Viewing Polynomials from k[x,y] as Elements of k[y][x]

[NoDoubts]

"Let f is a polynomial from k[x,y], where k is a field. Suppose that x appears in f with positive degree. We view f as an element of k(y)[x], that is polynomial in x whose coefficients are rational functions of y."

I think I am missing something...why do we need rational functions here? can't we represent any polynomial from k[x,y] as an element of k[y][x] i.e. polynomial in x whose coefficients are polynomials of y?

 

[jambaugh]

I'm guessing here given I don't know from where you are getting this but...

Recall that a polynomial is also a rational function just as an integer is also a rational number.

I think the idea is to keep the format of k(y)[x] as a ring of polynomials (in x) over a field. So the y-polynomial coefficients are treated as elements of the larger ring of rational functions of y.

 

[LorenzoMath]

Yes, k[X,Y]=k[X][Y]=k[Y][X].

k[X,Y] is NOT equal to k(Y)[X].

 

[NoDoubts]

yes, later on it says that "if f (polynomial from k[x,y]) is irreducible then it remains irreducible as an element of k(y)[x]"

can someone explain to me why it is so?

...I guess, if we assume f = gh where g,h are from k(y)[x] then we can multiply both sides by the product of common denominators of coefficients of g and h, So that we get f c(y) = g_1 h_1, where c(y), g_1, and h_1 are polynomials from k[x,y].

Now, how does this imply that f is not irreducible??

 

[NoDoubts]

hmmm...OK. looks like Herr Gauss has solved it long time ago )

http://books.google.com.hk/books?id...a=X&oi=book_result&resnum=4&ct=result#PPA8,M1