Splitting Fields. Prove Q(alpha^4)=Q(alpha)

In summary: K] = [M:L][L:K] = [M(\alpha):M][M:K] = [M(\alpha):K]$. This means that $M$ is a splitting field for $g$ over $K$. Since $g$ is irreducible over $K$, this means that $M = K(\alpha)$.In summary, we have shown that $M = K(\alpha)$, and since $\alpha \in M$, we have $K(\alpha) \subseteq M$. Since $M \subseteq K(\alpha)$, we must have $M = K(\alpha)$. Therefore, $\mathbb{Q}(\alpha^4) = K(\alpha) = \mathbb{
  • #1
caffeinemachine
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Let $f(x)\in \mathbb Q[x]$ be an irreducible polynomial of degree $n\geq 3$. Let $L$ be the splitting field of $f$, and let $\alpha\in L$ be a zero of $f$. Given that $[L:\mathbb Q]=n!$, prove that $\mathbb Q(\alpha^4)=\mathbb Q(\alpha)$.
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Attempt:

Lemma:
Let $F$ be any field and $f,g \in F[x]$. Let $K$ and $L$ be splitting fields for $f$ and $g$ over $F$ respectively. Write $h=fg\in F[x]$. Let $E$ be the splitting field for $h$ over $F$. Then $[E:F]\leq [K:F][L:F]$

Proof: Let $f=(x-\alpha_1)\cdots(x-\alpha_k)$ in $K[x]$ and $g=(x-\beta_1)\cdots(x-\beta_l)$ in $L[x]$. Its easy to see that $K=F(\alpha_1,\ldots,\alpha_k)$, $L=F(\beta_1,\ldots,\beta_l)$ and $E=F(\alpha_1,\ldots,\alpha_k,\beta_1,\ldots,\beta_l)$. From here its easy to conclude that \begin{equation*}[E:K]\leq [L:F]\tag{1}\end{equation*}Now, $[E:F]=[E:K][K:F]\leq [K:F][L:F]$. Hence the lemma is settled.$\blacksquare$

Corollary: Let $F$ be any field and $f\in F[x]$ with deg $f=n$. Let $E$ be the spliting field for $f$ over $F$. Assume that $[E:F]=n!$. Then $f$ is irreducible over $F$.

Assuming there are no mistakes in my argument..
In the light of the corollary I think the question redundantly confers irreducibility to $f$.

Now. Let $f=(x-\alpha_1)\cdots (x-\alpha_n)=(x-\alpha_1)g$ be the representation of $f$ in $L[x]$. Define $F_1=\mathbb Q(\alpha_1)$. Then $L$ is the splitting field for $g$ over $F_1$ satisfying $[L:F_1]=(n-1)!=\text{ deg }g$ and hence $g$ is irreducible over $F_1$. This piece of information suggests that induction might be helpful. I can settle the base case, that is, $n=3$. If you require then I can post the proof for the base case in my following posts.
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Can anybody see how to settle the question?
 
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  • #2

Let $[L: \mathbb{Q}] = n!$ and let $K = \mathbb{Q}(\alpha)$. Then from the given information, we have $[K:\mathbb{Q}] = n$. Since $\alpha$ is a zero of the irreducible polynomial $f$, we know that $K$ is a splitting field for $f$ over $\mathbb{Q}$. By the fundamental theorem of Galois theory, we have $[L:K] = [L:\mathbb{Q}] / [K:\mathbb{Q}] = (n!)/n = (n-1)!$. This means that $L$ is a splitting field for $f$ over $K$ with degree $(n-1)!$. Since $f$ is irreducible over $K$, we must have $f = (x-\alpha)g$ for some irreducible polynomial $g \in K[x]$ with degree $n-1$. This implies that $g$ is the minimal polynomial for $\alpha$ over $K$.
Now, let $M = K(\alpha^4)$. We want to show that $M = K(\alpha)$. Since $\alpha \in L$, we have $M \subseteq L$. Also, since $\alpha^4 \in K$, we have $K \subseteq M$. Therefore, we have $K \subseteq M \subseteq L$. We want to show that $[M:K] = 1$, which will imply that $M = K(\alpha)$.
Since $M$ is a finite extension of $K$, we have $[M:K] = [M:L][L:K]$. Since $L$ is a splitting field for $g$ over $K$, we know that $g$ splits completely in $L$. This means that $g$ has $n-1$ distinct roots in $L$. Since $g$ has degree $n-1$, this means that $g$ has all of its roots in $L$. Therefore, $g$ splits completely in $L$ and has all of its roots in $M$. This implies that $g$ is the minimal polynomial for $\alpha$ over $M$. Since $g$ has degree $n-1$, we have $[M(\alpha):M] = n-1$.
Now, we have $[M:
 

FAQ: Splitting Fields. Prove Q(alpha^4)=Q(alpha)

1. What is a splitting field?

A splitting field is a field extension of a given field that contains all the roots of a given polynomial. It is the smallest field that contains all the roots of the polynomial.

2. How do you prove that Q(alpha^4)=Q(alpha)?

To prove that Q(alpha^4)=Q(alpha), we need to show that every element in Q(alpha^4) is in Q(alpha) and vice versa. This can be done by showing that alpha^4 is a root of every polynomial in Q(alpha) and that alpha is a root of every polynomial in Q(alpha^4).

3. What is the significance of Q(alpha^4)=Q(alpha)?

This equation means that the field Q(alpha) is generated by alpha^4. This allows us to simplify calculations involving alpha by using its fourth power instead. It also helps us to understand the structure and properties of the field Q(alpha).

4. Can you provide an example of Q(alpha^4)=Q(alpha)?

One example is the polynomial f(x) = x^3 - 2 over Q. The splitting field of this polynomial is Q(alpha), where alpha is the real cube root of 2. In this case, Q(alpha^4)=Q(alpha) because alpha^4 is a root of f(x) and every polynomial in Q(alpha) can be expressed as a polynomial in alpha^4.

5. How is the concept of splitting fields used in mathematics and science?

Splitting fields are used in many areas of mathematics and science, including algebra, number theory, and physics. They are essential for solving polynomial equations and studying their roots. In algebra, splitting fields help us to understand the structure and properties of fields. In number theory, they are used to study the behavior of prime numbers. In physics, splitting fields are used to model and understand complex systems and phenomena.

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