Why is $M_2(2\mathbb{Z})$ a ring without unity?

Question

I can understand that The set $M_2(2\mathbb{Z})$ of $2 \times 2$ matrices with even integer entries is an infinite non commutative ring. But why It doesn't have any unity? I think $I(2\times2)$ is a unity for this ring. Where is my misunderstanding in subject of $I(2\times2)$ being a unity?

Answer 1

$I(2\times 2)=\left[\begin{matrix}1&0\\0&1\end{matrix}\right]\not\in M_2(2\mathbb{Z})$ since $M_2(2\mathbb{Z})$ has only even integers as its entries. It can be shown using an arbitrary matrix $B$ that $I(2\times 2)$ is the only possible unity in $M_2(2\mathbb{Z}),$ but since it is not an element of $M_2(2\mathbb{Z}),$ there is no unity. There are other rings without a unity, such as $2\mathbb{Z},$ the set of even integers.

Btw, the notation $2\mathbb{Z}$ is equivalent to the ideal $\langle 2\rangle := \{2x : x\in\mathbb{Z}\},$ if you didn't know.

Answer 2

The $2\times 2$ identity has entries $1,0,0,1$. Are all of those are even numbers?

Answer 3

If $I(2\times2)=\left[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right]$, then $I(2\times2)\notin M_2(2\mathbb Z)$ since $1$ is odd.

Answer 4

If your matrix ring had an identity element $I,$ then the square of its determinant would have to be $1$ (or $0$), since $I^2=I.$ The determinant cannot be $0$ since $I$ is invertible (or, non-singular, if you prefer). But the determinant of a $2 \times 2$ matrix over $2\mathbb{Z}$ is even ($ad-bc$ is a difference of a product of even numbers). So, you cannot have an identity element in this ring.

Answer 5

As everyone pointed out right away, your candidate is not an element of your ring.

Here is a slightly different way to prove that $M_2(2\mathbb Z)$ can't have an identity.

Notice that because your entries are always sums of products of things in $2\mathbb Z$, you're always accumulating powers of two.

In particular,

1. if $I$ is a nonzero element such that $I^2=I$ (this must hold for an identity!), then $I\in M_2(4\mathbb Z)$.

2. Then $IA\in M_2(8\mathbb Z)$ for any matrix $A\in M_2(2\mathbb Z)$.

3. But you know that you can pick $A\in M_2(2\mathbb Z)\setminus M_2(8\mathbb Z)$.

4. This contradicts the observation that $A=AI\in M_2(8\mathbb Z)$.

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Here is another argument reminiscent of Chris Leary's approach. Suppose $E$ is the identity of $M_2(2\mathbb Z)$.

Then the adjugate matrix $\mathrm{adj}(E)\in M_2(2\mathbb Z)$ also, and satisfies $E\mathrm{adj}(E)=det(E)I_2\in M_2(2\mathbb Z)$. In particular, $\det(E)$ is an even number. Also observe that both $E$ and $\mathrm{adj}(E)$ are both nonzero.

Since $E^2=E$, we have $\det(E)=\det(E)^2$, and the only two integers that satisfy $x^2=x$ are $0$ and $1$, and $1$ is obviously not even.

So $\det(E)=0$: but then $\mathrm{adj}(E)=E\mathrm{adj}(E)=0$, contradicting the earlier observation that $\mathrm{adj}(E)\neq 0$.