a Hilbert irreducible is not necessarily prime. Choose positive primes $p \equiv q \equiv 3 \pmod 4$; these are clearly in $\mathbb Z$ but not in $\mathcal H$. Then $pq$, $p^2$, $q^2$ and $p^2 q^2$ are all numbers in $\mathcal H$, and all four but the last are also irreducible.

From this we readily find that $pq \mid p^2 q^2$ but $pq \nmid p^2$, $pq \nmid q^2$ either. Example: choose 3 and 7, and thus we find 21, 9, 49, 441; the first three are irreducible but not prime.

But what you really want to prove is that all primes in $\mathcal H$ are irreducible. If $p$ is prime, then, as you well know, $p \mid ab$ means $p \mid a$ and/or $p \mid b$.

If $p$ is prime but reducible, we should be able to find numbers $m$ and $n$ in $\mathcal H$ such that $mn = p$. And $m \neq 1$, $n \neq 1$, to be perfectly clear. And it goes without saying that $-1 \not \in \mathcal H$.

So, if $mn = p$, then $p \mid m^2 n^2$ but $p \nmid m^2$ and $p \nmid n^2$ either, contradicting the assertion that $p$ is prime. As $p$ being prime but reducible leads to a contradiction, this must mean that if $p$ is prime, it is also irreducible.

For example, let's say that 45 is prime but not irreducible (in fact it's neither, but please play along). We have $5 \times 9 = 45$, and we readily find that $45 \mid 2025$ but $45 \nmid 25$ and $45 \nmid 81$ either, contradicting the assertion that 45 is prime.

Maybe the foregoing feels like sophistry. Remember that in $\mathcal H$, the irreducibles are congruent to 1 modulo 4. If an irreducible is not divisible by any prime congruent to 3 modulo 4 in $\mathbb Z$, it must also be prime, in both $\mathcal H$ and $\mathbb Z$.

To prove that a Hilbert prime is a prime number, show that a composite member of $H$ is not a Hilbert prime, as follows:

If $h\in H$ is composite, there exist odd $c,d,$ each greater than $1,$ with $h=cd$ and $\gcd(c,d)=1.$ Let $a=c^2$ and $b=d^2.$

Then $a,b\in H$ (because $c,d$ are odd and greater than $1$),and $h|ab=h^2$. But $h$ does not divide $a$ or $b.$ (As $a/h=c/d$ and $b/h=d/c$ are not integers, because $c, d$ are greater than $1$ and are co-prime.)

One way to restore unique factorization is to simply allow negative prime factors. Then, all integers $$\equiv 1\bmod 4$$, including negative ones, become uniquely factored. The absolute values of the prime factors are guaranteed by the usual unique factorization of natural numbers, and the signs are fixed by the factors being $$\equiv 1\bmod 4$$. For instance, $$441$$ mentioned in the comments is $$21×21$$ and $$9×49$$ when we allow only positive factors, but both factorizations collapse to $$(-3)^2×(-7)^2$$ when negative factors are allowed. The corresponding ordinary natural number factorization is $$441=3^2×7^2$$. A typical example for a negative input would be $$-231=(-3)×(-7)×(-11)$$ corresponding to the ordinary factorization $$231=3×7×11$$.