# What does the following notation $\mathbb{Q}(\sqrt{3})$ mean?

I came across the notation while looking at solutions to a problem. So without going through the entire problem, I solved some polynomials and found solutions to be

$$(\pm1,\pm1),(\pm \sqrt{3},0)$$

So 4 solutions. Then I am asked "What is the smallest field $k$ that contains $\mathbb{Q}$ such that all solutions lie $k^2$?"

I thought $\mathbb{R}^2$ would do so the field $\mathbb{R}$. But apparently, the answer is $\mathbb{Q}(\sqrt{3})$.

I'm just bewildered as I've never seen this notation, and it doesn't explain so I guess it's supposed be some standard, common way to express some field, but I don't know what it is. So I can't even google it, as I don't know what it's called or read as.

Does anyone know?

• Note, $\mathbb{R}^2$ is not a field. – Michael Albanese Dec 18 '15 at 20:19
• But $\mathbb{R}$ is a field though, yes? I'll re-write the question for clarity – Kydo Dec 18 '15 at 20:20
• Wait what? was the polynomial in $\mathbb R^2$ or in $\mathbb R$. Why are the solutions in $\mathbb R^2$? – Jorge Fernández Hidalgo Dec 18 '15 at 20:21
• See adjunction. – Alex Provost Dec 18 '15 at 20:23

$\mathbb{Q}(\sqrt{3}) = \{a + b \sqrt{3} \mid a, b \in \mathbb{Q}\}$. This is the smallest subfield of $\mathbb{R}$ that contains both $\mathbb{Q}$ and $\sqrt{3}$. In particular, $\mathbb{Q}(\sqrt{3})$ contains $0$, $\pm 1$, and $\pm\sqrt{3}$, but $\mathbb{Q}(\sqrt{3})$ doesn't contain numbers like $\sqrt{2}$, for example.
Edit. In general, if $K \subseteq F$ are fields and $\alpha \in F$, then $K(\alpha)$ denotes the smallest subfield of $F$ containing both $K$ and $\alpha$. Moreover, if $\alpha$ is algebraic over $K$, with minimal polynomial of degree $n$, then $$K(\alpha) = \{x_0 + \alpha x_1 + \cdots + \alpha^{n-1} x_{n-1} \mid x_0,\ldots, x_{n-1} \in K\}.$$
• So $\mathbb{Q}(i)=\{a+bi|a,b \in \mathbb{Q}\}$ for some general $i \in \mathbb{R}$? Does this apply also for other fields...say $\mathbb{R}(\sqrt{3})=\{a+b \sqrt{3}|a,b \in \mathbb{R}\}$? Are there such things too? – Kydo Dec 18 '15 at 20:23
• @Kydo: No, that particular expansion of $\mathbb Q(\alpha)$ only holds for $\alpha$s that can be written with a single square root. The general (for this purpose) defintion of $\mathbb Q(\alpha)$ is "the smallest subfield of $\mathbb C$ that contains both all of $\mathbb Q$ and $\alpha$". (Containing all of $\mathbb Q$ is redundant, though, because every subfield of $\mathbb C$ does that). – hmakholm left over Monica Dec 18 '15 at 20:36