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Let $R=\left\{\begin{pmatrix} a_1&a_2\\a_3&a_4 \end{pmatrix} \mid a_i\in \mathbb{Z} \right\}$ and let $I$ be the subset of $R$ with even entries. Show that $I$ is an ideal of $R$. What is the cardinality of $R/I$?

This is how I prove it:

Let $x=\{\left[\begin{array}{rr} 2b_1&2b_2\\2b_3&2b_4 \end{array}\right]|b_i\in Z\}\in I$ and let $y=\{\left[\begin{array}{rr} a_1&a_2\\a_3&a_4 \end{array}\right]|a_i\in Z\}\in R$.

Note: To show that I is and ideal of R , $xy\subseteq I$ and $yx\subseteq I$ for all $y\in R$.

First, we try $xy\subseteq I$, that is

$xy=\left[\begin{array}{rr} 2b_1&2b_2\\2b_3&2b_4 \end{array}\right]\left[\begin{array}{rr} a_1&a_2\\a_3&a_4 \end{array}\right]=\left[\begin{array}{rr} 2(b_1a_1+b_2a_3)&2(b_1a_2+b_2a_4)\\2(b_3a_1+b_4a_3)&2(b_3a_2+b_4a_4) \end{array}\right]\in I$

Now for $yx\subseteq I$.

$yx=\left[\begin{array}{rr} a_1&a_2\\a_3&a_4 \end{array}\right]\left[\begin{array}{rr} 2b_1&2b_2\\2b_3&2b_4 \end{array}\right]=\left[\begin{array}{rr} 2(a_1b_1+a_2b_3)&2(a_1b_2+a_2b_4)\\2(a_3b_1+a_4b_3)&2(a_3b_2+a_4b_4) \end{array}\right]\in I$

Since its satisfy the condition, therefore I is an ideal of R.

Now for finding the cardinality of R/I, I'm having a hard time solving for it. Can someone please show me how to get it. Thank you so much!

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  • $\begingroup$ Do you mean to say the cardinality of R/I? $\endgroup$ – Quinn Greicius Feb 23 '16 at 2:14
  • $\begingroup$ Well its one of the questions in my problem set given by my teacher and R/N is the cardinality being mentioned to find. Do you think its R/I?? If its R/I can you show me how to get it ?? thanks! $\endgroup$ – J. De Vera Feb 23 '16 at 2:23
  • $\begingroup$ It would be strange to have you go to all the trouble of showing $I$ to be an ideal, only to quotient out by a different, undefined ideal, so I'm inclined to believe it should be $R/I$... $\endgroup$ – Quinn Greicius Feb 23 '16 at 2:30
  • $\begingroup$ Okay, can you show me the cardinality of R/I then ? it would be a big help! $\endgroup$ – J. De Vera Feb 23 '16 at 2:33
  • $\begingroup$ See my answer below. I'm happy to elaborate, but I think you'll be able to finish it. $\endgroup$ – Quinn Greicius Feb 23 '16 at 2:56
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To compute the cardinality of the quotient, simply consider $R,I$ as abelian groups. $R$ is isomorphic to $\Bbb{Z}^4$, while $I$ is isomorphic to $2\Bbb{Z}^4$.

Every coset has a representative of the form $$\left[ \begin{matrix} a_1 & a_2 \\ a_3 & a_4 \end{matrix} \right]$$ where $a_1 , a_2 , a_3 , a_4 \in \{ 0,1 \}$. This means that the cosets are exactly $2^4=16$.

Another way to see this is to find a group isomorphism $$\Bbb{Z}^4/2\Bbb{Z}^4 \cong (\Bbb{Z}/2\Bbb{Z})^4$$ which can be found using the first isomorphism theorem on the projection $\Bbb{Z}^4 \to (\Bbb{Z}/2\Bbb{Z})^4$.

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We can see that $I=2R$, and taking the quotient sends the matrices with all even entries to zero. The quotient ring consists of all the cosets of $I$, so all you need to do is identify the different ways a matrix can fail to have all even entries. For example, $\left[\begin{array}{rr} 2&3\\7&7 \end{array}\right]$ would be in the coset defined by $I+\left[\begin{array}{rr} 0&1\\1&1 \end{array}\right]$. Can you see how to find the other cosets?

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It suffices to only consider $R/I$ as an abelian group. As an abelian group, we have $R=\mathbb Z^4$ and $I=2R$, i.e. $R/I=(\mathbb Z/(2))^4$ with $16$ elements.

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