Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $S \subset R$, $R$ ring.

Is $S$ a field, knowing that $\displaystyle{R=M_2(\mathbb{R}), \text{and } S= \begin{Bmatrix} \bigl(\begin{smallmatrix} 0&0 \\ 0&a \end{smallmatrix}\bigr): a \in \mathbb{R} \end{Bmatrix}}$ ?

I have shown that $S$ is an integral domain, so to check if $S$ is a field, don't I have just to check if each nonzero element of $S$ is invertible ?

So, don't I have to check if the determinant is equal to $0$ or not?

Let $A \in S$, $A=\bigl(\begin{smallmatrix} 0&0 \\ 0&a \end{smallmatrix}\bigr): a \in \mathbb{R}$

$\det(A)=0$, so $A$ is not invertible, right?

But, according to my notes, we can always find an invertible $A'= \bigl(\begin{smallmatrix} 0&0 \\ 0&\frac{1}{a} \end{smallmatrix}\bigr), \ a\in \mathbb{R}^{*} $

How can this happen?

share|improve this question
To talk about invertibility, you need an identity element. Can you find $E\in S$ such that $\forall A\in S(AE=EA=A)$? –  Git Gud May 18 '14 at 22:01
Yes, $E= \bigl(\begin{smallmatrix} 0&0 \\ 0&1 \end{smallmatrix}\bigr) $ –  evinda May 18 '14 at 22:03
Good. Now given $A\in S$, can you find $B$ auch that $AB=BA=E$? –  Git Gud May 18 '14 at 22:04
Yes, $A'$ that I have written in my post above.. –  evinda May 18 '14 at 22:07

2 Answers 2

up vote 2 down vote accepted

$A$ is not invertible in the space of $2 \times 2$ matrices. It is invertible in the set you have chosen. The $A'$ you show will multiply by $A$ to make the identity element in your set. In fact, there is a natural bijection between $R$ and $\Bbb R$ that preserves the field operations.

share|improve this answer
A ok...I got it!!! Thank you very much!!!! –  evinda May 18 '14 at 22:06

The point is that, though both $S$ and $R$ does have an identity elements, and $S\subseteq R$ is a subring, but it is not a substructure of rings with identity (in other words, the inclusion $S\hookrightarrow R$ does not preserve identity).

So, invertibility in $R$ means a different thing than invertibility in $S$.

Now $S$ is isomorphic to the field $\Bbb R$, so it is also a field, but no elements of it are invertible in $R$, indeed.

share|improve this answer
I understand...thank you!! –  evinda May 18 '14 at 22:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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