Meaning of $\mathbb{Z}\left[\sqrt{3}\right]$?

Question

I've found here (first answer) this expression: $\mathbb{Z}\left[\sqrt{3}\right]$, where $\mathbb{Z}$ stands for the set of integers. What is the meaning of the expression?

Answer 1

$$\mathbb Z[X]=\{a_0+a_1X+a_2X^2+...\mid a_i\in\mathbb Z\}$$ then $$\mathbb Z[\sqrt 3]=\{a_0+a_1\sqrt{3}+a_2\sqrt{3}^2+a_3\sqrt{3}^3+...\mid a_i\in\mathbb Z\}=\{\underbrace{(a_0+3a_2+9a_4+...)}_{\in\mathbb Z}+\sqrt{3}\underbrace{(a_1+3a_3+9a_5+...)}_{\in\mathbb Z}\mid a_i\in\mathbb Z\}$$ $$=\{a+\sqrt 3 b\mid a,b\in\mathbb Z\}$$

Answer 2

If you have a commutative ring $R$, and an element $\alpha$ which might not necessarily be in the ring $R$, the symbol $R[\alpha]$ will denote the set of elements of the form:

$r_0+r_1\alpha+r_2\alpha^2+r_3\alpha^3+\dots+r_k\alpha^k$

where each $r_i$ is an element of your ring $R$.

In your specific example, it would be elements of the form $z_0+z_1\sqrt{3}+z_2(\sqrt{3})^2+\dots+z_k(\sqrt{3})^k$, but notice that $\sqrt{3}$ to an even power is just an integer, so it simplifies to simply being an element of the form $a+b\sqrt{3}$.

Other important related concepts are the polynomial ring, where instead of a specific element, you use an unknown. $R[x]=\{r_0+r_1x+r_2x^2+\dots+r_kx^k~:~r_i\in R,~k\in\mathbb{N}\}$ where each polynomial has only finitely many nonzero terms.

Also of interest is $R[\![x]\!]=\{\sum\limits_{i=0}^\infty r_ix_i~:~r_i\in R\}$, the formal power series where we instead allow infinitely many nonzero terms.

Answer 3

It is the image of the ring of polynomials $\mathbf Z[X]$ under tha mappting \begin{align*} \mathbf Z&\longrightarrow \mathbf C\\ p(x)&\longmapsto p(\sqrt 3) \end{align*} As $(\sqrt3)^2=3$, all such $p(\sqrt 3)$ can be ultimately written in the form $\;a+b\sqrt3\enspace (a,b\in\mathbf Z)$.

Its field of fractions is $\;\mathbf Q[\sqrt 3]=\bigl\{a+b\sqrt3\mid a,b\in\mathbf Q\bigr\}$. It is indeed a field since one checks $$(a+b\sqrt3)^{-1}=\frac{a-b\sqrt3}{a^2-3b^2}.$$

Answer 4

This is a ring of integers, augmented with an additional value $\sqrt{3}$.

Answer 5

We have an (abstractly given) field extension $\mathbb{Q}(\sqrt{3}) = \frac{\mathbb{Q}[x]}{\langle x^2 - 3 \rangle}$. By identifying $\mathbb{Q}$ with its image under the map $$ \mathbb{Q} \hookrightarrow \mathbb{Q}[x] \to \frac{\mathbb{Q}[x]}{\langle x^2 - 3 \rangle} $$ we can consider $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{3})$. Then $\mathbb{Z}[\sqrt{3}]$ is defined as the smallest (by inclusion) subring $R \subseteq \mathbb{Q}(\sqrt{3})$ containing $\mathbb{Z}$ and $\sqrt{3}$ (where again we identify $\mathbb{Z}$ with its image in $\mathbb{Q}(\sqrt{3})$). As other answers have stated, one can show that any element of $\mathbb{Z}[\sqrt{3}]$ is of the form $a + b \sqrt{3}$ with $a,b \in \mathbb{Z}$.