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I came across an example in Stewart's Galois Theory wherein we have field $K$ and a simple extension $K\subset K(t)$.

In an example we are asked to find the degree $[\mathbb{Z}_5(t):\mathbb{Z}_5]$.

Initially I said okay $\mathbb{Z}_5(t)=\{a+ bt:a,b\in \mathbb{Z}_5\}$, and so a basis for the vector space $\mathbb{Z}_5(t)$ over $\mathbb{Z}_5$ is $\{1,t\}$ thus the degree $[\mathbb{Z}_5(t):\mathbb{Z}_5]$ is 2. However the answer is "infinite". I then thought that the set describing this field will be a generic combination of powers of $t$?

In another example, we are given the case when $K=\mathbb{Z}_2$ and are asked to describe the subfields of $K(t)$ of various forms, e.g. $K(t^2)$ and $K(t+1)$. I'm not even sure where to start with this!

To summarise my queries: Is there anything we can say about a generic extension $K\subset K(t)$? How do you describe the field whenever t is an unknown? How do we obtain that the degree is infinite? How can we go about describing the subfields of such a field (without applying actual Galois theory methods - these are introductory field theory exercises).

Thanks

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    $\begingroup$ $\mathbb{Z}_5(t)$ denotes the field of fractions over $\mathbb Z_5$. $\endgroup$ Apr 26 '20 at 14:39
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    $\begingroup$ The usual "degree" is the degree of an algebraic extension. In this case, we are dealing with an transcendental extension, since $t$ is not algebraic over $K$. $\endgroup$
    – rae306
    Apr 26 '20 at 14:42
  • $\begingroup$ How do you know $t$ isn't algebraic? Is it because it is an indeterminate? $\endgroup$
    – Natasha
    Apr 26 '20 at 15:15
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    $\begingroup$ @Natasha Yes... $\endgroup$
    – user26857
    Apr 26 '20 at 21:31
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    $\begingroup$ Degree is the vector space degree. For instance, you can't write $t^2$ as a $K$-linear combination of $1$ and $t$ so those elements don't give a $K$-basis, and hence the degree is definitely not $2$. You can see that by proceeding in this fashion the degree must be infinite. $\endgroup$
    – user208649
    Apr 27 '20 at 7:26

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