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If R[x] is a P.I.D. and R commutative, then R is a field

The Mathematician — Tue, 27 Jun 2023 13:00:25 +0000

How to prove that if R[x] is a P.I.D. and R commutative then R is a field

We will use the fact that every prime ideal in a P.I.D. is a maximal ideal, which we have proved here.

Prove that if R[x] is a P.I.D. and R commutative then R is a field

Proof: given $R[x]$ is a principal ideal domain. We have proven here $R$ is an integral domain since it is a subring of $R[x]$ . Further, we have that $R[x]/(x) \cong R$ and since $R$ is an integral domain, we do know that $(x)$ is a prime ideal. But every prime ideal is a maximal ideal and therefore $R[x]/(x)$ must be a field, which implicates that $R$ is a field.

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A subring R of the PID R[x] is an integral domain

The Mathematician — Fri, 23 Jun 2023 13:00:23 +0000

How to prove that a subring R of the PID R[x] is an integral domain

The reader should know that only proving that the subring has no zero divisors is not enough. A subring should also contain the identity element.

Prove that a subring R of the PID R[x] is an integral domain

Proof: we take the principal ideal domain $R[x]$ and let $R$ be its subring. Let’s assume that we have the non-zero elements $a,b \in R$ such that $ab = 0$ , i.e., $a$ and $b$ are zero divisors. Since $R \subset R[x]$ , we have that $ab = 0$ , but that is a contradiction since $R[x]$ has no zero divisors since $R[x]$ is an integral domain.

Now we need to figure out if $R$ contains the identity element. But, $R[x]$ contains the identity element since it is a principal ideal domain, and therefore, $R$ too (so $1_R \neq 0_R$ ).

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Any two nonzero elements of a principal ideal domain have a least common multiple

The Mathematician — Wed, 21 Jun 2023 13:00:27 +0000

How to prove that any two nonzero elements of a principal ideal domain have a least common multiple

In order to prove that statement, let’s recall what actually a least common multiple is. Let $a,b \in R$ nonzero elements, where $R$ is a P.I.D. A least common multiple of $a$ and $b$ is an element $e$ of $R$ such that:

$a \mid e$ and $b \mid e$
if $a \mid e'$ and $b \mid e'$ , then $e \mid e'$

Prove that any two nonzero elements of a principal ideal domain have a least common multiple

Proof: let $(a) \cap (b) = (e)$ . Then $(e) \subseteq (a)$ and $(e) \subseteq (b)$ . This means that there exists $r_1,r_2 \in R$ such that $a=r_1e$ and $b=r_2e$ . This implies that $a \mid e$ and $b \mid e$ , which proves the first statement.

If $a \mid e'$ and $b \mid e'$ , then $(e') \subseteq (a)$ and $(e') \subseteq (b)$ . Since $(a) \cap (b) = (e)$ , we must have that $e' \in (e)$ , which means there exists $r \in R$ such that $er = e'$ , and therefore, $e \mid e'$ .

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Two ideals (a) and (b) in PID are comaximal iff gcd(a,b) = 1

The Mathematician — Mon, 19 Jun 2023 13:00:27 +0000

How to prove that two ideals (a) and (b) in PID are comaximal iff gcd(a,b) = 1

The best way to tackle this problem is by using the definition of a comaximal.

Prove that any two ideal (a) and (b) in PID are comaximal iff gcd(a,b)=1

Proof: let

R

be a PID. We will start with the right implication:

“ $\Rightarrow$ “: let $d$ be the generator for the principal ideal generated by $a$ and $b$ , i.e.,

\begin{equation*} (d) = (a,b) = \{ax+by \ | \ x,y\in R\}. \end{equation*}

We have given that

(a) + (b) = R

since

(a)

and

(b)

are comaximal. So this means that

1 \in (a) + (b)

and therefore we have an

R

-linear combination

ax + by = 1

for some

x,y\in R

. This means that

gcd(a,b) = 1

since

ax + by \in (a,b)

“ $\Leftarrow$ “: reverse the previous proof.

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A quotient of a principal ideal domain by a prime is again a P.I.D.

The Mathematician — Thu, 15 Jun 2023 13:00:56 +0000

How to prove that quotient of a principal ideal domain by a prime is again a P.I.D.

The best way to tackle this problem is by using the fact every prime ideal is a maximal ideal in a P.I.D.

Prove that a quotient of a P.I.D. by a prime ideal is again a P.I.D.

Proof: let $P$ be a prime ideal of the principal ideal domain $R$ . Every prime ideal is a maximal ideal in $R$ , which we have proven here. So, we get that $R/P$ is a field, again, which we have seen earlier here. Since $R/P$ , it is definitely a P.I.D. too, which completes this proof.

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Every nonzero prime ideal in a P.I.D. is a maximal ideal

The Mathematician — Tue, 13 Jun 2023 13:00:46 +0000

How to show that every prime ideal in a P.I.D. is a maximal ideal

The easiest way to prove this is to use the definitions of a prime ideal, maximal ideal, and P.I.D. In the proof below, we want to show that $P$ is either a maximal ideal or a P.I.D.

Prove that every prime ideal in P.I.D. is a maximal ideal

Proof: Let $P = (p)$ and $M = (m)$ be a prime ideal and a maximal ideal of the principal ideal domain $R$ , respectively. Take for the assumption that $P$ is contained in $M$ . Then there exists an element $r \in R$ such that $p = rm$ . Since $P$ is a prime ideal, we have that either $r \in P$ or $m \in P$ . If $m \in P$ , then $P$ is the maximal ideal. If $r \in P$ , then $r = px$ for some $x \in R$ . Then $p = rm = pxm$ which means that $xm = 1$ , so this implies that $m$ is a unit. So $P = R$ .

Therefore, we have that P = M) or P = R).

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Derivative of sin^4(x)

The Mathematician — Fri, 09 Jun 2023 13:00:03 +0000

What is the derivative of sin^4(x)?

The derivative of $\sin^4(x)$ is $4\sin^3(x)\cos(x)$ .

Solution of the derivative of sin^4(x)

Solution: let $F(x) = g(f(x)) = \sin^4(x)$ where $g(u) = u^4$ and $f(x) = \sin(x)$ . To find the derivative of $\sin^4(x)$ , we need to apply the chain rule on $F(x)$ :

\begin{equation*} F'(x) = g'(f(x))f'(x) \end{equation*}

The derivative of $u^4$ is $4u^3$ and the derivative of $\sin(x)$ is $\cos(x)$ , which we have seen here earlier. Therefore, we get:

\begin{equation*} g'(f(x)) = g'(\sin(x)) = 4\sin^3(x) \text{ and } f'(x) = \cos(x). \end{equation*}

Substituting everything, we get:

\begin{align*}F'(x) &= g'(f(x))f'(x) \\&= 4\sin^3(x)\cos(x)\end{align*}

Therefore, the derivative of $\sin^4(x)$ is $4\sin^3(x)\cos(x)$ .

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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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Integral of 1/(x^2 – a^2)

The Mathematician — Mon, 05 Jun 2023 13:00:39 +0000

What is the integral of 1/(x^2 – a^2)?

The integral of

\frac{1}{x^2 - a^2}

\frac{1}{2a}\ln\lvert \frac{x - a}{x + a} \rvert + C

Solution of the integral 1/(x^2 – a^2)

Solution: We want to determine the next integral:

\begin{align*} \int \frac{dx}{x^2 - a^2}. \end{align*}

We will apply an integral technique such that the denominators will be linear functions:

\begin{align*} \frac{1}{x^2 - a^2} &= \frac{1}{(x-a)(x+a)} \\ &= \frac{A}{x - a} + \frac{B}{x + a} \\ &= \frac{Ax + Aa + Bx - Ba}{x^2 - a^2}. \end{align*}

We do have the following set of equalities:

\begin{align*} A + B &= 0 \\ Aa - Ba &= 1. \end{align*}

We see that

A = -B,

and therefore

B = -\frac{1}{2a}

and

A = \frac{1}{2a}

. Substituting everything above, we get:

\begin{align*} \int \frac{1}{x^2 - a^2} dx &= \int \frac{A}{x - a} dx + \int \frac{B}{x + a} dx \\ &= \frac{1}{2a} \int \frac{dx}{x - a} - \frac{1}{2a} \int \frac{dx}{x + a} \\ &= \frac{1}{2a} \ln \lvert x - a \rvert - \frac{1}{2a} \ln \lvert x + a \rvert + C \\ &= \frac{1}{2a} \ln \bigg\lvert \frac{x - a}{x + a} \bigg\rvert + C. \end{align*}

Conclusion

In conclusion, the integral of

\frac{1}{x^2 - a^2}

\frac{1}{2a}\ln \lvert \frac{x - a}{x + a} \rvert + C

, or in mathematical notation:

\begin{equation*} \int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln \bigg\lvert \frac{x - a}{x + a} \bigg\rvert + C. \end{equation*}The post Integral of 1/(x^2 – a^2) appeared first on Epsilonify.

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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