# Do these 25 gaussian integers make a finite field?

$$\{ a + b i \mid a \in \{0, 1, 2, 3, 4\}, b \in \{0, 1, 2, 3, 4\} \}$$

With calculation done in $$\pmod{5}$$, I'm wondering if this makes a finite field.

I thought the answer is yes at first.

Then the 24 elements (excluding $$0$$) must be a multiplicative group (should be closed under multiplication).

$$(3 + 4i)(4 + 3i) = 25 i \equiv 0 \pmod{5}$$

Now I'm confused.

Context:

I was reading this blog post (the extension fields part) . I guess my lesson learned is that adding $$\sqrt{-1}$$ is not the general method.

• I'm not sure why you are confused. You correctly noticed that it is not a field. In fact $\mathbb{Z}_5$ already has square roots of $-1$, namely $2$ and $3$. And so by adding more square roots of $-1$ you cannot obtain a field. Because that would violate the fact that a polynomial of degree $n$ over a field can have at most $n$ roots. Commented Apr 10, 2022 at 8:47
• I thought there is a finite field of 25 elements and this is the construction. Is this construction wrong? Or a finite field of 25 elements doesn't exist?
– zjk
Commented Apr 10, 2022 at 8:54
• There is a field of 25 elements. There's a field for any prime power. But that's not the construction. The correct construction is $\mathbb{Z}_5[X]/(X^2+X+1)$. You did $\mathbb{Z}_5[X]/(X^2+1)$ but $X^2+1$ is reducible over $\mathbb{Z}_5$. That's why it failed. Commented Apr 10, 2022 at 8:56
• @zjk:There is a field with 25 elements, unique up to isomorphism. It is the unique qudratic extension of teh field with five elements $\mathbb F_5$. It's just that one does not get that quadratic extension by adjoining a root of the polynomial $x^2+1$, because as pointed out, that polynomial is already reducible (has two roots) in $\mathbb F_5$. Commented Apr 10, 2022 at 8:57
• It may be simpler to use $$(2+i)(2-i)=5=0.$$ This reflects the observation in the comment by @freakish. Depending on the meaning of $i$ in any extension field of $\Bbb{Z}_5$ you must have either $2-i=0$ or $2+i=0$. Commented Apr 10, 2022 at 9:00

You are correct, this is not a field.

I assume that $$i$$ stands for square root of $$-1$$. Which means that your construction actually is

$$\mathbb{Z}_5[X]/(X^2+1)$$

and thus it cannot be a field, because $$X^2+1=(X-2)(X-3)$$ is reducible over $$\mathbb{Z}_5$$. Or in other words $$\mathbb{Z}_5$$ already has square roots of $$-1$$. Adding more cannot result in a field (which can have at most two square roots of $$-1$$).

I guess my lesson learned is that adding $$\sqrt{-1}$$ is not the general method.

Indeed. To obtain a field of order $$25$$ one way is to make a quotient by an irreducible polynomial of degree $$2$$, e.g.

$$\mathbb{Z}_5[X]/(X^2+X+1)$$

Adding $$\sqrt{-1}$$ works only when the field does not have it to begin with. And $$\mathbb{Z}_p$$ has a square root of $$-1$$ if and only if $$p\equiv 1\text{ (mod 4)}$$, which I think is one of Gauss' theorems.

The answer is not "yes", because the quotient ring of the ring of Gaussian integers modulo $$5$$ is given by $$\Bbb Z[i]/(5),$$ and $$(5)$$ is not a prime ideal. So the quotient has zero divisors and is not a field. Indeed, $$(3i-1)(3i+1)=(3i)^2-1=0$$ in this quotient ring, as you have found out.

References:

What's are all the prime elements in Gaussian integers $\mathbb{Z}[i]$

Classification of prime ideals in $\mathbb{Z}[i]$

Like you point out, the quotient ring $$\Bbb Z[i] / \langle 5 \rangle$$ has zero divisors so cannot be an integral domain, let alone a field: $$\langle 5 \rangle$$ is not even a prime ideal let alone a maximal one.

Indeed, since $$5 \equiv 1 \pmod 4$$, $$\langle 5 \rangle$$ factors as a product of prime ideals, namely $$\langle 5 \rangle = \langle 2 + i \rangle \langle 2 - i \rangle$$. So, we can identify $$\Bbb Z[i] / \langle 5 \rangle \cong \Bbb Z[i] / \langle 2 + i \rangle \oplus \Bbb Z[i] / \langle 2 - i \rangle \cong \Bbb Z_5 \oplus \Bbb Z_5$$ as rings. It's easier to see that there are zero divisors in the latter reprsentation, e.g., $$(1, 0) \cdot (0, 1) = (1 \cdot 0, 0 \cdot 1) = 0$$.

Since, on the other hand, $$p(x) := x^2 + 2$$ is irreducible modulo $$5$$, $$\langle p(x) \rangle$$ is a maximal ideal in $$\Bbb F_5[x]$$, so we can realize the field of $$25$$ elements concretely as $$\Bbb F_5[x] / \langle p(x)\rangle$$, that is, as the set of pairs $$a + b \zeta$$, where $$a, b \in \Bbb F_5$$, subject to the relation $$\zeta^2 = -2 = 3$$.

Another way to see that it's not a field is that $$-1$$ has square roots mod $$5$$, since $$2^2 = 4 = -1$$ and $$3^2 = 9 = -1$$. Therefore, if $$i$$ is another element not in $$\{0, 1, 2, 3, 4\}$$, then that would be a third square root of $$-1$$, but in a field $$-1$$ can only have two square roots.

We have seen that $$\mathbb Z[i]/<5>$$ fails to be a finite field, due essentially to $$5=(2+i)(2-i)$$ not actually being prime in $$\mathbb Z[i]$$.

But we can construct a $$5×5$$ finite field using a different imaginary quadratic field. The set $$\mathbb Z[\sqrt{-2}]/<5>$$, that is $$a+bi\sqrt2$$ where $$a$$ and $$b$$ are residues modulo $$5$$, works perfectly well.

Going through all the unique-factorization imaginary quadratic domains we find that in six out of nine cases the "two-dimensional" extension of $$\mathbb Z/<5>$$ is a finite field. We were just unlucky with the Gaussian integers.