Recently I have been reading about proving the algebraic integers form a ring, so let me get started about some simple definitions.
A field is called a number field if it contains $ \mathbf{Q} $.
A number if called an algebraic integer for a number field $ A $ if it can be written as a root of a monic polynomial in $ A $ with coefficient in $ \mathbf{Z} $.
The fact that the algebraic integer form a ring is surprising (to me), it is obvious that $ \sqrt{2} $ and $ \sqrt{3} $ are algebraic integer in $ Q(\sqrt{2}, \sqrt{3}) $, but why $ \sqrt{2} + \sqrt{3} $ an algebraic integer, or in other words, what is the monic polynomial that give root $ \sqrt{2} + \sqrt{3} $?
Turn out the key idea is this polynomial:
$ (x - \sqrt{2} - \sqrt{3})(x - \sqrt{2} + \sqrt{3})(x + \sqrt{2} - \sqrt{3})(x + \sqrt{2} + \sqrt{3}) $.
Obviously, this polynomial is monic and has $ \sqrt{2} + \sqrt{3} $ as a root, but is it in integer, the answer is YES! If we expand this thing, it becomes:
$ x^4 - 10x + 1 $
To verify that, we can put this in wolfram, and wolfram solve the solution as $ \pm\sqrt{5 \pm \sqrt{6}} $, does this value sound familiar. If you read my post about denesting the radical you can just recognize this is basically $ \pm \sqrt{2} \pm \sqrt{3} $
So we have some luck? No, turn out this trick will ALWAYS work because of the theory of symmetric polynomial.
Consider for a moment that we replace square root by variables instead, we have something like this.
$ (x - a_1 - b_1)(x - a_1 - b_2)(x - a_2 - b_1)(x - a_2 - b_2) $
Expanding this thing is messy, so we will not do that, the key idea is that we can consider this as a multi-variable polynomial in $ a_1 $ and $ a_2 $, and more importantly, it is symmetric, such that swapping $ a_1 $ and $ a_2 $ does not change the polynomial itself. Notice the $ x $ and the $ b $ are not going away, we are simply considering them as coefficients by now.
By the theorem of symmetric polynomial, we can rewrite this polynomial as a polynomial of the elementary symmetric polynomial, and viola, because the value of elementary symmetric polynomial is basically the coefficients of the minimal polynomial, they are integers. After substituting the equation with the value of $ a $, we get an expression with integer coefficients of $ X $ and $ b $.
We do the same thing to the $ b $ and finally we get an integer polynomial equation in $ X $ containing the root we wanted!
A field is called a number field if it contains $ \mathbf{Q} $.
A number if called an algebraic integer for a number field $ A $ if it can be written as a root of a monic polynomial in $ A $ with coefficient in $ \mathbf{Z} $.
The fact that the algebraic integer form a ring is surprising (to me), it is obvious that $ \sqrt{2} $ and $ \sqrt{3} $ are algebraic integer in $ Q(\sqrt{2}, \sqrt{3}) $, but why $ \sqrt{2} + \sqrt{3} $ an algebraic integer, or in other words, what is the monic polynomial that give root $ \sqrt{2} + \sqrt{3} $?
Turn out the key idea is this polynomial:
$ (x - \sqrt{2} - \sqrt{3})(x - \sqrt{2} + \sqrt{3})(x + \sqrt{2} - \sqrt{3})(x + \sqrt{2} + \sqrt{3}) $.
Obviously, this polynomial is monic and has $ \sqrt{2} + \sqrt{3} $ as a root, but is it in integer, the answer is YES! If we expand this thing, it becomes:
$ x^4 - 10x + 1 $
To verify that, we can put this in wolfram, and wolfram solve the solution as $ \pm\sqrt{5 \pm \sqrt{6}} $, does this value sound familiar. If you read my post about denesting the radical you can just recognize this is basically $ \pm \sqrt{2} \pm \sqrt{3} $
So we have some luck? No, turn out this trick will ALWAYS work because of the theory of symmetric polynomial.
Consider for a moment that we replace square root by variables instead, we have something like this.
$ (x - a_1 - b_1)(x - a_1 - b_2)(x - a_2 - b_1)(x - a_2 - b_2) $
Expanding this thing is messy, so we will not do that, the key idea is that we can consider this as a multi-variable polynomial in $ a_1 $ and $ a_2 $, and more importantly, it is symmetric, such that swapping $ a_1 $ and $ a_2 $ does not change the polynomial itself. Notice the $ x $ and the $ b $ are not going away, we are simply considering them as coefficients by now.
By the theorem of symmetric polynomial, we can rewrite this polynomial as a polynomial of the elementary symmetric polynomial, and viola, because the value of elementary symmetric polynomial is basically the coefficients of the minimal polynomial, they are integers. After substituting the equation with the value of $ a $, we get an expression with integer coefficients of $ X $ and $ b $.
We do the same thing to the $ b $ and finally we get an integer polynomial equation in $ X $ containing the root we wanted!
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