# 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.

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
5 years, 11 months ago

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Nice bro! :D

- 5 years, 6 months ago

@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 - 5 years, 1 month ago