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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.

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    $\begingroup$ $1+2i$ 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. $\endgroup$ Commented Mar 29, 2018 at 18:54
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    $\begingroup$ In other words the mapping $$a+bi+(8+11i)\mapsto (a+bi+(1+2i),a+bi+(6-i))$$ is an isomorphism. $\endgroup$ Commented Mar 29, 2018 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 $\oplus$. $\endgroup$ Commented Mar 29, 2018 at 19:00
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    $\begingroup$ Aren't those the same thing? At least up to isomorphism (assuming you use one for inner direct sum and the other for outer). $\endgroup$ Commented Mar 29, 2018 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. $\endgroup$ Commented Mar 30, 2018 at 4:36

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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...)

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