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Fulton's "Algebraic Curves" says the following:

Let $F$ and $G$ be polynomials belonging to $k[X,Y]$, where $k$ is a field. Let $F$ and $G$ not have a single common factor in $k[X][Y]$. Then they do not have a common factor in $k(X)[Y]$ either. As $k(X)[Y]$ is a PID, we have $(F,G)=1$.

What is $k(X)[Y]$, and how is it a PID? Also, is $k[X][Y]=k[X,Y]$?

freebird
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1 Answers1

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If $K$ is a field then $K[X]$ the ring of polynomials in $X$, is a Euclidean Domain. $K(X)$ however, is the field of rational 'functions' (polynomials) in $X$.

So for example $\frac{1}{X} \in K(X)$ but $\frac{1}{X}\not\in K[X]$, using $1$ as the multiplicative identity in $K$.

Now $K(X)$ is a field. Therefore, $K(X)[Y]$ is an E.D. and hence a P.I.D.

Also $K[X][Y]$ and $K[X,Y]$ can be shown to be equivalent (isomorphic).

$(K[X])[Y]$ is the set of polynomials in $Y$ with co-efficients in $K[X]$. So expressions of the form $\sum\limits_{i} a_i(x)y^i$.

$K[X,Y]$ is the set of polynomials in $X$ and $Y$ with coefficients in $K$.

e.g. of the form $\sum\limits_i \sum\limits_j a_{i,j}x^iy^j$.

These are finite sums of course. Only a finite number of the co-efficients are non-zero.

snulty
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