# Thread: 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 $\displaystyle E = \mathbb{Q} ( \sqrt{2} , \sqrt{5} )$.

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

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In the solution we read:

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

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

My problem is the following:

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

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