Proof that $\kappa(A[t])=\kappa(A)(t)$

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Let $A$ be an integrally closed ring. Is the fraction field of $A[t]$ just $\kappa(A)(t)$, where this denotes the fraction field $(A-\{0\})^{-1}A$ field extended by $t$.

Why I think this is so: $\kappa(A[t])$ easily includes into $\kappa(A)(t)$ (because the former consists of elements of the form $f/g$ where $g\ne 0$ and $f,g\in A[t]$, and the latter consists of rational functions in $t$ with coefficients in $A\subset \kappa(A)$), where the latter is by definition the smallest field generated by $\kappa(A)$ and $t$, and hence they are equal.

Is that a sound argument?

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I think your argument is fine. Here is a more explicit approach:

Elements of $\kappa(A)(t)$ look like $$ \frac{\sum\frac{a_i}{b_i}t^i}{\sum\frac{c_j}{d_j}t^j} $$ and clearing denominators gives us the following, with $B_i=\prod_{k\neq i}b_k$ and similarly for $D_j$:

$$ \frac{\prod d_i \sum a_iB_i t^i}{\prod b_i \sum c_j D_j t^j} $$

which is clearly in $\kappa(A[t])$.