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Problem: (a) Show $\mathbb{Z}[i]/\langle 5\rangle$ is a product of two fields. (b) Show $\mathbb{Z}[i]/\langle 3\rangle$ is a field. (c) Show $\mathbb{Z}[i]/\langle 2\rangle$ is neither a field nor a product of two fields.

My Attempt: Now from first glance, it is clear (a) and (c) will not be fields since both $5$ and $2$ are not irreducible in $\mathbb{Z}[i]$ and thus will not general maximal ideals and hence the quotient ring cannot be a field. It can also be shown that $\langle 3\rangle$ forms a maximal ideal, thus (b) is a field.

What I am having difficultly with is showing (a) is a product of 2 fields and that (c) is not a product of 2 fields.

For (a), I initially thought of the homomorphism, $\phi: \mathbb{Z}[i] \rightarrow \mathbb{Z}_5 \times \mathbb{Z}_5$ given by $a+bi \rightarrow (a \mod (5), b \mod (5))$ and then thought to show $\ker (\phi ) = \langle 5\rangle$ to prove $\mathbb{Z}[i]/\langle 5\rangle \cong \mathbb{Z}_5 \times \mathbb{Z}_5$. However I soon realised that $\phi$ is not a ring homomorphism since: $$\phi(a+bi)\star \phi(c+di) \neq \phi((a+bi)\cdot (c+di))$$

Question: Could anyone please point me in the right direction for parts (a) and (b).

Also more generally, is there any statement that can be made regarding if $\mathbb{Z}[i]/\langle a\rangle$ (where $a$ is an integer) can be written as some product of field?

Thank you.

Community
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Nanoputian
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  • Those aren’t quotient groups, they are quotient rings – Arturo Magidin Jan 31 '20 at 22:59
  • Thanks for pointing that out, I have fixed that now – Nanoputian Jan 31 '20 at 23:03
  • Hint: One of the statements of the Chinese Remainder Theorem is: if $I, J$ are ideals of a commutative ring $R$ such that $I + J = \langle 1 \rangle$, then $IJ = I \cap J$ and $R / (IJ) = R / (I \cap J) \simeq (R / I) \times (R / J)$ (it's an isomorphism of rings, and even an isomorphism of $R$-algebras). – Daniel Schepler Jan 31 '20 at 23:09
  • You might also try showing that a field has exactly two ideals, and a product of two fields has exactly four ideals. – Daniel Schepler Jan 31 '20 at 23:16
  • @DanielSchepler Thanks for the hint. So for (a), let $I = \langle 1+2i\rangle, J = \langle 1-2i \rangle$, then we have $IJ = \langle 5 \rangle$ and thus the required answer follows. Is this approach correct? If it is, I am stuck on how to show $\langle 1+2i\rangle + \langle 1-2i\rangle = \mathbb{Z}[i]$. For example, I cant seem to show that $1 \in \langle 1+2i\rangle + \langle 1-2i\rangle$. – Nanoputian Jan 31 '20 at 23:42
  • Since $\mathbb{Z}[i]$ is a PID, $\langle 1 + 2i, 1 - 2i \rangle = \langle \gcd(1+2i, 1-2i) \rangle = \langle 1 \rangle$. (And if you really wanted to, you could trace through the gcd algorithm to find the coefficients.) – Daniel Schepler Jan 31 '20 at 23:47
  • For part $(a)$ see this post. – Dietrich Burde Jan 31 '20 at 23:49
  • Another possible approach to (c) (aside from the ideal counting argument): in any field, or in any product of two fields, if $x^2 = 0$ then $x = 0$. – Daniel Schepler Jan 31 '20 at 23:53
  • thanks very much, that was very helpful. I can see now why (c) cant be a product of 2 fields, but why cant the same argument for (a) be applied to (c). So let $I = \langle 1+i \rangle, J = \langle 1-i \rangle, IJ = \langle 2 \rangle$. We also have $\langle 1+i, 1-i\rangle = \langle \gcd(1+i, 1-i)\rangle = \langle 1 \rangle$ and then apply the Chinese Remainder Theorem? – Nanoputian Feb 01 '20 at 00:35
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    In fact, $1-i = -i(1+i)$ where $-i$ is a unit so $\gcd(1+i,1-i) = 1+i$. – Daniel Schepler Feb 01 '20 at 01:21

1 Answers1

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Hint:

a) In $\mathbf Z[i]$, one has $5=(2+i)(2-i)$ and these factors are irreducible since their norm is equal to $5$. Furthermore, $\mathbf Z[i]$ is a P.I.D., so each generates a maximal ideal, and you can apply the Chinese remainder theorem:

$$\mathbf Z[i]/5\mathbf Z[i]\simeq \mathbf Z[i]/(2+i)\times\mathbf Z[i]/(2-i).$$

b) $$\mathbf Z[i]/2\mathbf Z[i]\simeq\mathbf Z[X]/(X^2+1)\!\Bigm/\!\!2\cdot\mathbf Z[X]/(X^2+1)\\\simeq \mathbf Z/2\mathbf Z[X]/(X^2+1)\simeq \mathbf Z/2\mathbf Z[X]/\bigl((X+1)^2\bigr).$$

Bernard
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    sorry, could you please explain in b) how you went from step 2 to step 3, and step 3 to step 4? I have only just started reading about this topic so I am not very familiar with all the theorems. Thanks. – Nanoputian Feb 01 '20 at 00:40
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    Step 2 to step 3: I used the standard isomorphism $\mathbf Z[X]/n\mathbf Z[X]\simeq (\mathbf Z/n\mathbf Z)[X]$ and the fact it is compatible with quotienting by a polynomial ideal. Step 3 to step 4: $\mathbf Z/2\mathbf Z$ is a field with characteristic $2$, and the Frobenius map is a ring homomorphism. Is that clear? – Bernard Feb 01 '20 at 00:46
  • yep, thanks that makes sense. – Nanoputian Feb 01 '20 at 01:03