# Subfields and degree of $[K(t):K]$ for a given field $K$?

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

• $\mathbb{Z}_5(t)$ denotes the field of fractions over $\mathbb Z_5$. Apr 26 '20 at 14:39
• 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 $K$. Apr 26 '20 at 14:42
• How do you know $t$ isn't algebraic? Is it because it is an indeterminate? Apr 26 '20 at 15:15
• @Natasha Yes... Apr 26 '20 at 21:31
• 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.
– user208649
Apr 27 '20 at 7:26