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A field F is algebraically closed if every non constant polynomial in F[x] has a root in F.

Is this the right definition? I am wondering if only one root in F and the rest of the roots not in F can also be considered 'algebraically closed'?

I remember reading somewhere that if a root is in F then all roots are in F. Is this a special case of something? If so, could someone enlighten me on what it is?

Thank you

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2 Answers

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1) The definition is right.

2) No. That not the same.

3) That's wrong. Consider $x(x^2+1)$ over the reals, which is not algebraically closed.

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Yes, every non-constant polynomial needs to have a root in the field; equivalently, it factors completely over the field. The second statement is not saying the same thing, because if only one root $\alpha$ is in $\mathrm{F}$, then we can factor out $(x-\alpha)$ to obtain a new (non-constant) polynomial, which must also have a root in $\mathrm{F}$.

The third statement is maybe something you read about quadratics; certainly if one root of a quadratic over $\mathrm{F}$ is in $\mathrm{F}$, then the other root is as well. But this statement is not true in general.

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