6
$\begingroup$

I am attempting to solve Ch 14 Problem 7.7 from Artin's algebra book.

Let $R=\mathbb{Z}[i]$ and let $V$ be the R-module generated by elements $v_1$ and $v_2$ with relations $(1+i)v_1+(2-i)v_2=0$ and $3v_1+5iv_2=0$. Write this module as a direct sum of cyclic modules.

Attempt

I have obtained $V\cong R^2/ \begin{bmatrix} 1+i & 3 \\ 2-i & 5i \end{bmatrix} R^2 \cong R/[8+11i]R=\mathbb{Z}[i]/(8+11i)$.

Now, I see that $(1+2i)(6-i)=8+11i.$ Now, I would like to show that $\mathbb{Z}[i]/(1+2i) \oplus\mathbb{Z}[i]/(6-i)\cong\mathbb{Z}[i]/(8+11i)$, so I can have $V$ as a direct sum of cyclic modules as needed, but how can I show this?

I have already shown that $(1+2i,6-i)=(1)=\mathbb{Z}[i]$ and thus $(1+2i)+(6-i)=\mathbb{Z}[i]$. Intuition would suggest that $(1+2i)\oplus(6-i)=\mathbb{Z}[i]$, although I think this is false since $(i-6)(1+2i)+(1+2i)(6-i)=0$.

I must confess that I am very new to module theory so please be patient with me. I don't even how it would be possible to have $\mathbb{Z}[i]/(1+2i) \oplus\mathbb{Z}[i]/(6-i)\cong\mathbb{Z}[i]/(8+11i)$ since $\mathbb{Z}[i]/(1+2i)$and $\mathbb{Z}[i]/(6-i)$ aren't even submodules of the same set.

$\endgroup$
  • 4
    $\begingroup$ $1+2i$ and $6-i$ are coprime elements of the Euclidean domain $\Bbb{Z}[i]$. Their product is $8+11i$ then also their "least common multiple". So this is just another instance of the Chinese Remainder Theorem. Compare with isomorphisms of $\Bbb{Z}$-modules $$\Bbb{Z}_2\oplus\Bbb{Z}_3\simeq\Bbb{Z}_6, \quad\Bbb{Z}_3\oplus\Bbb{Z}_5\simeq\Bbb{Z}_{15}$$ et cetera. $\endgroup$ – Jyrki Lahtonen Mar 29 '18 at 18:54
  • 1
    $\begingroup$ In other words the mapping $$a+bi+(8+11i)\mapsto (a+bi+(1+2i),a+bi+(6-i))$$ is an isomorphism. $\endgroup$ – Jyrki Lahtonen Mar 29 '18 at 18:55
  • $\begingroup$ Thank you. If I can show that that mapping is an isomorphism, this will show $\mathbb{Z}[i]/(8+11i) \cong\mathbb{Z}[i]/(1+2i) \times \mathbb{Z}[i]/(6-i)$. However, I am not trying to show $\times$, but $\oplus$. $\endgroup$ – Pascal's Wager Mar 29 '18 at 19:00
  • 1
    $\begingroup$ Aren't those the same thing? At least up to isomorphism (assuming you use one for inner direct sum and the other for outer). $\endgroup$ – Jyrki Lahtonen Mar 29 '18 at 19:24
  • $\begingroup$ For an answer following the method described in the problem statement: compute the Smith normal form of the relations matrix. Some relevant posts: 1, 2, 3. $\endgroup$ – André 3000 Mar 30 '18 at 4:36
1
$\begingroup$

In addition to everything said in the comments, I just wanted to remark one more thing that might be helpful concerning your last paragraph about the fact that $\mathbb{Z}[i]/(1+2i)$ and $\mathbb{Z}[i]/(6-i)$ are not submodules of some common module.

As was stated in the comments, the product of rings obtained from the chinese remainder theorem is in particular a product of modules, i.e. a direct sum, so this works perfectly fine. The point I wanted to add: If you want, you might still view both factors as submodules of $\mathbb{Z}[i]/(8+11i)$ by the following observation:

Consider a direct product of rings (commutative with $1$), $R=R_1\times R_2$. Then you might view any $R$-module $M$ as a direct sum of submodules $M_1,M_2\subset M$, where $M_i$ is a $R_i$-module (so in particular again a $R$-module) for $i=1,2$, namely set $M_1:=(1,0)\cdot M\subset M$ and $M_2:=(0,1)\cdot M\subset M$, then $$M=M_1\oplus M_2$$ as $R$-module. (By the way, the inverse procedure works as well, if $\tilde M_1$ and $\tilde M_2$ are $R_1$- resp. $R_2$-modules, then they in particular are both $R$-modules, via $(r_1,r_2)\cdot m_1:=r_1m_1$ for $m_1\in \tilde M_1$ and $(r_1,r_2)\cdot m_2:=r_2m_2$ for $m_2\in\tilde M_2$ and so one may build the $R$-module $\tilde M=\tilde M_1\oplus\tilde M_2$. This is also described in this question/answer.)

Applied to your situation: As you already proved $$R:=\mathbb{Z}[i]/(8+11i)\simeq \underset{=:R_1}{\underbrace{\mathbb{Z}[i]/(1+2i)}}\times\underset{=:R_2}{\underbrace{\mathbb{Z}[i]/(6-i)}},$$ we would now like to find elements $e_1,e_2\in R$ such that $e_1$ corresponds to $(1,0)\in R_1\times R_2$ and $e_2$ corresponds to $(0,1)$. To find those, we use the Euclidean algorithm to get $$1=\underset{=:e_2}{\underbrace{(1-3i)(1+2i)}}+\underset{=:e_1}{\underbrace{(-1)(6-i)}}.$$ So we might write $R$ as the direct sum of $R$-submodules $$R=e_1R\oplus e_2R=(7-i)R\oplus (i-6)R.$$ (Naturally, as a $R$-module this is of course isomorphic to the former version $R=R_1\oplus R_2$, as $e_i\cdot(\bullet)\colon R_i\rightarrow e_iR$ is an isomorphism of $R$-modules, $i=1,2$ - I just thought this slightly different point of view might possibly help to get more comfortable with the situation...)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.