Re: Fields and transcendentals



Richard Hayden wrote:
Hi,

It seems intuitively obvious to me that if t is some transcendental
over K, then K(t) (the smallest field containing both K and t) is
isomorphic to the field of rational functions over K.

What would be a formal proof of this?

Thanks,

Richard Hayden.


Isn't it trivial? Take what's intuitively obvious and run
with it:

Construct the polynomial algebra K[x], and the canonical
ring homomorphism from K[x] to K[t], the smallest ring
containing K and t (map x to t). Then show that if this
homomorphism fails to be an isomorphism it's because t is
the root of some polynomial P(x) in K[x].

That is, the homomorphism is *guaranteed* to be a surjection,
since K[t] is formed by the inclusion of all polynomials in t,
with coefficients in K; the only thing that could fail is that
there be some elements of K[x] that map to 0. Look at the
ideal that maps to 0, and find a polynomial having t as a root.

Finally, if K[x] is isomorphic to K[t], then their quotient
fields are isomorphic.

Dale.
.

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