Solved: "Prove kerϕ=⟨1+−5, 2⟩"

College
Algebra
Abstract algebra

osula9a

Answered question

2022-01-13

Prove

k e r ϕ = ⟨ 1 + \sqrt{- 5}, 2 ⟩

Answer & Explanation

Philip Williams

Beginner2022-01-14Added 39 answers

Step 1
Let

ϕ : Z [\sqrt{- 5}] \to Z_{2}

ϕ (p + q \sqrt{- 5}) = {[p + q]}_{2}

Want to prove

k e r ϕ = ⟨ 1 + \sqrt{- 5}, 2 ⟨

So what

m = p + q \sqrt{5} \in k e r ϕ \Leftrightarrow {[p + q]}_{2} = {[0]}_{2} \Leftrightarrow p + q \in 2 Z

Step 2
Only need to prove that every element

p + q \sqrt{5}

from the kernel is in

⟨ 1 + \sqrt{5}, 2 ⟩

.
Since

p + q

is even,

p - q

is also even,

= 2 n

for some integer n.
Then

p + q \sqrt{5} = n \times 2 + q (1 + \sqrt{5})

as desired.

William Appel

Beginner2022-01-15Added 44 answers

Step 1
Clearly
$ϕ (1 + \sqrt{5}) = 0$
and
$ϕ (2) = 0$ .
Hence
$⟨ 1 + \sqrt{5}, 2 ⟩ \subset k e r (ϕ)$
Suppose
$a + b \sqrt{5}$
is in kernal. Then
$a + b = 2 k$
for some k. If both a and b are even, then
$a + b \sqrt{5} = 2 (\frac{a}{2} + \frac{b}{2 \sqrt{5}}) \in ⟨ 1 + \sqrt{5}, 2 ⟩$
If both are odd, then
$a + \sqrt{5} = 1 + \sqrt{5} + (a - 1) + (b - 1) \sqrt{5}$
as
$(a - 1) \in ⟨ 1 + \sqrt{5}, 2 ⟩$
and $(b - 1)$ are even. In any case
$k e r (ϕ) \subset ⟨ 1 + \sqrt{5}, 2 ⟩$
And the conclusion follows.

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