Is $PGL_n(\mathbb{Q})=PGL_n(\mathbb{Z})$?

Question

Since $PGL_n(\mathbb{Q})=GL_n(\mathbb{Q})/Z(GL_n(\mathbb{Q}))$, I think I should be able to clear denominators in any matrix in $PGL_n(\mathbb{Q})$ to get a matrix in $PGL_n(\mathbb{Z})$. But then I am confused, because each embedding $\mathbb{Q} \hookrightarrow \mathbb{Q}_p$ induces an injective homomorphism $PGL_n(\mathbb{Q}) \rightarrow PGL_n(\mathbb{Q}_p)$. By my reasoning, the image would be in $PGL_n(\mathbb{Z}_p)$ which doesn't seem right, but I don't see the gap in my reasoning. Could someone clarify?

Answer 1

I figured it out. They are not equal. I had thought of $GL_n(Z)$ as matrices with nonzero determinant, when it should mean matrices in $M_n(Z)$ that are invertible, which is equivalent to having determinant invertible in $Z$. So the matrices with determinant not $\pm 1$ are in the complement.