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I have read numerous answers to similar questions, but I'm still stuck.

Here's my approach:

All elements in $\langle 3+i \rangle$ have the form $(a+bi)(3+i)=(a-b)+i(a+3b)= 2p + i 4q$, for $p, q\in \mathbb{Z}$, which implies that all other elements are not in $\langle 3+i \rangle$. But aren't there infinitely many such elements?

Can you please clarify what I'm not seeing in this problem?

sequence
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    When you say "All elements in ... are of the form $2p+i4q$, the ... ought to read "the ideal $\langle 3+i\rangle$". – ancient mathematician Apr 12 '17 at 06:34
  • And a hint: modulo the ideal, $i=-3$; but we also know $i^2=-1$. So modulo the ideal we know that [you can do the sums]. – ancient mathematician Apr 12 '17 at 06:35
  • @ancientmathematician Thanks, I've fixed the typo. – sequence Apr 12 '17 at 06:40
  • @ancientmathematician But again, $2p+i4q = 2p-12q = 2(p-6q)=2(p+54q)$. How does this help us to determine how many distinct elements there are in the quotient ring? – sequence Apr 12 '17 at 06:46
  • Where did you get that $(a+bi)(3+i)=2p+4qi$ from? When $a=1,b=0$ we have $(a+bi)(3+i)=3+i$, when $a=2,b=1$ we get $(a+bi)(3+i)=5+5i$ et cetera? Anyway, the former example implies that the cosets of $i$ and $-3$ coincide. – Jyrki Lahtonen Apr 12 '17 at 08:51
  • @JyrkiLahtonen We have that $i=-3$ and $i^2=-1$, so that $i^2=-1=9$. So we get that form of $2p+4qi$ by substitution. – sequence Apr 12 '17 at 08:55
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    In the ring $\Bbb{Z}$ there are infinitely many elements not in $\langle 2\rangle$. Yet all those elements (the odd integers) form a single coset, i.e. a single element in $\Bbb{Z}/\langle 2\rangle=\Bbb{Z}_2$. – Jyrki Lahtonen Apr 12 '17 at 08:56
  • Ok, but when you identify $i=-3$, you consequently identified $a+bi$ with $a-3b$, so only integers remain. – Jyrki Lahtonen Apr 12 '17 at 09:28
  • Sorry, I didn't get what you mean. Where do only integers remain? – sequence Apr 12 '17 at 09:44
  • Because you identified $i$ with $-3$, all the elements of $\Bbb{Z}[i]$ became identified with some integers. What I meant with "only integers remain" can more precisely be read "all the cosets have representatives that are integers". You also identified $-1$ with $i^2=9$, so we are down to modulo ten. – Jyrki Lahtonen Apr 12 '17 at 18:45

2 Answers2

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You say "$\dots$, which implies that all other elements are not in $(3+i)$".

There you are confusing the quotient ring ${\mathbb Z}[i]/(3+i)$ with the set difference ${\mathbb Z}[i] \setminus (3 + i)$. The former consists of the residue classes of ${\mathbb Z}[i]$ modulo the ideal $(3 + i)$, the latter consists of all elements of ${\mathbb Z}[i]$ that are not in the ideal $(3 + i)$.

To figure out how many elements the quotient ring ${\mathbb Z}[i]/(3 + i)$ has, you could do $${\mathbb Z}[i]/(3 + i) \cong {\mathbb Z}[x]/(x^2 + 1,x+3) \cong {(\mathbb Z}[x]/(x + 3))/(x^2+1) \cong {\mathbb Z}/(10).$$ In any case, the key is to realize that in the quotient ring, $i ^ 2 = -1$ and $i = -3$, which implies that $10 = 0$.

Magdiragdag
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  • How do you see that ${\mathbb Z}[i]/(3 + i) \cong {\mathbb Z}[x]/(x^2 + 1,x+3)$? The $x^2+1$ part is clear, but how is the $x+3$ part exactly justified? Also, ${(\mathbb Z}[x]/(x + 3))/(x^2+1) \cong {\mathbb Z}/(10)$. Why?

    For the $/$ vs. $\backslash$ in your first paragraph - my reasoning is that we need to find how many elements there are besides the zero element $\langle 3+i \rangle$, so we need to find all elements that are not in this generator set.

     – sequence Apr 12 '17 at 07:52
  • Under the isomorphism ${\mathbb Z}[i] \cong {\mathbb Z}[x]/(x^2 + 1)$, $i$ corresponds to (the residue class of) $x$. So, the ideal $(3 + i)$ corresponds to the ideal $(3 + x)$. For the final isomorphism: use that ${\mathbb Z}[x]/(x + 3) \cong {\mathbb Z}$ by mapping $x$ to $-3$. For your confusion with the set difference: I think I understand where you're getting at, but do realize that some elements that are not in $(3 + i)$ represent the same element of the quotient ring ${\mathbb Z}[i]/(3 + i)$, so just looking at the number of elements that are not in $(3 + i)$ is not going to help you. – Magdiragdag Apr 12 '17 at 18:09
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Just try to find an isomorphism by using the fact that $\Bbb{Z}[i]\cong \Bbb{Z}[x]/(x^2+1)$.

Then

$$\Bbb{Z}[i]/(i+3)\cong \Bbb{Z}[x]/(x+3,x^2+1)\cong \Bbb{Z}[x]/(x+3,-3x+1)\cong\Bbb{Z}[x]/(x+3,10)\underset{\mbox{iso thm}}{\cong} (\Bbb{Z}[x]/(x+3))/(10)\cong \Bbb{Z}/(10)\cong \Bbb{Z}/10\Bbb{Z}$$

because $x^2+1=x(x+3)+(-3x+1)$ and $x+3=3(x^2+1)+x(-3x+1)$ shows the equivalence of the first two ideals. The equivalence of the second two ideals is shown by $-3x+1=-3(x+3)+10$ and $10=(-3x+1)+3(x+3)$.

Now the number of elements might be obvious.

Daniel Bernoulli
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