field Archives - Epsilonify

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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The ideal M of R is maximal iff the quotient ring R/M is a field

The Mathematician — Tue, 21 Mar 2023 13:00:13 +0000

Let $R$ be a commutative ring with identity $1 \neq 0$ . The ideal $M$ of $R$ is maximal iff the quotient ring $R/M$ is a field.

Proof. This proof can be easily done by using the Lattice Isomorphism Theorem of Rings.

“ $\Rightarrow$ “: Given maximal ideal $M$ of $R$ . By the Lattice Isomorphism Theorem of Rings, the ideals of $R$ containing $M$ correspond bijectively with the ideals of $R/M$ . The only ideals of $R$ that consist $M$ are $M$ and $R$ itself. So we have that $R/R \cong 0$ and $R/M$ are ideals of $R/M$ . We use a handy proposition which we have proven here which implies that $R/M$ is a field.

“ $\Leftarrow$ “: Given $R/M$ a field. Then by this proposition here, we have that the only ideals of $R/M$ are $0$ and $R/M$ . We use the Lattice Isomorphism Theorem of Rings again, which means that there are two ideals consisting $M$ . We know that at least it is $M$ and $R$ , which is already two. This implies there is no ideal $I$ of $R$ such that $M \subset I \subset R$ . Therefore, $M$ is a maximal ideal of $R$ .

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The commutative ring R is a field iff its only ideals are 0 and R

The Mathematician — Fri, 17 Mar 2023 13:00:53 +0000

Let $R$ be a ring with identity $1 \neq 0$ . The commutative ring $R$ is a field iff its only ideals are $0$ and $R$

Proof. We know that a commutative ring $R$ is a field if and only if every nonzero element of $R$ is a unit. So now can translate the question as the following:

Every nonzero element of $R$ is a unit iff its only ideals are $0$ and $R$

“ $\Rightarrow$ “: Assume we have a proper ideal $I$ of $R$ which is not equal to the zero ideal. Then that ideal $I$ contains a unit element of $R$ , say $u \in I$ . Since $I$ is an ideal of $R$ , we have that $Ru \subseteq I$ . This implies that $1 \in I$ since there exists a $v \in R$ such that $vu = 1$ . Therefore, $I = R$ and can’t be a proper ideal of $R$ .

“ $\Leftarrow$ “: Assume that the only ideals of $R$ are $0$ and $R$ . Take an arbitrary nonzero element $x\in R$ . We want to prove that $x$ is a unit. Since the only ideals are $0$ and $R$ , we have that the ideal $(x)$ is equal to $R$ . So we get that $1 \in (x)$ . Therefore, there exist an $y \in R$ such that $xy = 1$ , and so $x$ is a unit of $R$ . Since $x$ is taken arbitrarily, we have that every nonzero element of $R$ is a unit.

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Show that Q[sqrt(2)] is a field

The Mathematician — Wed, 09 Nov 2022 13:00:00 +0000

Show that $\mathbb{Q}[\sqrt{2}]$ is a field

Proof. We have that

\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2} \ | \ a,b \in \mathbb{Q} \}

. So we need to show that all elements in

\mathbb{Q}[\sqrt{2}]

do have an inverse. Indeed, if we take arbitrary

a,b \in \mathbb{Q}

such that

a + b \sqrt{2} \in \mathbb{Q}[\sqrt{2}]

, then

\begin{align*} 1 &= (a + b \sqrt{2}) \frac{a - b \sqrt{2}}{(a + b \sqrt{2})(a - b \sqrt{2})} \\ &= (a + b \sqrt{2}) \frac{a - b \sqrt{2}}{(a^2 - ab\sqrt{2} + ab\sqrt{2} - 2b^2)} \\ &= (a + b \sqrt{2}) \frac{a - b \sqrt{2}}{(a^2 - 2b^2)} \end{align*}

and since

a^2 - 2b^2 \in \mathbb{Q}

and

a - b \sqrt{2} \in \mathbb{Q}[\sqrt{2}]

, this implies that

\frac{a - b \sqrt{2}}{(a^2 - 2b^2)}

is an inverse of

a + b \sqrt{2}

. So this implies that

\mathbb{Q}[\sqrt{2}]

is a field. This completes the proof.

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