euclidean domain Archives - Epsilonify

An element of minimum norm in Euclidean Domain is a unit

The Mathematician — Wed, 07 Jun 2023 13:00:01 +0000

How to prove that an element of the minimum norm in Euclidean Domain is a unit

The best way to prove this is by taking the definition of a Euclidean Domain. We need to take into account that the minimum norm is nonzero.

Prove that an element of minimum norm in Euclidean Domain is a unit

Proof: let

R

be an Euclidean Domain and let

N(x)

be the be the nonzero minimum norm of

x

. By definition of the Euclidean Domain, we have the following:

\begin{equation*} a = qx + r, \quad \text{where } r = 0 \text{ or } N(r) < N(x). \end{equation*}

In a Euclidean domain, it must hold for any

a

and

x

, so we can take

a = 1

. Now if

r = 0

, then we have that

x

is a unit. If

N(r)

N(x)

, then

N(r)

is smaller than the nonzero minimum norm

N(x)

. This implies that

N(r)

must be zero. Therefore,

x

is a unit.

Conclusion

For this type of questions, it is important to use definitions.

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If R is a Euclidean domain, then there are universal side divisors in R

The Mathematician — Fri, 02 Jun 2023 13:00:00 +0000

How to prove that a Euclidean Domain has universal side divisors

The best way to prove that is to use the definition of a Euclidean Domain straightly and by using the definition of universal side divisors.

Prove that if R is a Euclidean domain and not a field, then there are universal side divisors in R

Proof: let

R

be a Euclidean domain and let

b \in R - \widetilde{R} = R - R^{\times} \cup \{0\}

be a universal side divisor of minimal norm. This means that we took a universal side divisor which we are sure that

R

has no smaller norms of universal side divisors than the norm of

b

. By the definition of the Euclidean Domain, we can say

\begin{align*} x = qb + r \quad \text{where} \quad r = 0 \text{ or } N(r) < N(b). \end{align*}

Now, we need to show that

r \in R^{\times} \cup \{0\}

. It is obviously true that if

r = 0

, then

r \in R^{\times} \cup \{0\}

. If

N(r)

N(b)

, then we do know that

r

can't be a universal divisor since

b

is the smallest norm that is a universal divisor, which implicates that

r \not \in R - \widetilde{R}

. So

r

must be a unit and therefore

r \in R^{\times} \cup \{0\}

, which proves this proof.

Conclusion

It is crucial to notice that we are curious if there are universal side divisors, so showing one is perfectly fine.

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The integers are a Euclidean Domain

The Mathematician — Sun, 28 May 2023 13:00:37 +0000

How to prove that the integers are a Euclidean Domain

To show that the integers are a Euclidean domain (or possess a Division Algorithm), finding one norm only to verify the conditions to be a Euclidean Domain is enough.

Proof that the integers are Euclidean Domain

Proof: Define the following norm:

\begin{equation*} N(x) = \lvert x \rvert, \end{equation*}

where

x \in \mathbb{Z}

. We need to show that for any two elements

a,b \in \mathbb{Z}

with

b\neq 0

there exists elements

q

and

r

\mathbb{Z}

such that

\begin{equation*} a = qb + r \quad \text{with} \quad r = 0 \quad \text{or} \quad N(r) < N(q). \end{equation*}

We let

a

and

b

be nonzero elements and we will check two cases:

b

0

and

b

0

For $b$ > $0$ , let $a \in [qb, (q + 1)b)$ . Then $a = qb + r$ where $r \in [0, \lvert b \rvert)$ since $r = a - qb$ < $(q + 1)b - qb = b$ . Therefore, $N(r)$ < $N(b) = \lvert b \rvert$ or $r = 0$ .

For $b$ < $0$ , we take $a \in [-qb, -(q + 1)b)$ since $-b$ > $0$ . Then $a = q\cdot (-b) + r$ where $r = a + qb$ < $-(q + 1)b - qb = -b$ . Obviously, since -b > 0, we have that $\lvert r \rvert = N(r)$ < $N(-b) = \lvert -b \rvert$ or $r = 0$ .

We haven't included the case when the remainder is negative in all the above cases. The approach is similar to the above. Take the elements $b$ > $0$ and use the same properties we used there and $r > 0$ (note that we don't take this remainder to be negative!). Now let $a = q'b + r'$ , where $q' = q + 1$ and $r' = r - b$ . Then $r'$ < $0$ since $r \in [0, \lvert b \rvert)$ . Now we can perform the rest as $r - b \in (-b, b)$ :

\begin{align*} r - b < 0 &\iff r < b \\ &\iff r - b < b \\ &\iff r' < b \\ &\iff \lvert r' \rvert < \lvert b \rvert \\ &\iff N(r') < N(b), \end{align*}

which ends the proof by showing that the integers are indeed a Euclidean Domain. We could also prove everything with induction, but that is something that the reader can verify (but you can always leave a comment if you want know it).

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Fields are Euclidean Domains

The Mathematician — Wed, 24 May 2023 13:00:14 +0000

Are the fields Euclidean domains?

The fields are Euclidean domains. To see why, we need to find a norm that satisfies the Division Algorithm.

Proof that the fields are Euclidean Domains

Let $F$ be an arbitrary field. We can take the norm $N(a) = 0$ for all $a \in F$ . Take $a = qb + r$ . Since $F$ is a field, each element has an inverse. Therefore, if we take $r = 0$ and $q = ab^{-1}$ for every $a$ and $b \neq 0$ , we see that $F$ possess a Division Algorithm. So, $F$ is an Euclidean Domain.

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Every ideal in a Euclidean Domain is principal

The Mathematician — Sat, 20 May 2023 13:00:44 +0000

How to show that every ideal in a Euclidean Domain is principal

We need to show any ideal

I

of the Euclidean Domain

R

is principal, i.e.,

I = (x)

where

x

is any nonzero element of

I

of minimum norm.

Prove that every ideal in a Euclidean Domain is principal

Let

R

be an Euclidean Domain and

I

its arbitrary ideal. It is obvious that when

I

is zero ideal, then

I = (0)

. Now assume

I

is a non-empty ideal and let

b \neq 0 \in I

be its minimal norm. To show that

b

exists, we need to show that the following set has a minimal element:

\begin{align*} \{N(y) \ | \ y \in I\}. \end{align*}

But this set has a minimal element by the definition of norm. To see the full proof of this, check here. Now what is left is that

(b)

is contained or equal to

I

, and vice versa. “

(b) \subseteq I

“: This is obvious since

b

is an element of

I

.

“

I \subseteq (b)

“: Since the assumption was that

R

is an Euclidean Domain,

R

also possess the Division Algorithm. So take the element

a \in I

such that

a = qb + r

with

r = 0

N(r)

N(b)

. We get that

r = a - qb

and

a,qb \in I

, which implies that

r \in I

. Now we assumed that

b

is minimal, so this means that

r = 0

since we assumed that

b \neq 0

. So this means that

a = qb

and therefore

a \in (b)

Conclusion

Finally,

I = (b)

and so every ideal in a Euclidean Domain is principal.

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