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An Algebraist's View: Generalizing Vieta's Formula

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

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

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

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

Theorem: Suppose

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

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

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

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

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
2 years, 10 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 · 2 years ago

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Nice bro! :D Finn Hulse · 2 years, 6 months ago

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