All elements of factor ring $M_2(\mathbb{Z})/ M_2(2\mathbb{Z})$

abstract-algebraring-theory

I came across this example in the book Abstract Algebra by Gregory T. Lee. I found it hard to understand.

Let $R= M_2(\mathbb{Z})$ and let $I$ be the ideal consisting of all matrices whose entries are even. Then notice that for any $a_{ij} \in \mathbb{Z}$, we have
$$\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}+I=\begin{pmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{pmatrix}+I$$ where $b_{ij}$ is if $a_{ij}$ is even and 1 if $a_{ij}$ is odd. Thus, $R/I$ consists of the sixteen different elements $ \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} + I, \, b_{ij} \in \{0,1\}$.

$M_2(\mathbb{Z})$ is the set of 2 x 2 matrices over $\mathbb{Z}$.

My thoughts:

For $a_{ij} \in \mathbb{Z}$ and $ r, s, t, u \in 2 \mathbb{Z}$, we have:

$$\underbrace{a_{ij}}_{\text{even}} + r = \overbrace{0 + s}^{\text{even}}$$

and

$$\underbrace{a_{ij}}_{\text{odd}} + t = \overbrace{1 + u}^{\text {odd}}. $$

The elements of the $ R / I $ quotient ring are of the type $ \begin{pmatrix} b_ {11} & b_ {12} \\ b_ {21} & b_ {22} \end{pmatrix} + I $ with $ b_{ij} \in \{0, 1\}. $ Since there are $ 2 ^ 4 = 16$ arrangements with repetition of elements 0 and 1, then the set $ R / I $ has 16 elements.

I really need a more detailed explanation for this example.

Question:

How to prove that by double inclusion? Thanks in advance!

Best Answer

Let $$S = \bigg\{ \begin{pmatrix} b_ {11} & b_ {12} \\ b_ {21} & b_ {22} \end{pmatrix} + I : b_{ij} \in \{0,1\} \bigg\}$$ and we claim that $R/I = S$.

  • For every $a_{ij} \in \mathbb Z$ define $b_{ij}=0$ if $a_{ij}$ is even and $b_{ij}=1$ if $a_{ij}$ is odd. Thus $a_{ij}-b_{ij}$ is always even, and then $$\begin{pmatrix} a_{11} & a_ {12} \\ a_ {21} & a_ {22} \end{pmatrix} + I = \begin{pmatrix} b_ {11} & b_ {12} \\ b_ {21} & b_ {22} \end{pmatrix} + I$$ because $$\begin{pmatrix} a_{11} & a_ {12} \\ a_ {21} & a_ {22} \end{pmatrix} - \begin{pmatrix} b_ {11} & b_ {12} \\ b_ {21} & b_ {22} \end{pmatrix} = \begin{pmatrix} a_{11}-b_{11} & a_ {12}-b_ {12} \\ a_ {21}-b_ {21} & a_ {22}-b_ {22} \end{pmatrix} \in I.$$ This proves that every element of $R/I$ is equal to an element of $S$, and then $R/I \subseteq S$.
  • Clearly every element of $S$ is also an element of $R/I$, so $S \subseteq R/I$.
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