Quick question.
An author defines a normal extension $L:K$ as one in which every irreducible polynomial over $K$ that has at least one zero in $L$ splits over $L$.
Is this definition equivalent to: a normal extension $L:K$ is one in which each irreducible polynomial over $K$ has all its zeroes in $L$?
Which also seems to be equivalent to: a normal extension $L:K$ is that in which $L$ contains all the zeroes of every polynomial over $K$.
Is definition 1 = to definition 2 = definition 3?
On
Clearly not. For example, $L = \mathbb{Q}(i)$ is a normal extension since $x^2+1$ has a root in $L$ and splits there, but $x^3+2$ does not.
What is true is the following: Call $L\supset K$ a splitting field over $K$ if there is some polynomial $f$ such that (1) $f$ factors completely in $L$, and (2) $L = K(u_1, \dotsc, u_n)$ where the $u_i$ are all the roots of $f$.
Then
Theorem: $L$ is a splitting field over $K$ if and only if whenever an irreducible polynomial over $K$ has a root in $L$, it factors completely in $L$.
Not quite. The difference between definition 1 and definitions 2 and 3 is that definition 1 only requires polynomials with at least one root in $L$ to split over $L$.
As an example, consider the normal extension $\mathbb Q[\sqrt{2}]/\mathbb Q$. Clearly, there are polynomials with coefficients in $\mathbb Q$ that have no roots in $\mathbb Q[\sqrt{2}]$, such as $x^2+1$. However, every irreducible polynomial over $\mathbb Q$ that has a zero in $\mathbb Q[\sqrt{2}]$ splits. An example of such a polynomial is $x^2-2$.
If we instead look at the extension $\mathbb Q[\sqrt[3]{2}]/\mathbb Q$, we find that this is not normal. This is because the polynomial $x^3-2$, irreducible over $\mathbb Q$, has a root in $\mathbb Q[\sqrt[3]{2}]$, namely $\sqrt[3]{2}$, but it doesn't split. It factors as $(x-\sqrt[3]{2})(x^2+\sqrt[3]{2}x+\sqrt[3]{4})$, and the second term is irreducible.