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

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.


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.

  • 3
    $\begingroup$ 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. $\endgroup$
    – freakish
    Commented Apr 10, 2022 at 8:47
  • $\begingroup$ 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? $\endgroup$
    – zjk
    Commented Apr 10, 2022 at 8:54
  • 2
    $\begingroup$ 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. $\endgroup$
    – freakish
    Commented Apr 10, 2022 at 8:56
  • 2
    $\begingroup$ @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 $\mathbb F_5$. $\endgroup$ Commented Apr 10, 2022 at 8:57
  • 1
    $\begingroup$ 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 $2+i=0$. $\endgroup$ Commented Apr 10, 2022 at 9:00

5 Answers 5


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


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.


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.


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.


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