Vtyjkyuk

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 $a+b√2:a,bin Bbb Q$. But the same cannot be said for $Bbb Q[π]$. Could someone explain it to me?

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.

Let $alpha$ be a real number.

$mathbbQ[alpha]$ is the polynomial ring genereated by $alpha$. In other words, $mathbbQ[alpha]=f(alpha):finmathbbQ[X]$

On the other hand,

$mathbbQ(alpha)$ is the field genereated by $alpha$. In other words, $mathbbQ(alpha)=f(alpha):finmathbbQ(X)$

(The difference is that $mathbbQ[X]$ is the polynomial ring and $mathbbQ(X)$ is the field of rational functions so you can think of it as $mathbbQ(X)=FracmathbbQ[X]$

The point is that if $alpha$ is algebraic $mathbbQ[alpha]=mathbbQ(alpha)$. So in your case since, $sqrt2$ is algebraic, it is equivalent to think of it as the ring genereated by $sqrt2$ or the field genereated by $sqrt2$.

However, if $alpha$ is not algebraic (i.e. $pi$ is not a solution to a polynomial with rational coefficients), then $mathbbQ[alpha]subsetneqmathbbQ(alpha)$

The difference between the two is due to the fact that $sqrt2$ is an algebraic number and $pi$ is a trascendental number.

Indeed $mathbbQ[sqrt2]$ is also the ring of polynomials in $sqrt2$ with rational coefficients, but of course degrees $2$ and higher "collapse" because for example $sqrt2^3=2sqrt2$ which is so to say of degree $1$. For $pi$ this is never the case.

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

If $Kleq L$ is a field extension and $alphain L$ 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 $K$.