u,v are integral over R, then so is uv

Question

Let $R′ ⊃ R$ be $R$ algebra. Assume $u, v$ ∈ $R′$ satisfy the polynomial relation $X^2 − aX + b = 0$ and $X^2 − cX + d = 0$ respectively. Show that $uv$ is integral over $R$ as well and compute the polynomial relation satisfied. I know $uv$ is integral over $R$ because the integral closure forms a ring. But I have no clue whatsoever how to find the explicit polynomial relation satisfied by $uv$ over $R$. Can anyone suggest anything? I have a feeling theory of symmetric functions might come into picture. But I am not sure.

Answer 1

Well, you can just modify the standard proof to give you a polynomial. In particular, consider the module generated by $1,u,v,uv$. Multiplying by $uv$ and simplifying gives you a 4 terms which are linear combinations of $1,u,v,uv$. I.e $1\times uv=uv$, $u \times uv=-bv+auv$ etc. this can be turned into a matrix (first row is $0,0,0,1$ to represent $1 \times uv$, second is $0,0,-b,a$ to represent $uv^2=-bv+auv$ etc.). This gives the below matrix:

\begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -b & a \\ 0 & -d & 0 & c \\ bd & -ad & -bc & ac \end{bmatrix}

The characteristic polynomial of this matrix is then the desired degree 4 polynomial:

$$X^4 - a c X^3 + (a^2 d + c^2 b - 2 b d) X^2 - a b c d X + b^2 d^2$$

Answer 2

There is really not much to work with here. You are given that $$u^2-au+b=0\qquad\text{ and }\qquad v^2-cv+d=0,$$ and asked to find a polynomial that has $w:=uv$ as a zero. The two relations above already give $$w^2-avw+bv^2=v^2(u^2-au+b)=0,\tag{1}$$ $$w^2-cuw+du^2=u^2(v^2-cv+d)=0,\tag{2}$$ but also give $$u^2=au-b\qquad\text{ and }\qquad v^2=cv-d,$$ and hence plugging this into $(1)$ and $(2)$ respectively yields $$w^2-bd=v\big(aw-bc\big),$$ $$w^2-bd=u\big(cw-ad\big).$$ Taking products then shows that $$(w^2-bd)^2=w(aw-bc)(cw-ad),$$ which shows that $w=uv$ satisfies the quartic relation $$w^4-acw^3+(a^2d+bc^2-2bd)w^2-abcdw+b^2d^2=0.$$

Answer 3

Yes, elementary symmetric functions can be used. Let the first polynomial have roots $u$ and $u'$ and the second polynomial have roots $v$ and $v'$, so $$ a = u+u', b = uu', c = v + v', d = vv'. $$ [Edit: These expressions show $a$ and $b$ are symmetric polynomials in $u$ and $u$', while $c$ and $d$ are symmetric polynomials in $v$ and $v'$.]

Thinking symmetrically, consider the polynomial with roots $uv$, $uv'$, $u'v$, and $u'v'$, where we replace $u$ or $v$ in the product with $u'$ or $v'$ in all possible ways. This means we expand out $$ (X - uv)(X-uv')(X-u'v)(X-u'v'). $$ We can take advantage of the known symmetries between $u$ and $u'$ and between $v$ and $v'$ by multiplying these linear polynomials carefully: don't multiply all the polynomials together at the same time, but instead multiply the first two together and the last two together to get two quadratics. Then use the relations linking $a$, $b$, $c$, and $d$ to the roots in order to simplify how each intermediate quadratic looks. Then multiply the quadratics together and use symmetry again: you will see terms like $u^2 + u'^2$, which is symmetric in $u$ and $u'$ and write that as a polynomial in $u+u'$ and $uu'$.

As an example to check your work, when the initial polynomials are $X^2 - 2X + 3$ and $X^2 - 4X + 5$, $uv$ is a root of $$ X^4 - 8X^3 + 38X^2 - 120X + 225. $$