abstract algebra - $[L_1 L_2 : k] = [L_1 : k] [L_2 : k]$ for two finite field extensions $L_1/k$ and $L_2/k$ with $L_1 cap L_2 = k$

I want to prove or disprove the following:

Let $L_1$ and $L_2$ be two finite extensions of field $k$ inside of an

extension $L/k$. Moreover, $L_1 \cap L_2 = k$. Then the degree of the
composite field $L_1 L_2$ is $[L_1 L_2 : k] = [L_1 : k] [L_2 : k]$.

I want to solve this problem with basic field theory (I haven't studied Galois theory). Thanks in advance.

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Below is what I've done so far.

Let $[L_1 : k] = m$, $[L_2 : k] = n$, and write $L_1$, $L_2$ as $L_1 = k(\alpha_1, \dots, \alpha_m)$, $L_2 = k(\beta_1, \dots, \beta_n)$, where $\alpha_1, \dots, \alpha_m$ is a $k$-basis for $L_1$, and $\beta_1, \dots, \beta_n$ is a $k$-basis for $L_2$. Then we can easily show that

$$[L_1 L_2 : k] \le [L_1 : k] [L_2 : k],$$

where the equality is achieved iff $\beta_1, \dots, \beta_n$ are linearly independent over $L_1$.

I tried to use $L_1 \cap L_2 = k$ to prove the linear independence but failed.

Answer

The answer turns out to be negative.

Consider $k = \mathbb{Q}$, $L_1 = \mathbb{Q}(\alpha)$, and $L_2 = \mathbb(\alpha \zeta_3)$, where $\alpha$ is the real root of $x^3 - 2$, and $\zeta_3$ is a primitive root of $x^3 - 1$. Then $[L_1 : k] = [L_2 : k] = 3$, and it is easy to show that $L_1 \cap L_2 = k$ (one way is to consider the imaginary parts). However, $L_1 L_2 = \mathbb{Q}(\alpha, \zeta_3)$ is the splitting field of $x^3 - 2$, which has degree 6 over $\mathbb{Q}$.

It turns out that if one of $L_1/k$ and $L_2/k$ is Galois, then the argument would be true.

elementary set theory - Showing a function $H:mathbb R^{mathbb R}to mathbb R^{mathbb R}$ is onto and finding its inverse

Let $H:\mathbb R^{\mathbb R}\to \mathbb R^{\mathbb R} \\ H(f)=\begin {cases}f^{-1} & \text {$f$ is a bijection} \\ f & else \end {cases}$

Prove that $H$ is onto (I already showed it's 1-1) and find $H^{-1}$ and prove it.

Onto: Let $f\in \mathbb R^{\mathbb R}$.

Suppose $f$ is a bijection, then there's a single $f^{-1}$ and from the defintion of $H$ we'll get $H(f^{-1})=f$.

Suppose $f$ isn't a bijection, then from $H:$ $H(f)=f$ either way for every $f$ there's a source, therefore, $H$ is onto.

I think it's inverse is simply: $H(f)=\begin {cases}f^{-1} & else \\ f & \text {$f$ is a bijection} \end {cases}$

But I can't seem to be able to compose $H\circ H^{-1}$ and vice versa correctly...

Answer

By definition $\;H\circ H(f)=H(H(f))\;$:

$$H(H(f))=\begin{cases}H(f^{-1})&,\;\;f\;\;\text{is a bijection}=\left(f^{-1}\right)^{-1}=&f&,\;f\;\text{is a bijection}\\{}\\
H(f)&,\;\;f\;\;\text{else}=& f&,\;f\;\text{else}\end{cases}$$

sequences and series - Evaluate $sum_{n=1}^infty frac{n}{2^n}$

This is a homework question; I'm supposed to use power series to find the following sum: $$\sum_{n=1}^\infty \frac{n}{2^n}$$ I took the geometric series $$\frac{1}{1-x}=\sum_{n=0}^\infty {x^n}$$ and differentiated and multiplied both sides by x to get $$\frac{x}{(1-x)^2}=\sum_{n=1}^\infty {nx^n}$$ I'm stuck because I'm not sure how to make the $$\frac{1}{2^n}$$ term appear.

real analysis - Is a periodic function differentiable? uniformly continuous?

Let $f : \mathbb R
\rightarrow \mathbb R$ be a function such that $f(x + 1) = f(x)$ for all $x \in \mathbb R.$ Which of the
following statement(s) is/are true?

(A) $f$ is bounded.
(B) $f$ is bounded if it is continuous.

(C) $f$ is differentiable if it is continuous.
(D) $f$ is uniformly continuous if it is continuous.

It is a periodic function with period 1. I don't know how to proceed further...

Any Hints will be appreciated...

modular arithmetic - GCD of two polynomials in Mod 2

Let $p$ and $q$ be distinct primes.

I wonder is the following statement always true?
$$\gcd(x^p-1, x^q-1) \stackrel{?}{=} x-1$$

Answer

More generally, we have $\gcd(x^n-1,x^m-1)=x^{\gcd(n,m)}-1$.

integration - $int^infty_0frac{sin x}{x} , dx = frac{1}{2i}int^infty_{-infty} frac{e^{ix}-1}{x} , dx$, why?

How comes this true?

$$\int^\infty_0\frac{\sin x} x \, dx = \frac{1}{2i}\int^\infty_{-\infty} \frac{e^{ix}-1} x \, dx$$

sequences and series - What must be the simplest proof of the sum of first $n$ natural numbers?

I was studying sequence and series and used the formula many times $$1+2+3+\cdots +n=\frac{n(n+1)}{2}$$ I want its proof.

Thanks for any help.

Answer

Let the sum be $$S_n=1+2+3+\cdots +n\tag1$$ on reversing the same equation we get $$S_n=n+(n-1)+(n-2)+\cdots +1\tag2$$ On adding $(1)$ and $(2)$ we have each term equal to $n+1$ which will occur $n$ times i.e. $$2S_n=(n+1)+(n+1)+(n+1)\cdots\{n times\}+(n+1)$$ $$2S_n=n(n+1)$$ $\therefore$ $$S_n=\frac{n(n+1)}{2}.$$ Hope it helps!!!