Elements in the quotient ring $\mathbb Z \left [ \sqrt 2 \right ] \big /\langle 5+2\sqrt 2\rangle$

Question

I am trying to find the elements in $\mathbb Z \left [ \sqrt 2 \right ] \big /\langle5+2\sqrt 2\rangle$.

$\textbf{My attempts}$:

We know that there is a ring homomorphism $\mathbb Z[X] \to \mathbb Z \left [\sqrt 2 \right ] \big /\langle5+2\sqrt 2\rangle$. Because there is a ring homomorphism $\mathbb Z[X] \to \mathbb Z \left [ \sqrt 2 \right ]$ by $X\to \sqrt 2$ which is surjective and $\mathbb Z \left [\sqrt 2 \right ] \to \mathbb Z \left [ \sqrt 2 \right ] \big /\langle5+2\sqrt 2\rangle$ is also a surjective ring homomorphism. So, there is a surjective ring homomorphism $\mathbb Z[X] \to \mathbb Z \left [\sqrt 2 \right ] \big /\langle5+2\sqrt 2\rangle$, whose kernel is $\langle X^2-2, 5+2X\rangle$. We can conclude that $$\mathbb Z[X]/\langle X^2-2,5+2X\rangle \simeq \mathbb Z \left [\sqrt 2 \right ] \big /\langle5+2\sqrt 2\rangle.$$

Now, I am stuck. Any hints will be helpful. Thanks in advance.

Answer 1

Since $(5+2X)(5-2X)=25-4X^2= 17-4(X^2-2)$, we may rewrite the ideal you are modding $\mathbb Z[X]$ by as $$ I = \langle X^2-2,5+2X\rangle = \langle X^2-2, 5+2X, 17\rangle. $$ Since $17\in I$ is prime, you can rewrite the quotient $\mathbb Z[X]/I$ as a quotient of the polynomial ring over the field $\mathbb F_{17} = \mathbb Z / \langle 17\rangle$: $$ \mathbb Z[X]/I \cong \mathbb F_{17}[X]/\langle X^2-\bar2,\bar5+\bar2X\rangle. $$ Now $\mathbb F_{17}[X]$ is a PID and we can use the Euclidean algorithm to calculate a generator of the ideal: $$ \gcd(X^2-\bar2,\bar5+\bar2X)= X+\overline{11}. $$ Indeed, the square roots of $\bar 2$ in $\mathbb F_{17}$ are $\bar 6$ and $-\bar 6=\overline{11}$. We conclude that $$ \mathbb Z[X]/I \cong \mathbb F_{17}[X]/\langle X+\overline{11}\rangle \cong \mathbb F_{17}. $$

Answer 2

Welcome to MSE!

To clarify the hint in the comments, notice

$$17 = (5+2\sqrt{2})(5-2\sqrt{2}) \in \langle 5 + 2 \sqrt{2} \rangle.$$

Then if we look at the map $\varphi : \mathbb{Z} \to \mathbb{Z} \left [ \sqrt{2} \right ] \big / \langle 5 + 2 \sqrt{2} \rangle$ defined by

$$ \mathbb{Z} \hookrightarrow \mathbb{Z} \left [ \sqrt{2} \right ] \twoheadrightarrow \mathbb{Z} \left [ \sqrt{2} \right ] \big / \langle 5 + 2 \sqrt{2} \rangle $$

we must have $17 \in \text{Ker}(\varphi)$. (Do you see why this shows $\text{Ker}(\varphi) = \langle 17 \rangle$?)

If we can show that $\varphi$ is surjective, then we'll be done. It's quite late, so I'm not seeing a slick way to do this, but you can do it by hand without too much hassle.

Notice $\varphi(n) = n + 0 \sqrt{2} + \langle 5 + 2 \sqrt{2} \rangle$. (do you see why?)

So to show it's surjective, you should show that each $a + b \sqrt{2} + \langle 5 + 2 \sqrt{2} \rangle$ is actually of this form. That is, we need to find a member of the same coset whose coefficient on $\sqrt{2}$ is $0$.

If $b$ is even, then this is easy: just subtract $\frac{b}{2} \left ( 5 + 2 \sqrt{2} \right )$.

If $b$ is odd, this requires a bit more thought. We need to find a way to subtract an odd coefficient from the $\sqrt{2}$. But notice

$$(5 + 2 \sqrt{2})(x + y \sqrt{2}) = (5x + 4y) + (5y + 2x) \sqrt{2}.$$

So if we take $y=1$ and $x=0$ (or any number of other things), we'll get something which is in our ideal (thus doesn't change the coset) but which lets us subtract an odd number from the $\sqrt{2}$ coefficient.

Do you see how to put these pieces together to show that $\varphi$ is surjective, and thus that $\mathbb{Z} \left [ \sqrt{2} \right ] \big / \langle 5 + 2 \sqrt{2} \rangle \cong \mathbb{Z} / 17$?

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I hope this helps ^_^