$\mathbb{F}_q(t)$ vs $\mathbb{F}_q[t]$?

Question

I am having certain troubles understanding certain manuscript. What is the difference between $$ \mathbb{F}_q(t) \quad \text{ and } \quad \mathbb{F}_q[t]? $$

Answer 1

For any field $\Bbb K$, $\Bbb K[t]$ is the algebra of polynomials in the indeterminate $t$, having coefficients in the field $\Bbb K$:

$\Bbb K[t] = \left \{ \displaystyle \sum_0^n a_i t^i, \; a_i \in \Bbb K, \; 0 \le n \in \Bbb Z \right \}; \tag 1$

we note, as is well-known, that $\Bbb K[t]$ is an integral domain; that is, it has no zero-divisors:

$\forall a(t), b(t) \in \Bbb K[t], \; a(t) \ne 0 \ne b(t) \Longrightarrow a(t)b(t) \ne 0; \tag 2$

this follows easily from the fact that the product of the leading terms of

$a(t) = \displaystyle \sum_0^{\deg a} a_i t^i, \; b(t) = \displaystyle \sum_0^{\deg b} b_i t^i, \tag 3$

which is

$a_{\deg a} b_{\deg b} t^{\deg a + \deg b} \ne 0, \tag 4$

since

$a_{\deg a} \ne 0 \ne b_{\deg b}, \tag 5$

by the definition of "leading term". In fact, it is also well-known that $\Bbb K[t]$ is a principal ideal domain, but this need not concern us further here.

Since $\Bbb K[t]$ is an integral domain, it is possessed of well-defined field of fractions $\mathcal{Frac}(\Bbb K[t])$, which prima facie may be taken to be the set of quotients of elements of $\Bbb K[t]$:

$\mathcal{Frac}(\Bbb K[t]) = \left \{ \dfrac{a(t)}{b(t)}, \; a(t), 0 \ne b(t) \in \Bbb K[t] \right \}; \tag 6$

$\mathcal{Frac}(\Bbb K[t])$ is of course easily recognized as the field of rational functions in the indeterminate $t$, with coefficients in the field $\Bbb K$.

In the light of these remarks, we consider the field $\Bbb K(t)$, the smallest field, in the sense of set inclusion, containing both $\Bbb K$ and $t$, that is,

$\Bbb K \subset \Bbb K(t), \; t \in \Bbb K(t), \tag 7$

and therefore, since $\Bbb K(t)$ is closed under the algebraic operations addition and multiplication,

$\Bbb K[t] \subset \Bbb K(t); \tag 8$

furthermore, since $\Bbb K(t)$ is a field, it must also contain the quotients of elements of $K[t]$, that is,

$\mathcal{Frac}(\Bbb K[t]) \subset K(t) \tag 9$

as well. Now since $K(t)$ is by definition the smallest field $K[t]$, it follows that in fact

$\mathcal{Frac}(\Bbb K[t]) = K(t), \tag{10}$

and thus $K(t)$ is in fact the field of rational functions in $t$ over $\Bbb K$. We now see that the difference 'twixt $\Bbb K[t]$ and $\Bbb K(t)$ is concisely expressed by (10).