# Math Help - Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 15

1. ## Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 15

I am reading Nicholson: Introduction to Abstract Algebra, Section 6.2 - Algebraic Extensions.

Example 15 on page 282 (see attachment) reads as follows:

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Example 15.

Let $E = \mathbb{Q} ( \sqrt{2} , \sqrt{5} )$.

Find $[E \ : \ \mathbb{Q} ]$ , exhibit a $\mathbb{Q}$-basis of E, and show that $E = \mathbb{Q} ( \sqrt{2} + \sqrt{5} )$. Then find the minimum polynomial of $\sqrt{2} + \sqrt{5}$ over $\mathbb{Q}$.

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Solution: We write $L = \mathbb{Q} ( \sqrt{2} )$ for convenience so that $E = L(\sqrt{5})$ ... ... etc

... ... ... We claim that $X^2 - 5$ is the minimal polynomial of $\sqrt{5}$ over L. Because $\sqrt{5}$ and $- \sqrt{5}$ are the only roots of $X^2 - 5$ in $\mathbb{R}$, we merely need to show that $\sqrt{5} \notin L$. ... ... etc

My problem is the following:

How does showing $\sqrt{5} \notin L$ imply that $X^2 - 5$ is the minimal polynomial of $\sqrt{5}$ over L?

If $\sqrt{5} \in L$ then $X^2-5 = (X+\sqrt{5})(X-\sqrt{5})$, which would imply that the minimal polynomial would be $X-\sqrt{5}$. So, $\sqrt{5} \notin L$ implies $X-\sqrt{5}\notin L[X]$, which implies the minimal polynomial must be at least degree 2. Since $X^2-5 \in L[X]$ is a degree 2 polynomial with $\sqrt{5}$ a root, it is the minimal polynomial.