I want to show that $\mathbb Z[\sqrt{-2}] / (5) \simeq \mathbb F_{25} = Z_5[x] / (x^2+x+1)$.

my first attempt:

First I see that all elements in the ideal $(5)$ are given by $\{ (a+b \sqrt{-2}) 5 | a,b \in Z \}$ so a generic element has the form $5a+ 5 \sqrt{-2} b $.

Now I want to show that there are 25 equivalence classes in $\mathbb Z[\sqrt{-2}] / (5) $ where two elements $x,y$ are equivalent $\iff (y_1 +y_2 \sqrt{-2}) -(x_1+x_2 \sqrt{-2}) \in (5) $.

So I obtain the condition that $y_1 - x_1$ must be of the form $5a$ and $y_2 -x_2$ must be of the form $5b$ Thus

$$y_1 + y_2 -x_1 -x_2 = 5(a+b)$$ so I must have that $y_1 + y_2 -x_1 -x_2$ is a multiple of 5 for the system to have solution this seems to lead me to the solution that $\mathbb Z[\sqrt{-2}] / (5) \simeq \mathbb Z_{5}$ that is not correct.

my second attempt:

$$\mathbb Z[\sqrt{-2}] \cong \frac{\mathbb Z[x]}{(x^2+2)} \implies \frac{\mathbb Z[\sqrt{-2}]}{(5)} \cong \frac{\mathbb Z[x]}{(x^2+2,5)} \cong \frac{\mathbb Z}{(3,5)}$$

Where the last isomorphism is given by the map $f(x) \rightarrow f(1)$ this still seems incorrect.

Could someone please explain to me the flaws in my reasoning and try to rectify them?

  • $\begingroup$ In your $2^{nd}$ attempt $\mathbb Z[x]/(x^2+2,5)\cong \Bbb{F}_5[x]/(x^2+2)$. The map you wrote is not an isomorphism. $\endgroup$ – sharding4 Jul 2 '17 at 1:17
  • $\begingroup$ @sharding4 Thank you, I imagined there was a problem in that step, so the map given by $f(x) \rightarrow f(1)$ is not always isomorphic? $\endgroup$ – Monolite Jul 2 '17 at 1:27
  • 1
    $\begingroup$ The evaluation map isn't going to give you an isomorphism with the ring of polynomials. That would be $\Bbb{Z}[x]\cong \Bbb{Z}$. The map factors through $\Bbb{Z}[x]/(x-1)$ which is isomorphic to $\Bbb{Z}$ $\endgroup$ – sharding4 Jul 2 '17 at 1:37
  • $\begingroup$ What's your $\epsilon$? $\endgroup$ – Orat Jul 2 '17 at 3:42
  • $\begingroup$ @Orat sorry $\epsilon = \sqrt{-2}$ $\endgroup$ – Monolite Jul 2 '17 at 13:21

Everything is easier if you know that $\mathbb{Z}[\sqrt{-2}]$ is a Euclidean domain with respect to the usual norm function (essentally the same proof as for $\mathbb{Z}[\sqrt{-1}]$).

We can now show that $5$ is irreducible in $\mathbb{Z}[\sqrt{-2}]$. If $5=xy$, then $5^2=|x|^2|y|^2$. A proper factorization would have $|x|^2=5$, but $$ |a+b\sqrt{-2}|^2=a+2b^2=5 $$ is satisfied by no pair of integers $a$ and $b$.

Therefore $\mathbb{Z}[\sqrt{-2}]/(5)$ is a field and it obviously has characteristic $5$. Since it is generated over $\mathbb{F}_5$ by an element whose square is $-2=3$, it is a proper quadratic extension of $\mathbb{F}_5$, because $3$ is not a square in it; hence it is $\mathbb{F}_{25}$.

About your first attempt, you are correct up to a certain point but jump to a wrong conclusion: you can arbitrarily choose $a$ and $b$ among $\{0,1,2,3,4\}$, which makes for $25$ elements. You indeed prove that $x_1\equiv x_2\pmod{5}$ and $y_1\equiv y_2\pmod{5}$.

In the second attempt, you also do a wrong step: the correct isomorphism is $$ \mathbb{Z}[x]/(x^2+2,5)\cong(\mathbb{Z}/5\mathbb{Z})[x]/(x^2+2) $$ which is essentially the same as the first attempt, but requires a (short) proof.


In your first attempt, at the line

$y_1+y_2 - x_1-x_2 = 5(a+b)$

you forgot the $\epsilon$. Also don't forget $a,b$ themselves must be in $\mathbb{Z}[\sqrt{-2}]$.

Here is a way to prove it using a counting argument:

Every element in $\mathbb{Z}[\sqrt{-2}]$ is of the form $x+y\sqrt{-2}$. If we quotient modulo 5, we will get that every element of $\frac{\mathbb{Z}[\sqrt{-2}]}{(5)}$ is of the form $\overline{x}+\overline{y} \overline{\sqrt{-2}}$. Where the overline represents the classes mod $(5)$. Now if you take $(x,y) \in \{0,1,2,3,4\}^2$ and you associate to it $\overline{x}+\overline{y}\overline{\sqrt{-2}}$, you can easily prove that this is a surjection and therefore your quotient has at most 25 elements.

I will assume that you have at some point proven that $(5)$ is maximal and therefore the quotient is a field. A field that has at most 25 elements and contains $\mathbb{F}_5$ can either have 5 or 25 elements, therefore it is enough to show, that it must have more than 5 elements. The classes $\overline{0}, \overline{1}, \cdots , \overline{4}$ are all distinct, since $5$ does not divide difference between either of these. Now if we can show that $\sqrt{2}$ is a class on its own, we are done. Now assume that $\overline{\sqrt{2}}$ is equal to one of these. We write $\overline{\sqrt{2}}=\overline{i} (0 \leq i \leq 4)$. Then we have: $5(a+b\sqrt{5})=\sqrt{2} -i$. In particular $5a=1$, which is a contradiction, therefore your field is indeed $\mathbb{F}_{25}$.

As for your second attempt, your evaluation in 1 is actually not an isomorphism. It is surjective, because your quotient is in fact the zero ring, since 3,5 are coprime, but then it obviously cannot be injective.

  • $\begingroup$ Thank you very much, I can't see the error in my first argument: I was summing $y_1 - x_1 = 5a$ and $y_2 - x_2 = 5b$. I love your argument and how simple it is, only point I have trouble with is why can a field with at most $25$ elements and that contains $\mathbb{F}_5$ have either $5 $or $25$ elements? $\endgroup$ – Monolite Jul 2 '17 at 14:20
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    $\begingroup$ If a field contains $\mathbb{F}_5$, then it is an $\mathbb{F}_5$-vector space. The number of its elements must therefore be a power of 5. (since it is isomorphic to some $\mathbb{F}_5^n$). There is no power of 5 between 5=5^1 and 25=5^2. $\endgroup$ – Keen Jul 2 '17 at 15:13

In your first attempt, you have essentially shown there are $25$ distinct cosets, namely:

$\{[a+b\sqrt{-2}]:a,b \in \{0,1,2,3,4\}\}$.

So, if nothing else, we know we have a commutative ring of $25$ elements.

To show this is a field, it suffices to show that $(5)$ is a maximal ideal in $\Bbb Z[\sqrt{-2}]$.

This is much easier if you already know $\Bbb Z[\sqrt{-2}]$ is a Euclidean domain, and thus a PID, in which case if $J$ is an ideal containing $(5)$ it must be of the form $(d)$ where $d|5$.

Arguing from norms, you should be able to show $5$ is irreducible (in $\Bbb Z[\sqrt{-2}]$); hence, you have a field (because $(5)$ is thus a maximal ideal) of $25$ elements, which is thus isomorphic to $\Bbb F_{25}$.

Now if $u$ is a root of the polynomial $x^2 + x + 1 \in \Bbb Z_5[x]$, it should be clear that $\Bbb Z_5(u) \cong \dfrac{\Bbb Z_5[x]}{(x^2+x+1)}$. This is also a field, since $x^2 + x + 1$ is irreducible over $\Bbb Z_5$ (this is easy to show since we need merely test all five elements of $\Bbb Z_5$ as roots).

Considering $\Bbb Z_5(u)$ as a vector space over $\Bbb Z_5$, we see it has dimension $2$ with basis $\{1,u\}$. Since all elements are thus of the form:

$a+bu: a,b \in \Bbb Z_5$, this is again a field of $25$ elements, and thus isomorphic to $\Bbb F_{25}$.

One can also exhibit an isomorphism, but this is not as easy as sending:

$[a + b\sqrt{-2}] \mapsto a + bu$ (which is merely an abelian group isomorphism, because $u^2 = 4u + 4 \neq 3 = -2$).

As if by magic, we find that:

$(u+2)^2 = u^2 + 4u + 4 = (4u + 4) + (4u + 4) = 2(4u + 4) = 3u + 3 \neq 1$

$(u+2)^3 = (u+2)(u+2)^2 = (u+2)(3u + 3) = 3u^2 + 4u + 1 = 3(4u + 4) + 4u + 1 = (2u + 2) + (4u + 1) = u + 3 \neq 1$

$(u+2)^4 = ((u+2)^2)^2 = (3u + 3)^2 = 4u^2 + 3u + 4 = 4(4u + 4) + 3u + 4 = (u + 1) + (3u + 4) = 4u \neq 1$

$(u+2)^6 = ((u+2)^3)^2 = (u+3)^2 = u^2 + u + 4 = (4u + 4) + (u + 4) = 3 \neq 1$

$(u+2)^8 = ((u+2)^4)^2 = (4u)^2 = u^2 = 4u + 4 \neq 1$

$(u+2)^{12} = ((u+2)^6)^2 = 3^2 = 4 \neq 1$.

This shows that $u+2$ is a primitive element of $\Bbb Z_5(u)$ (since it generates $(\Bbb Z_5(u))^{\ast}$).

Now in $\Bbb Z[\sqrt{-2}]/(5)$, the coset of $\sqrt{-2}$ is an element of order $8$, which suggests an isomorphism might send this coset to $(u+2)^3 = u+3$, that is we set:

$[a+b\sqrt{-2}] \mapsto (a+3b) + bu$

I leave to you the exquisitely joyous task of showing this is indeed a ring-homomorphism, and thus is an isomorphism between the two versions of our field.


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