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A problem in Dummit and Foote states:

Let $k$ be a field and let $k(x)$ be the field of rational functions in $x$ with coefficients from $k$. Let $t \in k(x)$ be the rational function $\frac{P(x)}{Q(x)}$ with relatively prime polynomials $P(x)$, $Q(x) \in k[x]$, with $Q(x) \neq 0$. Then $k(x)$ is an extension of $k(t)$ and to compute its degree it is necessary to compute the minimal polynomial with coefficients in $k(t)$ satisfied by $x$.

By $k(t)$, do they mean $k$ adjoin $t$, i.e. the set of polynomials $k_0 + k_1\frac{P(x)}{Q(x)} + ... + k_n(\frac{P(x)}{Q(x)})^n$? Or do they mean the set of rational functions in $t$, e.g. $\frac{\frac{P(x)}{Q(x)} + 1}{2(\frac{P(x)}{Q(x)})^2 + 3}$?

Thanks!

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By $k(t)$, they mean the set of rational functions in $t$ (the second option). These form a field. The other, which might be denoted $k[t]$, do not form a field.

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  • $\begingroup$ Thanks! But why is $\mathbb{Q}$ adjoin $\sqrt{2}$ denoted $\mathbb{Q}(\sqrt{2})$ instead of $\mathbb{Q}[\sqrt{2}]$? $\endgroup$
    – badatmath
    Commented Apr 21, 2011 at 16:09
  • $\begingroup$ It happens that $\mathbb{Q}[\sqrt{2}] = \mathbb{Q}(\sqrt{2})$. Standard convention is to right the latter rather than the former, but either notation would be correct. $\endgroup$
    – Charles Staats
    Commented Apr 21, 2011 at 16:17
  • $\begingroup$ Please substitute "write" for "right" in the comment above. $\endgroup$
    – Charles Staats
    Commented Apr 22, 2011 at 3:30

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