Endomorphisms of the complex field

MathematicsAbstract AlgebraHard

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Find all the endomorphisms of the complex field preserving real numbers. $Q, Q[\sqrt{2}], Z[i] = \{a + bi \mid a, b \in Z\} .$

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Step 1

Hi nguyệt, let's find all the endomorphisms of the complex field that preserve real numbers.

Field Endomorphisms of $\mathbb{C}$ fixing $\mathbb{R}$

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Step 2

An endomorphism of the complex field is a ring homomorphism from the complex numbers to itself. We are looking for maps sigma such that sigma of r equals r for every real number r.

$$\sigma: \mathbb{C} \to \mathbb{C} \text{ s.t. } \sigma(r) = r, \forall r \in \mathbb{R}$$
$$\sigma(z_1 + z_2) = \sigma(z_1) + \sigma(z_2), \quad \sigma(z_1 z_2) = \sigma(z_1) \sigma(z_2)$$
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Step 3

Since every complex number z can be written as a plus b times i, where a and b are real, we can apply the properties of the homomorphism.

$$z = a + bi \quad (a, b \in \mathbb{R})$$
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Step 4

Applying sigma to z, and using the fact that sigma preserves addition and multiplication, we get sigma of a plus sigma of b times sigma of i.

$$\sigma(z) = \sigma(a + bi) = \sigma(a) + \sigma(b)\sigma(i)$$
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Step 5

Because the map preserves real numbers, sigma of a is simply a, and sigma of b is simply b.

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Step 6

This shows that the entire map is determined by what happens to the imaginary unit i.

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Step 7

We know that i squared equals negative one. Applying the homomorphism to both sides, we get sigma of i squared equals sigma of negative one.

$$i^2 = -1 \implies \sigma(i^2) = \sigma(-1)$$

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About This Question

Subject
Mathematics
Topic
Abstract Algebra
Difficulty
Hard
Question Type
Open Ended

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