$F(x)$ versus $F[x]$ where $F$ is a field

Question

What is the difference, if there is any, between $F(x)$ and $F[x]$, where $F$ is a field and $x$ is an element of a field extension $E/F$ that is not in $F$? I have some guesses, but I was never sure about this. For example, for one specific usage in a textbook, I see $\mathbb{Q}[i]$ and $\mathbb{Q}(\pi)$. And maybe a more general question: what happens when $F$ is just a ring? For example, for the ring of Gaussian integers, the notation I usually see is $\mathbb{Z}[i]$, with brackets instead of open parentheses. I guess I never properly learned the notation.

Answer 1

Short answer: given a ring $A$ and an element $x$, the notation $A[x]$ denotes the smallest ring that contains both $A$ and $x$. If $A$ happens to be a field, then $A(x)$ denotes the smallest field that contains both $A$ and $x$. That's all.

Longer answer: given a ring $A$, we denote by $A[x]$ the ring of polynomials in the variable $x$ with coefficients in $A$. Starting from the description in the above paragraph, if a ring contains $x$, it must also contain $x^2,x^3$, and so on. And sums of these. And the results of multiplying these with one another and with the elements of $A$. Without any more context, these are just polynomials the way you're used to them.

If $A\subseteq B$ and $x\in B$, then $x$ inherits any relations it may satisfy in $B$ when used in this polynomial ring. One example is the ring of Gaussian integers $\Bbb Z[i]$, where $i^2=-1$. If $x$ satisfies no relations that can be described using only elements of $A$, then it acts like in the previous paragraph, and you get regular polynomials with no special rules. Your $\Bbb Q[\pi]$ is an example.

If $F$ is a field, then $F(x)$ is the smallest field that contains both $F$ and $x$. Starting from the description in the first paragraph, this field must for the same reason contain all polynomials in the variable $x$ with coefficients in $F$. But it must also contain the reciprocal of any non-zero polynomial. Same things as in the above paragraph apply: If $x$ satisfies some algebraic relation that can be phrased using only coefficients from $F$ (we call this "$x$ is algebraic over $F$"), then that's baked into $F(x)$. Your $\Bbb Q(i)$ is an example of that, where $i^2=-1$ is the algebraic relation in question.

When $x$ is not algebraic over $F$, then $F(x)$ becomes the field of rational functions in the variable $x$ with coefficients in $F$. Which is to say, the elements of $F(x)$ are fractions where the numerators and the denominators are polynomials in $F[x]$. One concrete example is $\Bbb Q(\pi)$. But the fact that $\pi$ lies inside a well-known extension of $\Bbb Q$ is unnecessary. $\Bbb Q(x)$ works just as well, where $x$ is just an abstract variable.

If $x$ is algebraic over $F$, then $F[x]$ and $F(x)$ turn out to be isomorphic: fractions in $F(x)$ can always be expanded so that the denominators have no $x$ in them. Thus you will see either one used, which can be a point of confusion. For instance, $\Bbb Q(i)$ and $\Bbb Q[i]$ are, for all practical intents and purposes, the same thing.

Answer 2

Well, $F[x]$ means ring adjunction of $x$, while $F(x)$ means field adjunction of $x$. More concretely,

$F(x) = \left\{\frac{f(x)}{g(x)}\mid f(x),g(x)\in F[x], g(x)\ne 0\right\}$ is the field of quotients of $F[x]$.

$\Bbb Q(i) = \{a+ib\mid a,b\in\Bbb Q\}$ with $i^2=-1$ is the field adjunction of $i$ to $\Bbb Q$. It's a field and the same as the ring adjunction $\Bbb Q[i]$ (as its also a field).

$\Bbb Z[i] = \{a+ib\mid a,b\in\Bbb Z\}$ with $i^2=-1$ is the ring adjungation of $i$ to $\Bbb Q$. Its's a ring.

Beware of $\Bbb Q(\pi)$. As $\pi$ is tranzendential over $\Bbb Q$, this field extension of $\Bbb Q$ is infinite.

You need to learn about ring and field adjunction.

Answer 3

$F[x]$ is the ring of all polynomials in $x$ with coefficients from $F$. In contrast, $F(x)$ is the smallest field which extends $F$ and contains $x$: in other words, it contains all rational functions (i.e. quotients of polynomials) in $x$ with coefficients from $F$, rather than just all polynomials. Note that sometimes these can be the same, e.g. $\mathbb{Q}\left[\sqrt{2}\right]$ is the same as $\mathbb{Q}\left(\sqrt{2}\right)$, since any polynomial in $\sqrt{2}$ can be written in the form $a+b\sqrt{2}$, and if $a,b,c,d \in \mathbb{Q}$ then $$\frac{a+b\sqrt{2}}{c+d\sqrt{2}} = \frac{ac-2bd}{c^2-2d^2} + \frac{bc-ad}{c^2-2d^2}$$ where the coefficients are still in $\mathbb{Q}$.