For a field extension

of the field

, with a transcendental element

over

, my textbook says the following:

"

contains the field of fractions of

, which is the smallest subfield containing

and

, we denote this field by

."

It also says that

is defined to be the image of the evaluation homomorphism

. I think this corresponds to the usual notation of meaning polynomials in

, with coefficients form

, right?

Right
Anyway, my first question is how do we know that

is the smallest field containing

and

?

Let be any field containing (why? Transcendence of over is important here)
Next, in an example it says that

is transendental in

, the field

is isomorphic to the field

of rational functions over

in the indeterminate

.

This is very confusing. How does it conclude this?

by the definition should equal the fraction field of

. If

is a field

It can't possibly be a field: we know that if is any field and is an element that belongs to some extension field of , then is a field iff is algebraic over , and if this is the case then we have that
then we can conclude that its fraction field is itself, and by the evaluation homomorphism

above (with codomain restricted to its image, making it an isomorphism), we do see

, which is almost the result, I think

does actually equal

Impossible. Read above
, but I'm not sure why, can anyone explain? Also, I don't see how

is a field, in fact I think it's only a integral domain.

So can someone explain how this example works?

Define ; it's easy to check this is a well-defined homomorphism of rings (why?) , and since it isn't trivial and are fields it automatically is 1-1 and onto
Also, in the exercise section it asks, for a indeterminate

whether the statement:

is true or false. It should be similar to the above example, can someone explain this too?