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I am reading Section 6.4: Splitting Fields in Beachy and Blair: Abstract Algebra.

I am currently studying Example 6.4.2 on page 290 which concerns the splitting field of \(\displaystyle x^3 - 2 \text{ over } \mathbb{Q} \) .

In Example 6.4.2, B&B show that the splitting field of \(\displaystyle x^3 - 2 \text{ over } \mathbb{Q} \) is \(\displaystyle \mathbb{Q} ( \sqrt[3]{2} , \omega ) \) where \(\displaystyle \omega \) is any complex number such that \(\displaystyle \omega^3 = 1 \).

Then we read the following:

"We claim that the degree of the splitting field over \(\displaystyle \mathbb{Q}\) is 6. Since \(\displaystyle x^3 - 2 \) is irreducible over \(\displaystyle \mathbb{Q} \), we have \(\displaystyle [ \mathbb{Q} ( \sqrt[3]{2} \ : \ \mathbb{Q} ] = 3 \). The polynomial \(\displaystyle x^2 + x + 1 \) is irreducible over \(\displaystyle \mathbb{Q} \) and stays irreducible over \(\displaystyle \mathbb{Q} ( \sqrt[3]{2} ) \) since it has no root in that field. The degree of \(\displaystyle x^2 + x + 1 \) is not a divisor of \(\displaystyle [ \mathbb{Q} ( \sqrt[3]{2} \ : \ \mathbb{Q} ] \). Since \(\displaystyle \omega \) is a root of \(\displaystyle x^2 + x + 1 \), this implies that \(\displaystyle [ \mathbb{Q} ( \sqrt[3]{2} , \omega ) \ : \ \mathbb{Q} ( \sqrt[3]{2} ) ] = 2\)."

My question is as follows:

Why do B&B make the point that the degree of \(\displaystyle x^2 + x + 1 \) is not a divisor of \(\displaystyle [ \mathbb{Q} ( \sqrt[3]{2}) \ : \ \mathbb{Q} ] \)?

Why is this important to check before you apply the result that \(\displaystyle [ \mathbb{Q} ( \sqrt[3]{2}, \omega) \ : \ \mathbb{Q} ] = [ \mathbb{Q} ( \sqrt[3]{2}, \omega) \ : \ \mathbb{Q} ( \sqrt[3]{2}) ] [ \mathbb{Q} ( \sqrt[3]{2}) \ : \ \mathbb{Q} ] \)?

I would appreciate some help regarding this issue.

Peter

I am currently studying Example 6.4.2 on page 290 which concerns the splitting field of \(\displaystyle x^3 - 2 \text{ over } \mathbb{Q} \) .

In Example 6.4.2, B&B show that the splitting field of \(\displaystyle x^3 - 2 \text{ over } \mathbb{Q} \) is \(\displaystyle \mathbb{Q} ( \sqrt[3]{2} , \omega ) \) where \(\displaystyle \omega \) is any complex number such that \(\displaystyle \omega^3 = 1 \).

Then we read the following:

"We claim that the degree of the splitting field over \(\displaystyle \mathbb{Q}\) is 6. Since \(\displaystyle x^3 - 2 \) is irreducible over \(\displaystyle \mathbb{Q} \), we have \(\displaystyle [ \mathbb{Q} ( \sqrt[3]{2} \ : \ \mathbb{Q} ] = 3 \). The polynomial \(\displaystyle x^2 + x + 1 \) is irreducible over \(\displaystyle \mathbb{Q} \) and stays irreducible over \(\displaystyle \mathbb{Q} ( \sqrt[3]{2} ) \) since it has no root in that field. The degree of \(\displaystyle x^2 + x + 1 \) is not a divisor of \(\displaystyle [ \mathbb{Q} ( \sqrt[3]{2} \ : \ \mathbb{Q} ] \). Since \(\displaystyle \omega \) is a root of \(\displaystyle x^2 + x + 1 \), this implies that \(\displaystyle [ \mathbb{Q} ( \sqrt[3]{2} , \omega ) \ : \ \mathbb{Q} ( \sqrt[3]{2} ) ] = 2\)."

My question is as follows:

Why do B&B make the point that the degree of \(\displaystyle x^2 + x + 1 \) is not a divisor of \(\displaystyle [ \mathbb{Q} ( \sqrt[3]{2}) \ : \ \mathbb{Q} ] \)?

Why is this important to check before you apply the result that \(\displaystyle [ \mathbb{Q} ( \sqrt[3]{2}, \omega) \ : \ \mathbb{Q} ] = [ \mathbb{Q} ( \sqrt[3]{2}, \omega) \ : \ \mathbb{Q} ( \sqrt[3]{2}) ] [ \mathbb{Q} ( \sqrt[3]{2}) \ : \ \mathbb{Q} ] \)?

I would appreciate some help regarding this issue.

Peter

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