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Callum Dudley 2022-07-03 Answered
A B is a ring extension. Let y , z B elements which satisfy quadratic integral dependance y 2 + a y + b = 0 and z 2 + c z + d = 0 over A. Find explicit integral dependance relations for y + z and y z.
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Answers (1)

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 + a y + b = 0 y = a a 2 4 b 2  or  y = a + a 2 4 b 2 .
For a while choose y = a a 2 4 b 2 . Same for z 2 + c z + d = 0.
Then y + z = ( a + c ) ( a 2 4 b + c 2 4 d ) 2
( y + z ) + a + c 2 = a 2 4 b + c 2 4 d 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 [ 1 2 ] ) [ T ]. If 1 2 A then everything is finished without needing computing the exact relations but if you simplify it, no coefficient will have 1 2 .
The same method will work for x y but I have doubt if the second method give us a better monic relation than a relation at ( A [ 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 ) + ( a + c ) 2 4 = a 2 4 b + c 2 + 2 a 2 4 b c 2 4 d 4
( y + z ) 2 + ( a + c ) ( y + z ) + 2 a c 4 = b d + a 2 4 b c 2 4 d 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 ( a c 2 + b + d ) ( y + z ) 2 + 2 ( a + c ) ( a c 2 + b + d ) ( y + z ) + ( a c 2 + b + d ) 2 = ( a 2 4 b ) ( c 2 4 d ) 4
It's obvious that all coefficients will be in A and won't have 1 2 and we only need to pay attention to the constant coefficient.
( a 2 c 2 4 + b 2 + d 2 + a c b + a c d + 2 b d ) ( a 2 c 2 4 a 2 d b c 2 + 4 b d )
Now one can see the simplified form and for sure there is no fractions like 1 2 or 1 4 .
"For x y 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 + b + c + d = 0 do as following;
a + b + c + d = 0 a + b = ( c + d )
Take a square
a 2 + 2 a b + b = c + d + 2 c d
Then we are in case with two radicals;
a 2 + b c d = 2 b + 2 c d
Now let's be sure that we won't have coefficients with 1 2 and so on.
x y = a + a 2 4 b 2 c 2 + c 2 4 d 2 = a c + a c 2 4 d + c a 2 4 b + a 2 4 b c 2 4 d 4
x y a c 4 ( a 2 4 b ) ( c 2 4 d ) 4 = a 4 c 2 4 d + c 4 a 2 4 b
( x y ) 2 + a 2 c 2 16 + ( a 2 4 b ) ( c 2 4 d ) 16 a c 2 ( x y ) ( a 2 4 b ) ( c 2 4 d ) 2 ( ( x y ) a c 4 ) = a 2 16 ( c 2 4 d ) + c 2 16 ( a 2 4 b ) + a c 8 ( c 2 4 d ) ( a 2 4 b )
( x y ) 2 + a 2 c 2 16 + a 2 c 2 16 a 2 d 4 b c 2 4 + b d a c 2 ( x y ) a 2 c 2 16 + a 2 d 4 c 2 a 2 16 + b x 2 4 = ( a c 8 + 1 2 ( x y ) a c 8 ) ( c 2 4 d ) ( a 2 4 b )
( x y ) 2 a c 2 ( x y ) + b d = 1 2 ( x y ) ( c 2 4 d ) ( a 2 4 b )
( x y ) 4 a c ( x y ) 3 a 2 c 2 4 ( x y ) 2 + 2 b d ( x y ) 2 a c b d ( x y ) + b 2 d 2 = 1 4 ( x y ) 2 ( c 2 4 d ) ( a 2 4 b )
And in the end
( x y ) 4 a c ( x y ) 3 + ( c 2 b + d a 2 2 b d ) ( x y ) 2 a c b d ( x y ) + b 2 d 2 = 0
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