5

which of the following is/are true?

  1. $\sin 7^\circ$ is an algebraic over $\mathbb{Q}$

  2. $\sin^{-1}(1)$ is algebraic over $\mathbb{Q}$

  3. $\cos (\pi/7)$ is algebraic over $\mathbb{Q}$

  4. $\sqrt{2}+\sqrt{\pi} $ is algebraic over $\mathbb{Q}(\pi)$

an algebraic number is a number that is a root of a non-zero polynomial in one variable with rational coefficients, I am perfectly sure that option $1$, $3$ are algebraic number. could any one help me to find out others option?

Zhen Lin
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Myshkin
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1 Answers1

3

For option 2:

Evaluate $\sin^{-1}(1)$. It's easy

For $4$:

$\sqrt{2}+\sqrt{\pi}$ satisfies is a root of the equation:

$$(x^2-2-\pi)^2-2\pi=0$$ $$x^4-2(2+\pi)x^2+(\pi+2)^2-2\pi=0$$ $$x^4-(4+2\pi)x^2+\pi^2+2\pi+4=0$$

Hence $\sqrt{2}+\sqrt{\pi}$ is algebraic over $\mathbb{Q}(\pi)$

Here I will include the proofs of 2 useful facts:

Fact 1: Let $E$ be an extension field of a field $F$. Let $u\in E$. Then $u$ is algebraic over $F$ iff $F(u)$ is finite dimesnsional over $F$.

Proof:

Forward direction: Consider the elements $1,u,u^2,...u^{[F(u):F]}$. If these $[F(u):F]+1$ elements were linearly independent over $F$, then the dimension of $F(u)$ over $F$ must be greater than or equal to $[F(u):F]+1$ (contradiction). Thus, we conclude that they are linearly dependent. Hence there exists $a_0,a_1,....,a_{[F(u):F]}\in F$ such that not all of thm are zero and: $$a_{[F(u):F]}u^{[F(u):F]}+a_{[F(u):F]-1}u^{[F(u):F]-1}+...+a_1u+a_0=0$$ Hence $u$ is algebraic over $F$.

Backward direction: We start by showing that $F(u)=F[u]$. Clearly, $F[u]\subseteq F(u)$. Since $F(u)$ is the intersection of all subfields that contain $F\cup\{u\}$., therefore $F(u)\subseteq F[u]$ if $F[u]$. Now we verify that $F[u]$ is a field. Consider the ring homomorphism $\phi: F[x]\rightarrow F(u)$ that sends $p(x)$ to $p(u)$. It can be shown that $\phi$ is surjective. Hence $F[u]\cong F[x]/Ker\phi$. $Ker\phi=\{p(x)\in F[x]|p(u)=0\}$. Since $F$ is a field, therefore $F[x]$ is a PID. Since $Ker\phi$ is an ideal of $F[x]$, therefore $Ker\phi=(f(x))$ for some $f(x)\in F[x]$. It can be easily verified that $(f(u))$ is a prime ideal of $F[x]$. Since $F[x]$ is a PID, therefore $Ker\phi=(f(x))$ is a maximal ideal of $F[x]$. Hence $F[u]\cong F[x]/Ker\phi$ is a field. Thus $F(u)=F[u]$. Verify that $\{1,u,u^2,...,u^{deg(f)-1}\}$ is a basis for $F[u]$ over $F$. Hence, $F(u)=F[u]$ is a finite dimensional vector space over $F$.

Fact 2: Let $E$ be an extension field of a field $F$. Let $K:=\{x\in E|x$ is algebraic over$F\}$. Then $K$ is a field.

Proof: Let $x,y$ be algebraic over $F$. By fact 1, we deduce that $F(x)$ is finite dimensional over $F$. Since $y$ is algebraic over $F$, therefore $y$ is algebraic over $F(x)$. Thus, by fact 1 we deduce that $F(x)(y)$ is finite dimensional over $F(x)$. Since $F(x)$ is finite dimensional over $F$. Hence $F(x)(y)$ is finite dimensional over $F$. Since $x+y\in F(x)(y)$, therefore $F(x+y)$ is a subfield of $F(x)(y)$. Thus, $F(x+y)$ is finite dimensional over $F$. By fact 1, we deduce that $x+y$ is algebraic over $F$. Hence it is an element of $K$. The rest of the field axioms can be verified for $K$ similarly.

Since $\sqrt{2}$ is algebraic over $Q$, therefore it is algebraic over $Q[\pi]$. Similarly, $\sqrt{\pi}$ is algebraic over $\mathbb{Q}[\pi]$ (as it is the root of the equation $x^2-\pi=0$). Thus, we know that $\sqrt{2}+\sqrt{\pi}$ is algebraic over $\mathbb{Q}[\pi]$ by fact 2.

Amr
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