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Why must $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ -- the set of all polynomials in $\sqrt{2}$ and $\sqrt{3}$ with rational coefficients -- contain multiplicative inverses?

I have gathered that every element of $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ takes form $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{2}\sqrt{3}$ for some $a,b,c,d \in \mathbb{Q}$, and I have shown that $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ are both fields, but it's not clear to me why this allows for general multiplicative inverses to exist in $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$.

hd121
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    Hint: $\mathbb{Q}[\sqrt{2},\sqrt{3}] = \mathbb{Q}[\sqrt{2}][\sqrt{3}]$. You probably did not need that the base field was $\mathbb{Q}$ when you showed that $\mathbb{Q}[\sqrt{3}]$ was a field. – Tobias Kildetoft Sep 17 '13 at 12:00
  • Related to http://math.stackexchange.com/questions/397733/algebraic-field-extension-question. – lhf Sep 17 '13 at 12:50

5 Answers5

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Let $K$ be an extension field of the field $F$ and let $0\ne a\in K$ be algebraic over $F$; then $F[a]$ contains the inverse of $a$.

Indeed, if $f(X)=c_0+c_1X+\dots+c_{n-1}X^{n-1}+X^n$ is the minimal polynomial of $a$ over $F$, then it's irreducible, so $c_0\ne 0$ and $$ c_0+c_1a+\dots+c_{n-1}a^{n-1}+a^n=0. $$ Multiply by $c_0^{-1}a^{-1}$ to get $$ a^{-1}=-c_0^{-1}(c_1+\dots+c_{n-1}a^{n-2}+a^{n-1}) $$ so $a^{-1}\in F[a]$.

If now $0\ne b\in F[a]$, you also have $F[b]\subseteq F[a]$. Since $F[b]$ is an $F$-subspace of $F[a]$, it's finite dimensional over $F$; therefore $b$ is algebraic over $F$ and, by what we showed above, $b^{-1}\in F[b]\subseteq F[a]$.

Now you can apply this to $\mathbb{Q}[\sqrt{2},\sqrt{3}]$, which is equal to $F[\sqrt{3}]$ where $F=\mathbb{Q}[\sqrt{2}]$, which is a field. Then also $F[\sqrt{3}]$ is a field by the same reason. You can take $K=\mathbb{C}$, of course.

egreg
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Here is another take:

Let $\alpha \in \mathbb{Q}[\sqrt{2}, \sqrt{3}]$, $\alpha\ne0$, and consider $\phi: \mathbb{Q}[\sqrt{2}, \sqrt{3}] \to \mathbb{Q}[\sqrt{2}, \sqrt{3}]$ given by $\phi(x)=\alpha x$. Then $\phi$ is an injective $\mathbb{Q}$-linear transformation. Since $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ is a finite-dimensional vector space over $\mathbb{Q}$, the injectivity of $\phi$ implies its surjectivity. In particular, $1$ is in the image of $\phi$, which proves that $\alpha$ is invertible in $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$.

This argument works in general to prove that $ \mathbb{Q}[S]$ is a field when $S$ is a finite set of algebraic numbers.

lhf
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It comes from general facts. One way to do this is the following.

First note that $\sqrt{2}$ is algebraic over $\mathbb{Q}$, with minimal polynomial $m = x^{2} - 2$, which is irreducible in $\mathbb{Q}[x]$ for general reasons.

Then $K = \mathbb{Q}[\sqrt{2}]$ is a field. You can explicitly compute inverses in $K$. If $0 \ne f(\sqrt{2}) \in F$, for some $f \in \mathbb{Q}[x]$, then $(f, m) = 1$, so Euclid's algorithm yields $u, v \in \mathbb{Q}[x]$ such that $$ u f + v m = 1, $$ and now evaluate for $x = \sqrt{2}$ to get $$ u(\sqrt{2}) \cdot f(\sqrt{2}) = 1, $$ so $u(\sqrt{2})$ is the required inverse.

Now repeat with $\mathbb{Q}[\sqrt{2}, \sqrt{3}] = K[\sqrt{3}]$ to get that this is a field too.

Andreas Caranti
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The easy way to see this is to express an element of $\mathbb Q[\sqrt2,\sqrt3]$ as $A+B\sqrt3$, where A and B are members of $\mathbb Q[\sqrt2]$. Then $\dfrac{1}{A+B\sqrt3}=\dfrac{A-B\sqrt3}{A^2-3B^2}$. The denominator, $A^2-3B^2$, is a member of $\mathbb Q[\sqrt2]$, so you already know how to invert that.

Bill Kleinhans
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Do you remember proof of :

If $A$ is a nilpotent matrix the $I+A$ is invertibe...

what we do is somehow see that

$(I+A)(I-A+A^2-A^3\dots )=(I-A+A^2-A^3\dots )(I+A)=I$

and then, we say that : as $A$ is nilpotent, $A^n=0$ for some positive integer $n$

So, we end up with the case that $I-A+A^2-A^3\dots = I-A+A^2-A^3\dots +(-1)^nA^n$

and as $I-A+A^2-A^3\dots +(-1)^nA^n$ is "very well defined" than $I-A+A^2-A^3\dots$ to be a "MATRIX"

we say that $I+A$ is invertible.

I would suggest you to repeat the same process in case of $\mathbb{Q}[\sqrt{2},\sqrt{3}]$

please make sure there are no polynomials in $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ with degree more than $2$ (are you sure why???) as we "kill" coefficients of $x^2$ and above by using a result:

"$\sqrt{2},\sqrt{3}$ are algebraic with polynomials $x^2-2$ and $x^2-3$"

if you can intuitively see that now, then we can go further..

please make your self comfortable with :

given any "polynomial" i have an expression $g(x)$ such that $f(x)g(x)=g(x)f(x)=1$

but then $g(x)=\frac{1}{f(x)}$ need not be a polynomial.

But in our case, "Thanks to algebraic nature of $\sqrt{2},\sqrt{3}$" we end up with "killing" of higher powers.

Thus, any polynomial is "invertible" and thus we are done.