Let $R=k[x^4,x^3y,xy^3,y^4]$ be a polynomial ring. We can see $R$ is not integrally closed (since $x^2y^2 \in Q(R)$ is integral over $R$ but $x^2y^2 \notin R$ ). Therefore $k[x^4,x^3y,x^2y^2,xy^3,y^4]\subseteq\bar{R}$ (where $\bar{R}$ is integral closure). How to show the other side?
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If we can show that $A:= k[x^4,x^3y,x^2y^2,xy^3,y^4]$ is integrally closed, then we will of course be done. Since we call a domain "integrally closed" if it is integrally closed inside its field of fractions, we note first that the field of fractions of $A$ is just $K_A:= k(x^4,x^3y,x^2y^2,xy^3,y^4)$.
Now, that field of fractions clearly lies in the field $k(x,y)$. And of course, $A$ lies in the polynomial ring $k[x,y]$. So, if an element $f \in K_A$ is integral over $A$, then it is a root of a monic polynomial in $A[t]$. By what we have said above, $f$ can actually be thought to lie in $k(x,y)$ and the coefficients of the monic polynomial it satisfies can be thought to lie in $k[x,y]$. Hence, since $k(x,y)$ is the field of fractions of $k[x,y]$, $f$ must also be integral over $k[x,y]$.
It is well known that $k[x,y]$ is integrally closed, and so we therefore have that $f \in k[x,y]$. So in looking for the integral closure of $A$, we need only look in $k[x,y]\cap K_A$, the intersection taken inside $k(x,y)$ (i.e. we need only look for elements of the fraction field which are actually mere polynomials).
Now with $f \in k[x,y]\cap K_A$ integral over $A$ we suppose first that $f$ is homogeneous. Then by the fact that it is in $K_A$ its degree must be divisible by $4$. Since $A$ contains all degree 4 homogeneous polynomials, $f$ must be in $A$. (Note that this is the part of the argument that fails if $x^2y^2$, or any other degree 4 monomial, is missing from the ring).
If instead $f$ were not homogeneous then let $f = f_0 + \cdots + f_d$ be its decomposition into homogeneous components. Then, as can be checked, $f_d$ will satisfy a monic polynomial given by taking the top degree pieces of the monic polynomial that $f$ satisfies. So $f_d$ is also integral over $A$. By the argument above $f_d$ will then be in $A$. Since the integral closure of $A$ is certainly a ring, integrality of $f$ and $f_d$ means that $f-f_d = f_0 + \cdots + f_{d-1}$ will be integral over $A$ as well. By induction all the homogeneous pieces $f_0,\ldots,f_d$ will be integral over $A$, hence in $A$ by the argument above. Since the integral closure is a ring, we then finally have $f = f_0 + \cdots + f_d \in A$.
