Is it a field or not?

Question

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?

Answer 1

$A$ 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.

Answer 2

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.