# Proof

• By hypothesis, $(F,+,\cdot)$ is a field with two absolute values $|\cdot|_1$ and $|\cdot|_2$ defined on it.
• Assume, $|\cdot|_1$ and $|\cdot|_2$ are dependent.
• Then for $x\in F$ we have $|x|_1 < 1\Longleftrightarrow |x|_2 < 1.$
• Let $x_0\in F$ with $|x_0|_1 > 1,$ then $|x_0|_2 > 1.$
• For an arbitrary $x\in F$ with $x\neq 0$, we have for some positive real number $\lambda > 0$ $$|x|_1=|x_0|_1^\lambda.$$
• For all real numbers $\epsilon > 0$ we get the approximations $$|x|_1^{\lambda-\epsilon} < |x_0|_1^\lambda< |x|_1^{\lambda+\epsilon},$$ implying $$\left|\frac{x^{\lambda-\epsilon}}{x_0^\lambda}\right|_1 < 1 < \left|\frac{x^{\lambda+\epsilon}}{x_0^\lambda}\right|_1.$$
• Since $|\cdot|_1$ and $|\cdot|_2$ are dependent, we get $$\left|\frac{x^{\lambda-\epsilon}}{x_0^\lambda}\right|_2 < 1 < \left|\frac{x^{\lambda+\epsilon}}{x_0^\lambda}\right|_2,$$ implying $$|x|_2^{\lambda-\epsilon} < |x_0|_2^\lambda< |x|_2^{\lambda+\epsilon}$$ for all $\epsilon > 0,$ implying $$|x|_2^{\lambda} = |x_0|_2^\lambda.$$
• Obviously, $\lambda$ exists and is a positive real number.

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### References

#### Bibliography

1. Lang, Serge: "Algebra - Graduate Texts in Mathematics", Springer, 2002, 3rd Edition