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
4 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.

Staff - 4 years ago

Nice bro! :D

- 4 years, 6 months ago