Consider the ring $M_2(\mathbb{R})$ of all $2 × 2$ matrices over $\mathbb{R}$.

  1. Is the zero ideal $\{0_{M_2(\mathbb{R})}\}$ maximal in $M_{2}(\mathbb{R})$?

We know, in the ring $\mathbb{Z}$, $\{0\}$ is not a maximal ideal, since, for example, $0\subset (2)\subset R$. So, can we use the same reasoning here? For example, $\{0_{M_2(\mathbb{R})}\}$ is not maximal in $M_{2}(\mathbb{R})$ since $\{0_{M_2(\mathbb{R})}\} \subset \begin{bmatrix} p & p\\ p & p \end{bmatrix}\subset M_{2}(\mathbb{R})$, where $p$ is a prime of $\mathbb{R}$?

  1. Is the quotient ring $M_2(\mathbb{R})/\{0_{M_2(\mathbb{R})}\}$ a division ring?

Since the zero ideal is not maximal in $M_{2}(\mathbb{R})$, $M_{2}(\mathbb{R})/\{0_{M_2(\mathbb{R})}\}$ is not a field. And, since all fields are division rings, can we say the vice versa is true? And, hence, $M_{2}(\mathbb{R})/\{0_{M_2(\mathbb{R})}\}$ is not a division ring because it is not a field. Would that be wrong?

Update: 1. To show that zero ideal is a maximal, can we show $M_{2}(\mathbb{R})/\{0_{M_2(\mathbb{R})}\}$ is a field and hence the conclusion. What does a general element in $M_{2}(\mathbb{R})/\{0_{M_2(\mathbb{R})}\}$ would look like in that case?

  • $\begingroup$ I was afraid of that. Could you please tell how can I prove $M_2(\mathbb{R})$ is a division ring? $\endgroup$
    – Jellyfish
    Commented Apr 20, 2015 at 23:59
  • 1
    $\begingroup$ @Jellyfish : $M_2(\Bbb R)$ is not a division ring. $\endgroup$
    – rschwieb
    Commented Apr 21, 2015 at 0:17
  • $\begingroup$ Is there any other simpler way to prove this? $\endgroup$
    – Jellyfish
    Commented Apr 21, 2015 at 1:44

1 Answer 1


The zero ideal is in fact maximal in $M_2(\mathbb{R})$, as it is for $M_n(F)$ for any $n>1$ and field $F$:

To prove the zero ideal is maximal, we take any non-zero ideal $\mathfrak{A}$ and we must prove it is the whole ring $M_n(F)$. To do this, it certainly suffices to show that $\mathfrak{A}$ contains the identity matrix. Let $0 \neq A \in \mathfrak{A}$ and let $\mathfrak{B}=\{e_1,...,e_n\}$ be the standard basis of $F^n$. Then $Ae_j \neq 0$ for some $j$. Now for $1 \leq i \leq n$, let $Q_i$ be the unique element of $M_n(F)$ such that $Q_ie_i=e_j$ and $Q_ie_l=0$ for $l \neq i$. Then extend $Ae_j$ to a basis $\mathfrak{C}$ of $F^n$ and let $P_i$ be the unique matrix such that $P_i(Ae_j)=e_i$ and $P_i$ maps all other members of $\mathfrak{C}$ to $0$. Then $P_iAQ_ie_j=\delta _{ij} e_j$, hence $P_iAQ_i$ is the matrix with a $1$ in the $i,i$ position and zeroes elsewhere. Therefore, \begin{gather*} I_n=\sum_{i=1}^nP_iAQ_i \end{gather*} which is clearly in the ideal $\mathfrak{A}$, and we are done.

For the second part of your question, note that for any ring $R$, $R \cong R/(0)$, so you are simply asking if $M_2(\mathbb{R})$ is a division ring. But it is not. I'm sure you can think of a nonzero matrix with zero determinant, for instance.


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