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.

  • 2
    $\begingroup$ Multiply $5 + 2X$ by $5 - 2X$ and see what happens. $\endgroup$ Jun 21, 2021 at 6:07
  • $\begingroup$ Same methods in the dupe work here, e.g. using $\!\bmod (17,2x\!+\!5)\!:\ 2x\equiv -5\equiv 12\iff x\equiv 6\,$ in my answer for $\rm\color{#0a0}{surjectivity}$. This is a FAQ - please search for answers before posting questions. $\endgroup$ Jun 21, 2021 at 18:03
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    $\begingroup$ @BillDubuque This question is different from the one you linked in many aspects: There $\sqrt{-11}\equiv -1$ renders it obvious that the homomorphism from $\mathbb Z$ is injective. Here $2\sqrt{2}\equiv -5$ doesn't allow that conclusion immediately. There $\mathbb Z/\langle 12\rangle$ is not a field and hence the method I used in my answer doesn't work either. IMHO this is not a duplicate of a question about $\mathbb Z[\sqrt{-11}]/\langle 1+\sqrt{-11}\rangle$. $\endgroup$
    – Christoph
    Jun 21, 2021 at 18:09
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    $\begingroup$ @BillDubuque Your prior comment just assumes we can reduce mod 17. Just closing a question as a duplicate of one that doesn't match and claiming my comment is "wrong" surely is discouraging for new users asking questions. If it is a duplicate, we should find a question that is a better match, you must have one on your list of hundreds. $\endgroup$
    – Christoph
    Jun 21, 2021 at 18:26
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    $\begingroup$ @Christoph Further, in addition to my first comment, my answer there explicitly links to answer showing how to get rid of that nonunit coef on the radical, viz. 'Same proof works for $\,w=a+bi$ when $\gcd(a,b)=1$'. Duplicates don't need to be identical - see abstract duplicates. $\endgroup$ Jun 21, 2021 at 18:28

2 Answers 2


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 $$ \mathbb Z[X]/I \cong \mathbb F_{17}[X]/\langle X+\overline{11}\rangle \cong \mathbb F_{17}. $$

  • $\begingroup$ Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. $\endgroup$ Jun 21, 2021 at 18:58

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 $\mathbb{Z} \left [ \sqrt{2} \right ] \big / \langle 5 + 2 \sqrt{2} \rangle \cong \mathbb{Z} / 17$?

I hope this helps ^_^

  • $\begingroup$ thank you very much for your such nice explanations. $\endgroup$
    – Aritra
    Jun 21, 2021 at 11:12
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    $\begingroup$ Once we have $17\in\operatorname{Ker}(\varphi)$, we can rewrite this as a quotient of $\mathbb F_{17}[X]$, which is a PID and where gcd-calculations can be done using the Euclidean algorithm! $\endgroup$
    – Christoph
    Jun 21, 2021 at 15:07
  • $\begingroup$ Do we need to calculate gcd in this approach? Because in this approach it follows from the ring homomorphism, surjection etc. Right? @Christoph $\endgroup$
    – Aritra
    Jun 21, 2021 at 16:29
  • $\begingroup$ You're approach is fine, of course! I was just suggesting an alternative that doesn't need a case distinction to prove surjectivity of this ring homomorphism. $\endgroup$
    – Christoph
    Jun 21, 2021 at 16:39
  • $\begingroup$ Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. $\endgroup$ Jun 21, 2021 at 18:58

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