If you understood the above this last must be easier now to attack.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?
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?