Is $\Bbb Q[\sqrt{2}]$ a field? Is $\Bbb Q[\sqrt{2}]$ a field? My guess is no because I think $$\Bbb Q[\sqrt{2}]\cong \Bbb Q[x]/(x^2-2)$$ and that polynomial is reducible. But I feel I am overcomplicating things and I'm not sure if I am correct.
What's a simpler way of seeing that this thing is not a field? (Or is it a field?)
 A: There are at least two ways to see that $\mathbb{Q}[\sqrt{2}]=\mathbb{Q}[x]/(x^2-2)$ is a field:


*

*$\mathbb{Q}$ is a field, so $\mathbb{Q}[x]$ is a PID. The element $x^2-2$ is irreducible (as it is a quadratic and has no root in $\mathbb{Q}$), so it is prime (irreducible $\Rightarrow$ prime in PID). So $(x^2-2)$ is a prime ideal. But again because we are in a PID, this ideal $(x^2-2)$ is actually maximal. And it is a standard fact that the quotient of a ring by maximal ideal is a field.

*One can also write $\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}: a, b\in\mathbb{Q}\}$ and show explicitly that it is a field. So show that it is closed under addition, and multiplication. In other words, show that the product of $a+b\sqrt{2}$ and $c+d\sqrt{2}$ (where $a,b,c,d\in\mathbb{Q}$) can again be written in the form $s+t\sqrt{2}$ with $s,t\in\mathbb{Q}$. Finally, let's show that it is closed under inverses: if $a+b\sqrt{2}$ is non-zero, then first check that $a^2-2b^2\neq 0$ (you will need the fact that $\sqrt{2}\notin\mathbb{Q}$) and then check that
$$
\frac{1}{a+b\sqrt{2}} = \frac{a-b\sqrt{2}}{a^2-2b^2} = \frac{a}{a^2-2b^2} + \left(\frac{-b}{a^2-2b^2}\right)\sqrt{2}\in\mathbb{Q}[\sqrt{2}]
$$

A: The polynomial $x^2-2$ is irreducible over $\mathbb{Q}$, because it has no roots (and has degree $2$).
In general, if $L$ is an extension field of $F$ and $b\in L$ is algebraic over the field $F$, then $F[b]=F(b)$. Here $F[b]$ is the image of the ring homomorphism $F[x]\to L$ fixing $F$ and sending $x$ to $b$; $F(b)$ is the minimal subfield of $L$ containing $F$ and $b$.
Since $F[b]\subseteq F(b)$ for obvious reasons, it is sufficient to see that $F[b]$ is a field, that is, every nonzero element has an inverse in $F[b]$.
A nonzero element $c\in F[b]$ is again algebraic over $F$, because $F(c)\subseteq F(b)$ and so the extension $F(c)/F$ is finite. If $g(x)=a_0+a_1x+\dots+a_{n-1}x^{n-1}+x^n$ is the minimal polynomial of $c$ over $F$, we have $a_0\ne0$ (or the polynomial would be reducible) and from
$$
a_0c^{-1}=a_1+a_2c+\dots+a_{n-1}c^{n-2}+c^{n-1}
$$
we get
$$
c^{-1}=\alpha_0+\alpha_1c+\dots+\alpha_{n-1}c^{n-1}\in F[c]\subseteq F[b]
$$
A: Thr maximal ideals of $\mathbf Z[x]$ have the for $(p, f(x))$, where $p$ is a prime, and $f(x)$ is an irreducible polynomial modulo $p$.
