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I come across following ring

$F[x,y] $ and $F(x)[y]$ where $F$ is a field.

I think both are same initially . But as $xy$ is irreducible in $F(x)[y]$ but reducible in $F[x,y] $.

Which is a bigger ring than another?

Please give me a reference where I can read more about the above type of ring.

Any Help will be appreciated.

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$F[x,y]$ are polynomials with variables $x,y$ and coefficients in $F$, while $F(x)[y]$ are polynomials with variable $y$ with coefficients from the field of the rational functions over $F$, noted $F(x)$.

The difference is that $F[x,y]=(F[x])[y]$, meaning that these are polynomials in $y$ with coefficients being polynomials in $F[x]$, but in $F(x)[y]$ the coefficients are not only polynomials in $F[x]$, but all rational functions $p/q$ with $p,q \in F[x], q\neq0$.

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We have $F[X,Y]=F[X][Y]$, but not $F[X,Y]=F(X)[Y]$, because $F(X)$ is the fraction field of the integral domain $F[X]$, which is "bigger".

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$F [x,y] $ is the polynomial ring in the variables $x $ and $y $ with coefficients in $F $.

$F (x) $ is the quotient field of the polynomial ring $F [x] $,that is, $F (x)$ contains polynomials of the form $\dfrac {f (x)}{g (x)} $, where $f (x),g (x)\in F [x] ,g (x)\neq 0$.

Now $F (x)[y] $ is the polynomial ring in the variable $y $ with coefficients in $F (x) $.

Let $f (y)=\dfrac{y}{x^2} \in F (x)[y]$. Does $f (y)$ belong to $F [x,y]$?

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