Is this definition of splitting field pleonastic?

Question

I found this definition of a splitting field but I am wondering if the second condition does not implies the first one.

If $L$ is generated over $K$ by the zeros of the polynomials of the family, does not it follows that such zeros belong to $L$ and so every $f_i$ splits into linear factors?

I don't think it's really redundant because "the zeros" might just refer to those zeros that happen to be in L, e.g. $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ is generated by the zeros of $(x^2+1)(x^2-2)$ which lie in $\mathbb{Q}(\sqrt{2})$. — Matthew Towers, Apr 28 '21 at 13:12

score 0 · Answer 1 · answered Apr 28 '21 at 13:14

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If (2) does not hold, there would be infinitely many "splitting fields". The condition (2) guarantees that L is the smallest field in which the function family factors into linear factors.

answered Apr 28 '21 at 13:14

Muses_China

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I asked if the first condition were superfluous. I think the second one it sufficente. – aleio1 Apr 28 '21 at 13:36

Is this definition of splitting field pleonastic?

1 Answers1