Why does $\Bbb Z[{1+\sqrt{-19}\over 2}]$ contain $1$?

Question

Why does $\Bbb Z[{1+\sqrt{-19}\over 2}]$ contain $1$? What multiples of ${1+\sqrt{-19}\over 2}$ form 1?

Edit: Why does a subring of $\Bbb Z$ that contains ${1+\sqrt{-19}\over 2}$ have to contain 1?

Answer 1

$1$ is in $\mathbb Z$, so $$ 1 = 1 + 0\cdot\frac{1+\sqrt{-19}}2 $$

More generally $R[\alpha]$ is by definition the smallest ring that contains both $R$ and $\alpha$. Therefore every element of $R$ is in $R[\alpha]$ no matter what $\alpha$ is.

Answer 2

There's a confusion here. The notation $\mathbb{Z}\left[\frac{1+\sqrt{-19}}2\right]$ does not mean the multiples of $\frac{1+\sqrt{-19}}2$. It's the set of all numbers $P\left(\frac{1+\sqrt{-19}}2\right)$ where $P(x)$ is a polynomial with interger coefficients.

Now, consider the polynomial $P(x)=1$.