Meaning of $\Bbb Q[√2]$

Question

I know that $\Bbb Q[x]$ is the set of all polynomials in $x$ with rational coefficients. And that $\Bbb Q[√2]$ is the smallest field containing $\Bbb Q$ and $√2$. But the other day our professor denoted by $\Bbb Q[π]$ the set of all polynomials in $π$ with rational coefficients. So what does $\Bbb Q[√2]$ mean? I see that either definition for $\Bbb Q[√2]$ leads to the same set {$a+b√2:a,b\in \Bbb Q$}. But the same cannot be said for $\Bbb Q[π]$. Could someone explain it to me?

Answer 1

You have been somewhat lied to. $\Bbb Q[\sqrt 2]$ is not, a priori, the smallest field containing $\Bbb Q$ and $\sqrt 2$. Its definition is the ring of polynomials in the variable $\sqrt 2$ (which has the property that $\sqrt 2^2 = 2$) with rational coefficients. It happens to be a field, but that's coincidental. The field is denoted as $\Bbb Q(\sqrt 2)$.

So $\Bbb Q[x]$ and $\Bbb Q[\pi]$ are rings of polynomials, while $\Bbb Q(x)$ and $\Bbb Q(\pi)$ are their respective fields of fractions, i.e. the fields of rational functions with rational coefficients.

Answer 2

Usually, $K(\alpha)$ denotes the FIELD generated by $K$ and $\alpha$, and $K[\alpha]$ denotes the RING generated by $K$ and $\alpha$.

If $K\leq L$ is a field extension and $\alpha\in L$ is algebraic over $K$, then $K[\alpha]= K(\alpha)$. However, if $\alpha$ is transcendental. then $K[\alpha]\neq K(\alpha)$. The former is obtained by substituting $\alpha$ into every polynomial over $K$, the latter is obtained by substituting $\alpha$ into every rational function over $K$.

Answer 3

The difference between the two is due to the fact that $\sqrt{2}$ is an algebraic number and $\pi$ is a trascendental number.

Indeed $\mathbb{Q}[\sqrt{2}]$ is also the ring of polynomials in $\sqrt{2}$ with rational coefficients, but of course degrees $2$ and higher "collapse" because for example $\sqrt{2}^3=2\sqrt{2}$ which is so to say of degree $1$. For $\pi$ this is never the case.

Answer 4

Let $\alpha$ be a real number.

$\mathbb{Q}[\alpha]$ is the polynomial ring genereated by $\alpha$. In other words, $\mathbb{Q}[\alpha]=\{f(\alpha):f\in\mathbb{Q}[X]\}$

On the other hand,

$\mathbb{Q}(\alpha)$ is the field genereated by $\alpha$. In other words, $\mathbb{Q}(\alpha)=\{f(\alpha):f\in\mathbb{Q}(X)\}$

(The difference is that $\mathbb{Q}[X]$ is the polynomial ring and $\mathbb{Q}(X)$ is the field of rational functions so you can think of it as $\mathbb{Q}(X)=Frac\mathbb{Q}[X]$

The point is that if $\alpha$ is algebraic $\mathbb{Q}[\alpha]=\mathbb{Q}(\alpha)$. So in your case since, $\sqrt{2}$ is algebraic, it is equivalent to think of it as the ring genereated by $\sqrt{2}$ or the field genereated by $\sqrt{2}$.

However, if $\alpha$ is not algebraic (i.e. $\pi$ is not a solution to a polynomial with rational coefficients), then $\mathbb{Q}[\alpha]\subsetneq\mathbb{Q}(\alpha)$