Callum Dudley
2022-07-03
Answered

$A\subset B$ 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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ladaroh

Answered 2022-07-04
Author has **11** answers

Follow the idea of the next exercise or you may want to start with writing

${y}^{2}+ay+b=0\u27f9y=\frac{-a-\sqrt{{a}^{2}-4b}}{2}\text{or}y=\frac{-a+\sqrt{{a}^{2}-4b}}{2}.$

For a while choose $y=\frac{-a-\sqrt{{a}^{2}-4b}}{2}$. Same for ${z}^{2}+cz+d=0$.

Then $y+z=\frac{-(a+c)-(\sqrt{{a}^{2}-4b}+\sqrt{{c}^{2}-4d})}{2}$

$\u27f9(y+z)+\frac{a+c}{2}=-\frac{\sqrt{{a}^{2}-4b}+\sqrt{{c}^{2}-4d}}{2}$

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 $(A[\frac{1}{2}])[T]$. 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 $(A[\frac{1}{2}])[T]$, you may want to compute it to see.

After making first square you will have;

$(y+z{)}^{2}+(a+c)(y+z)+\frac{(a+c{)}^{2}}{4}=\frac{{a}^{2}-4b+{c}^{2}+2\sqrt{{a}^{2}-4b}\sqrt{{c}^{2}-4d}}{4}$

$(y+z{)}^{2}+(a+c)(y+z)+\frac{2ac}{4}=-b-d+\frac{\sqrt{{a}^{2}-4b}\sqrt{{c}^{2}-4d}}{2}$

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.

$(y+z{)}^{4}+2(a+c)(y+z{)}^{3}+2(\frac{ac}{2}+b+d)(y+z{)}^{2}+2(a+c)(\frac{ac}{2}+b+d)(y+z)+(\frac{ac}{2}+b+d{)}^{2}=\frac{({a}^{2}-4b)({c}^{2}-4d)}{4}$

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.

$(\frac{{a}^{2}{c}^{2}}{4}+{b}^{2}+{d}^{2}+acb+acd+2bd)-(\frac{{a}^{2}{c}^{2}}{4}-{a}^{2}d-b{c}^{2}+4bd)$

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;

$a+\sqrt{b}+\sqrt{c}+\sqrt{d}=0\u27f9a+\sqrt{b}=-(\sqrt{c}+\sqrt{d})$

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.

$xy=\frac{a+\sqrt{{a}^{2}-4b}}{2}\frac{{c}^{2}+\sqrt{{c}^{2}-4d}}{2}=\frac{ac+a\sqrt{{c}^{2}-4d}+c\sqrt{{a}^{2}-4b}+\sqrt{{a}^{2}-4b}\sqrt{{c}^{2}-4d}}{4}$

$xy-\frac{ac}{4}-\frac{\sqrt{({a}^{2}-4b)({c}^{2}-4d)}}{4}=\frac{a}{4}\sqrt{{c}^{2}-4d}+\frac{c}{4}\sqrt{{a}^{2}-4b}$

$(xy{)}^{2}+\frac{{a}^{2}{c}^{2}}{16}+\frac{({a}^{2}-4b)({c}^{2}-4d)}{16}-\frac{ac}{2}(xy)-\frac{({a}^{2}-4b)({c}^{2}-4d)}{2}((xy)-\frac{ac}{4})=\frac{{a}^{2}}{16}({c}^{2}-4d)+\frac{{c}^{2}}{16}({a}^{2}-4b)+\frac{ac}{8}\sqrt{({c}^{2}-4d)({a}^{2}-4b)}$

$(xy{)}^{2}+\frac{{a}^{2}{c}^{2}}{16}+\frac{{a}^{2}{c}^{2}}{16}-\frac{{a}^{2}d}{4}-\frac{b{c}^{2}}{4}+bd-\frac{ac}{2}(xy)-\frac{{a}^{2}{c}^{2}}{16}+\frac{{a}^{2}d}{4}-\frac{{c}^{2}{a}^{2}}{16}+\frac{b{x}^{2}}{4}=(\frac{ac}{8}+\frac{1}{2}(xy)-\frac{ac}{8})\sqrt{({c}^{2}-4d)({a}^{2}-4b)}$

$(xy{)}^{2}-\frac{ac}{2}(xy)+bd=\frac{1}{2}(xy)\sqrt{({c}^{2}-4d)({a}^{2}-4b)}$

$(xy{)}^{4}-ac(xy{)}^{3}-\frac{{a}^{2}{c}^{2}}{4}(xy{)}^{2}+2bd(xy{)}^{2}-acbd(xy)+{b}^{2}{d}^{2}=\frac{1}{4}(xy{)}^{2}({c}^{2}-4d)({a}^{2}-4b)$

And in the end

$(xy{)}^{4}-ac(xy{)}^{3}+({c}^{2}b+d{a}^{2}-2bd)(xy{)}^{2}-acbd(xy)+{b}^{2}{d}^{2}=0$

${y}^{2}+ay+b=0\u27f9y=\frac{-a-\sqrt{{a}^{2}-4b}}{2}\text{or}y=\frac{-a+\sqrt{{a}^{2}-4b}}{2}.$

For a while choose $y=\frac{-a-\sqrt{{a}^{2}-4b}}{2}$. Same for ${z}^{2}+cz+d=0$.

Then $y+z=\frac{-(a+c)-(\sqrt{{a}^{2}-4b}+\sqrt{{c}^{2}-4d})}{2}$

$\u27f9(y+z)+\frac{a+c}{2}=-\frac{\sqrt{{a}^{2}-4b}+\sqrt{{c}^{2}-4d}}{2}$

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 $(A[\frac{1}{2}])[T]$. 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 $(A[\frac{1}{2}])[T]$, you may want to compute it to see.

After making first square you will have;

$(y+z{)}^{2}+(a+c)(y+z)+\frac{(a+c{)}^{2}}{4}=\frac{{a}^{2}-4b+{c}^{2}+2\sqrt{{a}^{2}-4b}\sqrt{{c}^{2}-4d}}{4}$

$(y+z{)}^{2}+(a+c)(y+z)+\frac{2ac}{4}=-b-d+\frac{\sqrt{{a}^{2}-4b}\sqrt{{c}^{2}-4d}}{2}$

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.

$(y+z{)}^{4}+2(a+c)(y+z{)}^{3}+2(\frac{ac}{2}+b+d)(y+z{)}^{2}+2(a+c)(\frac{ac}{2}+b+d)(y+z)+(\frac{ac}{2}+b+d{)}^{2}=\frac{({a}^{2}-4b)({c}^{2}-4d)}{4}$

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.

$(\frac{{a}^{2}{c}^{2}}{4}+{b}^{2}+{d}^{2}+acb+acd+2bd)-(\frac{{a}^{2}{c}^{2}}{4}-{a}^{2}d-b{c}^{2}+4bd)$

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;

$a+\sqrt{b}+\sqrt{c}+\sqrt{d}=0\u27f9a+\sqrt{b}=-(\sqrt{c}+\sqrt{d})$

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.

$xy=\frac{a+\sqrt{{a}^{2}-4b}}{2}\frac{{c}^{2}+\sqrt{{c}^{2}-4d}}{2}=\frac{ac+a\sqrt{{c}^{2}-4d}+c\sqrt{{a}^{2}-4b}+\sqrt{{a}^{2}-4b}\sqrt{{c}^{2}-4d}}{4}$

$xy-\frac{ac}{4}-\frac{\sqrt{({a}^{2}-4b)({c}^{2}-4d)}}{4}=\frac{a}{4}\sqrt{{c}^{2}-4d}+\frac{c}{4}\sqrt{{a}^{2}-4b}$

$(xy{)}^{2}+\frac{{a}^{2}{c}^{2}}{16}+\frac{({a}^{2}-4b)({c}^{2}-4d)}{16}-\frac{ac}{2}(xy)-\frac{({a}^{2}-4b)({c}^{2}-4d)}{2}((xy)-\frac{ac}{4})=\frac{{a}^{2}}{16}({c}^{2}-4d)+\frac{{c}^{2}}{16}({a}^{2}-4b)+\frac{ac}{8}\sqrt{({c}^{2}-4d)({a}^{2}-4b)}$

$(xy{)}^{2}+\frac{{a}^{2}{c}^{2}}{16}+\frac{{a}^{2}{c}^{2}}{16}-\frac{{a}^{2}d}{4}-\frac{b{c}^{2}}{4}+bd-\frac{ac}{2}(xy)-\frac{{a}^{2}{c}^{2}}{16}+\frac{{a}^{2}d}{4}-\frac{{c}^{2}{a}^{2}}{16}+\frac{b{x}^{2}}{4}=(\frac{ac}{8}+\frac{1}{2}(xy)-\frac{ac}{8})\sqrt{({c}^{2}-4d)({a}^{2}-4b)}$

$(xy{)}^{2}-\frac{ac}{2}(xy)+bd=\frac{1}{2}(xy)\sqrt{({c}^{2}-4d)({a}^{2}-4b)}$

$(xy{)}^{4}-ac(xy{)}^{3}-\frac{{a}^{2}{c}^{2}}{4}(xy{)}^{2}+2bd(xy{)}^{2}-acbd(xy)+{b}^{2}{d}^{2}=\frac{1}{4}(xy{)}^{2}({c}^{2}-4d)({a}^{2}-4b)$

And in the end

$(xy{)}^{4}-ac(xy{)}^{3}+({c}^{2}b+d{a}^{2}-2bd)(xy{)}^{2}-acbd(xy)+{b}^{2}{d}^{2}=0$

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