An Algebraist's View: Generalizing Vieta's Formula

It turns out that Vieta's formula applies to any algebraically closed field. Call a field FF algebraically closed if(f) every polynomial in F[x]F[x] splits over FF.

Theorem: C\mathbb{C} is algebraically closed.

Proof: It suffices to show that there is not a proper finite field extension of C\mathbb{C}. This is a healthy exercise in Galois theory, and is thus left to the reader. (Hint: Suppose, by way of contradiction, that L:CL:\mathbb{C} is a proper finite extension of C\mathbb{C}. What can we say about [L:C][L:\mathbb{C}]?) \boxed{ }

Thus, we see that Vieta's formula works in at least one algebraically closed field. Let's broaden the scope.

Theorem: Suppose

p(x)=0knλkxkF[x]\displaystyle p(x)=\sum_{0\leq k\leq n}\lambda_kx^k \in F[x],

where FF is an algebraically closed field, and that p(x)p(x) splits as

p(x)=λn1in(xαi)\displaystyle p(x)=\lambda_n\prod_{1\leq i\leq n}(x-\alpha_i).

Then, 1i1<i2<<imnαi1αi2αim=(1)mλnmλn\displaystyle \sum_{1\leq i_1<i_2<\cdots<i_m\leq n}\alpha_{i_1}\alpha_{i_2}\cdots\alpha_{i_m}=(-1)^m\frac{\lambda_{n-m}}{\lambda_n}.

Proof: A field is necessarily commutative and distributes over addition. Thus, the proof is simply a matter of noting the coefficients must match up. \boxed{ }

Thus, we can use Vieta's formula for arbitrary polynomials over fields.

[Edit: Sorry for the awful formatting. I'm used to writing things in pure LaTeX, so this is a little weird to me.]

Note by Jacob Erickson
6 years, 9 months ago

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Nice bro! :D

Finn Hulse - 6 years, 5 months ago

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@Jacob Erickson Can you add this to a suitable skill in the Vieta Formula Wiki? Let me know if you think a different skill would be suitable.

Calvin Lin Staff - 5 years, 11 months ago

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