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$A\subset B$ is a ring extension. Let $y,z\in B$ elements which satisfy quadratic integral dependance ${y}^{2}+ay+b=0$ and ${z}^{2}+cz+d=0$ over $A$. Find explicit integral dependance relations for $y+z$ and $yz$.
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For a while choose $y=\frac{-a-\sqrt{{a}^{2}-4b}}{2}$. Same for ${z}^{2}+cz+d=0$.
Then $y+z=\frac{-\left(a+c\right)-\left(\sqrt{{a}^{2}-4b}+\sqrt{{c}^{2}-4d}\right)}{2}$
$⟹\left(y+z\right)+\frac{a+c}{2}=-\frac{\sqrt{{a}^{2}-4b}+\sqrt{{c}^{2}-4d}}{2}$
By first, squaring both sides you will earn a monic relation for $y+z$ but it has (only) one sqrt so after squaring it for second time you will have a monic relation with no sqrt at $\left(A\left[\frac{1}{2}\right]\right)\left[T\right]$. If $\frac{1}{2}\in A$ then everything is finished without needing computing the exact relations but if you simplify it, no coefficient will have $\frac{1}{2}$.
The same method will work for $xy$ but I have doubt if the second method give us a better monic relation than a relation at $\left(A\left[\frac{1}{2}\right]\right)\left[T\right]$, you may want to compute it to see.
After making first square you will have;
$\left(y+z{\right)}^{2}+\left(a+c\right)\left(y+z\right)+\frac{\left(a+c{\right)}^{2}}{4}=\frac{{a}^{2}-4b+{c}^{2}+2\sqrt{{a}^{2}-4b}\sqrt{{c}^{2}-4d}}{4}$
$\left(y+z{\right)}^{2}+\left(a+c\right)\left(y+z\right)+\frac{2ac}{4}=-b-d+\frac{\sqrt{{a}^{2}-4b}\sqrt{{c}^{2}-4d}}{2}$
Now we go for the second squaring, but as we want to get rid of radicals we take $-b-d$ to the left hand side and then we square sides.
$\left(y+z{\right)}^{4}+2\left(a+c\right)\left(y+z{\right)}^{3}+2\left(\frac{ac}{2}+b+d\right)\left(y+z{\right)}^{2}+2\left(a+c\right)\left(\frac{ac}{2}+b+d\right)\left(y+z\right)+\left(\frac{ac}{2}+b+d{\right)}^{2}=\frac{\left({a}^{2}-4b\right)\left({c}^{2}-4d\right)}{4}$
It's obvious that all coefficients will be in $A$ and won't have $\frac{1}{2}$ and we only need to pay attention to the constant coefficient.
$\left(\frac{{a}^{2}{c}^{2}}{4}+{b}^{2}+{d}^{2}+acb+acd+2bd\right)-\left(\frac{{a}^{2}{c}^{2}}{4}-{a}^{2}d-b{c}^{2}+4bd\right)$
Now one can see the simplified form and for sure there is no fractions like $\frac{1}{2}$ or $\frac{1}{4}$.
"For $xy$ we will encounter 3 radicals! What can we do?"
Don't be afraid! 3 radicals is not a very scay case yet. For getting rid of 3 radicals in an equation like $a+\sqrt{b}+\sqrt{c}+\sqrt{d}=0$ do as following;
$a+\sqrt{b}+\sqrt{c}+\sqrt{d}=0⟹a+\sqrt{b}=-\left(\sqrt{c}+\sqrt{d}\right)$
Take a square
${a}^{2}+2a\sqrt{b}+b=c+d+2\sqrt{cd}$
Then we are in case with two radicals;
${a}^{2}+b-c-d=2\sqrt{b}+2\sqrt{cd}$
Now let's be sure that we won't have coefficients with $\frac{1}{2}$ and so on.
$xy=\frac{a+\sqrt{{a}^{2}-4b}}{2}\frac{{c}^{2}+\sqrt{{c}^{2}-4d}}{2}=\frac{ac+a\sqrt{{c}^{2}-4d}+c\sqrt{{a}^{2}-4b}+\sqrt{{a}^{2}-4b}\sqrt{{c}^{2}-4d}}{4}$
$xy-\frac{ac}{4}-\frac{\sqrt{\left({a}^{2}-4b\right)\left({c}^{2}-4d\right)}}{4}=\frac{a}{4}\sqrt{{c}^{2}-4d}+\frac{c}{4}\sqrt{{a}^{2}-4b}$
$\left(xy{\right)}^{2}+\frac{{a}^{2}{c}^{2}}{16}+\frac{\left({a}^{2}-4b\right)\left({c}^{2}-4d\right)}{16}-\frac{ac}{2}\left(xy\right)-\frac{\left({a}^{2}-4b\right)\left({c}^{2}-4d\right)}{2}\left(\left(xy\right)-\frac{ac}{4}\right)=\frac{{a}^{2}}{16}\left({c}^{2}-4d\right)+\frac{{c}^{2}}{16}\left({a}^{2}-4b\right)+\frac{ac}{8}\sqrt{\left({c}^{2}-4d\right)\left({a}^{2}-4b\right)}$
$\left(xy{\right)}^{2}+\frac{{a}^{2}{c}^{2}}{16}+\frac{{a}^{2}{c}^{2}}{16}-\frac{{a}^{2}d}{4}-\frac{b{c}^{2}}{4}+bd-\frac{ac}{2}\left(xy\right)-\frac{{a}^{2}{c}^{2}}{16}+\frac{{a}^{2}d}{4}-\frac{{c}^{2}{a}^{2}}{16}+\frac{b{x}^{2}}{4}=\left(\frac{ac}{8}+\frac{1}{2}\left(xy\right)-\frac{ac}{8}\right)\sqrt{\left({c}^{2}-4d\right)\left({a}^{2}-4b\right)}$
$\left(xy{\right)}^{2}-\frac{ac}{2}\left(xy\right)+bd=\frac{1}{2}\left(xy\right)\sqrt{\left({c}^{2}-4d\right)\left({a}^{2}-4b\right)}$
$\left(xy{\right)}^{4}-ac\left(xy{\right)}^{3}-\frac{{a}^{2}{c}^{2}}{4}\left(xy{\right)}^{2}+2bd\left(xy{\right)}^{2}-acbd\left(xy\right)+{b}^{2}{d}^{2}=\frac{1}{4}\left(xy{\right)}^{2}\left({c}^{2}-4d\right)\left({a}^{2}-4b\right)$
And in the end
$\left(xy{\right)}^{4}-ac\left(xy{\right)}^{3}+\left({c}^{2}b+d{a}^{2}-2bd\right)\left(xy{\right)}^{2}-acbd\left(xy\right)+{b}^{2}{d}^{2}=0$