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?
$\begingroup$
$\endgroup$
asked Aug 15, 2018 at 7:33
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
3
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.
-
$\begingroup$ Oh I see! He also used the notation $\Bbb Q(x)$ for the field of fractions {$\frac{p(x)}{q(x)}:q(x)≠0$}. I realize now that it indeed is the smallest field containing $\Bbb Q$ and $x$. $\endgroup$ Commented Aug 15, 2018 at 7:51
-
1$\begingroup$ @HritRoy Exactly. Now because you can take any $\frac{f(\sqrt 2)}{g(\sqrt2)}$ and expand it to make the denominator a rational number, and therefore make the whole thing into a polynomial, we have $\Bbb Q(\sqrt 2)\cong \Bbb Q[\sqrt 2]$. Some like to use this isomorphism to write $\Bbb Q[\sqrt 2]$ when they really mean $\Bbb Q(\sqrt 2)$. $\endgroup$– ArthurCommented Aug 15, 2018 at 7:54
-
$\begingroup$ Oh! Now I see why we use the same notation for both in case of $√2$. $\endgroup$ Commented Aug 15, 2018 at 7:59
$\begingroup$
$\endgroup$
6
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 $\mathbb{Q}[\alpha]\subsetneq\mathbb{Q}(\alpha)$
-
$\begingroup$ While defining $\Bbb Q[\alpha]$ you used $\Bbb Q[X]$. Are the two defined differently then? $\endgroup$ Commented Aug 15, 2018 at 8:01
-
$\begingroup$ $\mathbb{Q}[X]$ has no relation between $X$ because $X$ is a "variable" (sometimes also called indeterminate). On the other hand $\alpha$ is a number and there may be "algebraic relations" such as $\alpha^2-2=0$ for $\alpha=\sqrt{2}$, whilst a variable does not have such relations. $\endgroup$– darumaCommented Aug 15, 2018 at 9:20
-
$\begingroup$ So we define $\Bbb Q[X]$ in the usual way as the ring of all polynomials in the indeterminate $X$ having rational coefficients and in case of $\Bbb Q[\alpha]$ we make use of any algebraic relation about $\alpha$ that we know of. $\endgroup$ Commented Aug 15, 2018 at 9:31
-
$\begingroup$ One thing I'd like to ask. You wrote the elements of $\Bbb Q[X]$ as simply $f$ and the elements of $\Bbb Q[\alpha]$ as $f(\alpha)$. Was there any particular reason behind this? $\endgroup$ Commented Aug 15, 2018 at 9:35
-
1$\begingroup$ To answer your first question, yes, $\mathbb{Q}[\alpha]$ means just plug in $\alpha$ into all polynomials in $\mathbb{Q}[X]$ and consider it as some subset of the reals. To answer your second question, you say $f=f(X)=X^2-2$ (it's just slopiness in notation, really) and $f(\alpha)$ just means "plug the real number $\alpha$ in your polynomial. $\endgroup$– darumaCommented Aug 15, 2018 at 9:39
$\begingroup$
$\endgroup$
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.
$\begingroup$
$\endgroup$
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 $K$.
answered Aug 15, 2018 at 7:46