Find all the units in the indicated rings.

Question

Find all the units in the indicated rings.

$Q[√3]$ : I know that this is somehow related to the set of rational numbers but my book doesn't define the elements of this group and I couldn't find anything on google aside from the definition of rational numbers.

$M(2, Z)$ : I suspect that for this group, if we take any element and its inverse, their commutative products would yield the identity matrix. But since the matrices draw from integers for elements within the matrices, either the determinate is $1$ or is a divisor of each element in the matrix. I don't know how to express that numerically.

$M(2, R)$: For this, as long as the matrix is invertible, it and it's product equal the identity. So this should just be GL(2,R) I think.

Any help solving this would be appreciated.

Answer 1

1) Because $\sqrt{3}$ is algebraic over $\mathbb{Q}$, the ring $\mathbb{Q}[\sqrt{3}]$ is actually the entire field $\mathbb{Q}(\sqrt{3})$, so the units are the nonzero elements.

2) A unit in $M_2(\mathbb{Z})$ is an invertible matrix in $M_2(\mathbb{Q})$ whose unique inverse in $M_2(\mathbb{Q})$ has integer entries. If $A$ is an invertible matrix in $M_2(\mathbb{Q})$, then $\det(A) \cdot \det(A^{-1}) = 1$. If $A$ and $A^{-1}$ both have integer entries, then both of these determinants are integers, so they must both be 1 or both be -1. Conversely, if the determinant of a matrix in $M_2(\mathbb{Z})$ is $\pm 1$, then the adjoint formula for its inverse shows that its inverse also has integer entries. It follows that the units in $M_2(\mathbb{Z})$ are precisely the matrices with determinant equal to $\pm 1$.

3) You are correct.

(2 and 3 both fall under the more general: the units of $M_n(R)$ where $R$ is an integral domain are the matrices with determinant equal to a unit in $R$)

Answer 2

The units in $\mathbb{Q}[\sqrt{3}]$ are the elements $a+b\sqrt{3}$ such that $(a+b\sqrt{3})(c+d\sqrt{3})=1$ for some $c,d \in \mathbb{Q}$. Expanding, and equating coefficients of $\sqrt{3}$ and $1$, $$ ac+3bd=1,ad+bc=0 $$ There are two solutions of this set of equations: $$ a=c=0,b \neq 0, d=\frac{1}{3b} $$ and $$ a \neq 0, a^2-3b^2 \neq 0,c=\frac{a}{a^2-3b^2},d=\frac{b}{3b^2-a^2} $$ Both work, the choice of $a$ and $b$ give the nature of the unit. I leave you to show that the whole set, barring zero, falls into the above category. Hence the set of units is $\mathbb Q[\sqrt{3}]$ without $0$.

For the second one, suppose our matrix is$ \begin{pmatrix} a\ b \\ c\ d \end{pmatrix}$ , then we know that the inverse is $\frac{1}{ad-bc}\begin{pmatrix} d\ -b \\ -c\ a \end{pmatrix}$ So the inverse exists and is an integer matrix if and only if the determinant is $\pm 1$.It follows that all such matrices make up the units of $M(2,\mathbb Z)$.

As for the third, once again we look at the above example, and realize that nothing special other than $ad-bc \neq 0$ needs to be assumed. It follows that all invertible matrices are units in $M(2,\mathbb R)$.