Why is $\mathbb{Q}[\sqrt{2}]$ a field?

Question

It seems to me that the hard part of this proof is multiplicative inverses. I know how to prove this by considering the multiplication of two arbitrary elements:

$$(a + b\sqrt{2})(a' + b'\sqrt{2}) = \dots$$

and seeing what values of $\{a', b'\}$ make the result equal to $1 + 0\sqrt{2}$. This is just solving a linear system, and at the end you get:

$$ \begin{aligned} a' &= -\frac{a}{2b^2 - a^2}\\ b' &= \frac{b}{2b^2 - a^2}\\ \end{aligned} $$

It so happens that $2b^2 - a^2 \ne 0$ when $a, b \in \mathbb{Q}$, so the inverse always exists. Magic!

But it's just magic. Where did that come from? What is going on here? And can it be generalized to prove that these sorts of things are fields?

• $\mathbb{Q}[\sqrt{3}]$
• $\mathbb{Q}[\sqrt[3]{q}] = \{a + b\sqrt{q} + c\sqrt[3]{q}^2\}$
• $\mathbb{Q}[\sqrt{q}, \sqrt{p}] = \{a + b\sqrt{q} + c\sqrt{p} + d\sqrt{pq}\}$

(I know that these are examples of "field extensions" from Wikipedia, but we've never talked about them in class. Also I imagine that $p$ and $q$ have to satisfy some conditions -- not being square is probably one.)

Answer 1

The reason is much more general: consider multiplication by $a +b\sqrt2$ in the $\mathbf Q$-vector space $\mathbf Q[\sqrt2]$. If is non-zero, it is an injective endomorphism of this space. As we're in finite dimension, an injective endomorphism is surjective. Hence $1$ is attained, i.e. there exists $a'+b'\sqrt2\in\mathbf Q[\sqrt2]$ such that $$(a +b\sqrt2)(a'+b'\sqrt2)=1.$$

The most general result is this:

Let $A\hookrightarrow B$ an injective ring homomorphism. Suppose $A, B$ are integral domains, and $B$ is a finitely generated $A$-module. Then $B$ is a field if and only if $A$ is a field.

Answer 2

$(a+b\sqrt2)(a-b\sqrt2) = a^2 - 2b^2$

If $a^2 - 2b^2 \ne 0, \frac{a}{a^2-2b^2} - \frac{b}{a^2-2b^2}\sqrt2$ is the multiplicative inverse of $a+b\sqrt2$

If $a^2 - 2b^2 = 0$ then $a$ or $b$ is irrational. (Or $a$ = $b$ = $0$)

Answer 3

Instead of blindly solving the linear system, you can also observe that $(a+b\sqrt2)(a-b\sqrt 2) = a^2-2b^2$, where the right-hand side is rational and can be divided away -- so $$\frac{a-b\sqrt2}{a^2-2b^2} = \frac{a}{a^2-2b^2} + \frac{-b}{a^2-2b^2}\sqrt2$$ must be an inverse of $a+b\sqrt 2$.

This principle works for $\mathbb Q[\sqrt D]$ in general (so-called quadratic field). With some modifications it can also be used for for $\mathbb Q[\sqrt D,\sqrt E]$ (eliminate the square roots one at a time), but heavier methods are needed $\mathbb Q[\sqrt[3]D]$.

Answer 4

I feel like the real question here is why all nontrivial solutions of $a^2-2b^2=0$ are irrational.

Assume $b\neq 0$. Then we get $\frac{a}{b}=\sqrt{2}$. But the RHS is irrational, so the LHS must have at least one irrational term.

Assume $b=0$. Then $a=0$

This proof works for the same problem where we replace $\sqrt{2}$ with $\sqrt{D}$ for any non-square $D$.

Answer 5

This may not be the answer you want, but here goes...

We call an element algebraic over a field (I'll stick to working over $\mathbb{Q}$) if it is the root of a non-zero polynomial over that field.

Say $\alpha$ is algebraic. If we take the set $I = \{ f(x) \in \mathbb{Q}[x] \;|\; f(\alpha)=0 \}$ (all polynomials with $\alpha$ as a root), $I$ forms an ideal in $\mathbb{Q}[x]$. Now $\mathbb{Q}[x]$ is a principal ideal domain (PID) so $I$ has a generator, say $I=(p(x))$ (i.e. everything in $I$ turns out to be a multiple of $p(x)$). Let's assume that $p(x)$ is monic (has a leading coefficient $1$).

$p(x)$ is then called the minimal polynomial of $\alpha$ (over $\mathbb{Q}$). This is the polynomial of least degree which has $\alpha$ as a root. Also, because of minimality it turns out that $p(x)$ is irreducible.

For example: $\sqrt[3]{2}$ is algebraic (over $\mathbb{Q}$) with minimal polynomial $x^3-2$. So any polynomial in $\mathbb{Q}[x]$ which has $\sqrt[3]{2}$ as a root must be divisible by $x^3-2$.

Ok. Some ring theory: Irreducible polynomials generate maximal ideals. In other words, if $J$ is an ideal and $I \subseteq J \subseteq \mathbb{Q}[x]$, then $J=I$ or $J=\mathbb{Q}[x]$ (there's nothing between $I$ and $\mathbb{Q}[x]$). Quotients by maximal ideals yield fields. Therefore, $\mathbb{Q}[x]/I = \mathbb{Q}[x]/(p(x))$ is a field.

What does this have to do with $\alpha$?

Consider the evaluation homomorphism: $\varphi:\mathbb{Q}[x] \to \mathbb{C}$ defined by $\varphi(f(x))=f(\alpha)$ (i.e. plug in $\alpha$). By definition, the image of $\varphi$ is $\mathbb{Q}[\alpha]$ (i.e. polynomials with $\alpha$ plugged in). The kernel of $\varphi$ is the set of all $f(x)$ such that $\varphi(f(x))=f(\alpha)=0$. This is exactly $I=(p(x))$! By the first isomorphism theorem: $\mathbb{Q}[x]/(p(x)) \cong \mathbb{Q}[\alpha]$. Since we know that the quotient is a field, $\mathbb{Q}[\alpha]$ is a field.

Now what if $\alpha$ is transcendental (i.e. not algebraic)? Then there is no non-zero polynomial with $\alpha$ as a root. This means that the kernel is trivial. In that case $\mathbb{Q}[x] \cong \mathbb{Q}[\alpha]$. Now $\mathbb{Q}[x]$ is not a field. So for transcendental elements, $\mathbb{Q}[\alpha] \not= \mathbb{Q}(\alpha)$ (parentheses thing = smallest field containing thing).

Now why does the magic trick work for $\mathbb{Q}[\sqrt{2}]$?

Basically Galois theory.

An image of an element under a Galois automorphism us called a conjugate (generalizing the complex conjugate). For $\mathbb{Q}[\sqrt{2}]$ you have 2 automorphisms: (1) The identity $a+b\sqrt{2} \mapsto a+b\sqrt{2}$ and (2) The other one $a+b\sqrt{2} \mapsto a-b\sqrt{2}$ (i.e. send $\sqrt{2}$ to $-\sqrt{2}$).

A product of an element and all of its conjugates yields an element of the base field so $(a+b\sqrt{2})(a-b\sqrt{2})$ must be rational. Using this you can then rig up the "magic trick" to rationalize the inverse of an element.

Answer 6

From the edit:

And can it be generalized to prove that these sorts of things are fields?

The "magic" that is going on to compute inverses is actually not necessary to prove these things are fields, i.e. that when complex numbers $x_1,x_2,\cdots, x_m$ are roots of (rational coefficient) polynomials then the ring $\mathbb{Q}[x_1,\cdots,x_m]$ (the set of all polynomial expressions in $x_1,\cdots,x_m$ with coefficients from $\mathbb{Q}$) is a closed under the operations of not just addition and multiplication but division as well. Firstly, one can induct, since we can write $\mathbb{Q}[x_1,\cdots,x_m]=\mathbb{Q}[x_1,\cdots,x_{m-1}][x_m]$. That is, we must show that when $K$ is some field comprised of complex numbers and $x$ is an algebraic number, then $K[x]$ is also a field. To do this, we just need some linear algebra.

Use the fact that $x$ is a root of a polynomial to determine that $K[x]$ is a finite-dimensional vector space over $K$. Observe there are no zero divisors, i.e. the zero product property $ab=0\Rightarrow a=0$ or $b=0$ holds. Then pick any nonzero element $a\in K[x]$, and consider the function $f(x)=ax$ which is "multiplication-by-$a$." This is a linear map, and by the zero product property its kernel is trivial, hence it is a bijection (as $K[x]$ is finite-dimensional), which means there exists a $b$ for which $f(b)=1$, i.e. $ab=1$. This shows all nonzero elements are invertible in $K[x]$.

From the comments:

So in order to apply that to the others, I guess I would need to know the conjugate of some arbitrary $a + b\sqrt[3]{q} + c\sqrt[3]{q}^2$?

Indeed, computing inverses in such contexts amounts to "rationalizing the denominator," which itself involves computing conjugates. However when our algebraic numbers are roots of polynomials beyond degree two, there will be more than one conjugate! What is a conjugate of an algebraic number anyway?

If $x$ is the root of some polynomial, then there is a unique monic polynomial of minimal degree having $x$ as a root, and this is termed the minimal polynomial of $x$. The conjugates of $x$ are precisely the other roots of this minimal polynomial. The norm of $x$ will be the product of all these conjugates, which will be rational since it is plus or minus the constant coefficient of the minimal polynomial. Thus, to compute $1/x$, simply multiply the top and bottom by all of $x$'s conjugates.

(To be clear: I am speaking of $x$ algebraic over $\mathbb{Q}$, and the norm in the relative extension $\mathbb{Q}(x)/\mathbb{Q}$.)

This broaches the subject of Galois theory. A number field is (essentially) any collection of complex numbers which is closed under the usual arithmetic operations ($+,-,\times,/$) and is a finite-dimensional vector space over the rationals $\mathbb{Q}$. For example, you've seen $K=\mathbb{Q}+\mathbb{Q}\sqrt{2}$ is such a number field.

A symmetry of a number field (relative to $\mathbb{Q}$ anyway) is a relabelling of its elements, i.e. a function $f:K\to K$, which preserves the truth of all equations upon relabelling. For example, the "conjugation" map $a+b\sqrt{2}\mapsto a-b\sqrt{2}$ is a relabelling of the elements of $\mathbb{Q}+\mathbb{Q}\sqrt{2}$ (treat $a+b\sqrt{2}$ as a single number), and this relabelling turns the true equation $(1+\sqrt{2})^2=3+2\sqrt{2}$ into $(1-\sqrt{2})^2=3-2\sqrt{2}$, which is another true equation. Since all our equations are essentially polynomial expressions built from the arithmetic operations, it suffices to check such a function preserves the arithmetic operations, i.e. that it satisfies the two properties $f(a+b)=f(a)+f(b)$ and $f(ab)=f(a)f(b)$.

The symmetry group of a number field is its Galois group $\mathrm{Gal}(K/\mathbb{Q})$ defined by

$$\mathrm{Gal}(K/\mathbb{Q})=\left\{f:K\to K~\mathrm{such~that}~ \begin{array}{c} f(a+b)=f(a)+f(b) \\ f(ab)=f(a)f(b) \end{array}~\mathrm{for~all~}a,b\in K\right\}. $$

(I am avoiding discussing the important caveat that some extensions are not "Galois," i.e. the number of symmetries is less than it what it's supposed to be, which is the dimension of $K$ as a vector space over $\mathbb{Q}$.)

If $K=\mathbb{Q}(x)$ and $\mathrm{Gal}(K/\mathbb{Q})=\{\sigma_1,\cdots,\sigma_n\}$ then the symmetries $\sigma_i$ shuffle the roots of polynomials, hence they restrict to permutations on $x$'s conjugates, and in particular the set of conjugates is precisely the "orbit" $\{\sigma_1(x),\cdots,\sigma_n(x)\}$!

Determining what the symmetries in a Galois group are is in general a very difficult problem. Sometimes you need a number system a little bigger than $\mathbb{Q}(x)$ in order to get all of $x$'s conjugates (the "Galois closure"). For instance, there are no nontrivial symmetries of $\mathbb{Q}(\sqrt[3]{2})\subset\mathbb{R}$, but if you allow a complex number into the mix (a cube root of unity $\omega=\cos(2\pi/3)+i\sin(2\pi/3)$), then the conjugates of $\sqrt[3]{2}$ are $\omega\sqrt[3]{2}$ and $\omega^2\sqrt[3]{2}$, and more generally the conjugates of $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ are

$$ a+b\sqrt[3]{2}+c\sqrt[3]{4}, \quad a+b\omega\sqrt[3]{2}+c\omega^2\sqrt[3]{4}, \quad a+b\omega^2\sqrt[3]{2}+c\omega\sqrt[3]{4}. $$