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Let $K$ be any subfield of $\mathbb{C}$ and let $m(t)$ be a quadratic polynomial over $K$. show that all zeros of $m(t)$ lie in an extension $K(\alpha)$ of $K$ where $\alpha^2=k\in K$. Thus allowing square roots $\sqrt k$ enables us to solve all quadratic equations over $K$.

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What have you tried? Do you have any thoughts? –  Ross Millikan Mar 30 '11 at 3:52

2 Answers 2

Obviously the quadratic formula works. But you might wonder why it works. Here's a nice way of understanding why. Suppose more generally that $\rm\:K\:$ is a field of characteristic $\ne 2\:$ (so that we can divide by $\:2\:$ below) and suppose that you have a calculator that can compute the field operations of $\rm\:K\:$ as well as square roots of elements in $\rm\:K\:$. Then I claim that given the sum $\rm\: b = r+s\:$ and product $\rm\:c = r\:s\:$ of any two elements $\rm\:r,s\in K\:,\:$ you can solve for $\rm\:r,s\:$ on your calculator. To do so it suffices to find their difference $\rm\:d = r-s\:$ since $\rm\ r = (r+s + r-s)/2 = (b+d)/2\ $ and $\rm\ s = (r+s)-r = b-r\:.\:$

But how can we find their difference $\rm\:d\:$? With a little help from symmetry. Note that $\rm\:d = r-s\ $ is almost symmetric, i.e. it merely changes sign when we swap $\rm\:r\:$ and $\rm\:s\:.\:$ We can eliminate this sign change by squaring it. Therefore, since $\rm\:d^2\:$ is a symmetric polynomial in $\rm\:(r,s)\:,\:$ by the fundamental theorem of symmetric polynomials, it can be expressed as a polynomial in the elementary symmetric polynomials on $\rm\:(r,s)\:,\:$ namely $\rm\ r+s,\ \ r\ s\:.\:$ While we could use Gauss's algorithm to find this polynomial, here it is rather obvious, namely $\rm\ (r+s)^2 - (r-s)^2 = \ 4\:r\:s\:,\ $ i.e. $\rm\ b^2 -d^2 = 4\:c\:,\ $ so $\rm\:d^2 = b^2-4\:c\:.\:$ Thus we can compute $\rm\:d\:$ by taking the square root of $\rm\:b^2-4\:c\:,\:$ and this yields $\rm\:r,s\:$ via the equations above. Therefore we can recover any two numbers from their sum and product - by way of a symmetry-derived equation relating the sum, product and difference.

This is the essence of the quadratic formula. Indeed, $\rm\:r,s\:$ are roots of $\rm\: (x-r)\ (x-s)\: =\: x^2 - b\ x + c\ $ with the discriminant $\rm\ b^2 - 4\:c\:,\ $ so the grade-school quadratic formula amounts to the same formula derived above. The advantage of the above viewpoint is that it serves to better reveal the innate symmetries -- something that will become much clearer when one studies Galois theory. For a taste see the section on Lagrange resolvents in the Wikipedia quadratic equation page.

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the above is of course true, but I also recall the existence of a 'purely algebraic' proof (in terms of field extensions etc.) for this fact. If anyone has any ideas, they'd be very welcome. –  Gerben Mar 30 '11 at 8:18
@Bruno: @Gerben: It's not simply Vieta's formulas. The point of my post is to encourage the reader to revisit the grade school proof of the quadratic formula from the more advanced viewpoint of Galois theory. –  Bill Dubuque Mar 30 '11 at 12:36
@Bill Dubuque: There are a number of Babylonian "word problems" that are equivalent to solving $x+y=a$, $xy=b$. The solutions (implicitly) use the identity $(x-y)^2+4xy=(x+y)^2$. –  André Nicolas Jul 3 '11 at 21:19

Hint: Use the quadratic formula.

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Thanks!!!!! no se porque no me hhabia dado cuenta!!!!! –  user8465 Mar 30 '11 at 4:10

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