- #1

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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?

- Thread starter NoDoubts
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- #1

- 20

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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?

- #2

- 2,238

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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

- #3

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

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

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

- #4

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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??

- #5

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

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