Let R be a ring with IBN. (i) If R^∞ is a free left R-module having an infinite basis, prove tha

Let R be a ring with IBN. (i) If R^∞ is a free left R-module...

Let $R$ be a ring with IBN. (i) If $R^{\infty}$ is a free left $R$-module having an infinite basis, prove that $R \oplus R^{\infty} \cong R^{\infty}$. (ii) Prove that $R^{\infty} \not R^{n}$ for any $n \in \mathbb{N}$. (iii) If $X$ is a set, denote the free left $R$-module $\bigoplus_{x \in X} R x$ by $R^{(X)}$. Let $X$ and $Y$ be sets, and let $R^{(X)} \cong R^{(Y)}$. If $X$ is infinite, prove that $Y$ is infinite and that $|X|=|Y|$; that is, $X$ and $Y$ have the same cardinal. Hint. Since $X$ is a basis of $R^{(X)}$, each $u \in R^{(X)}$ has a unique expression $u=\sum_{x \in X} r_{X} x ;$ define $$ \operatorname{Supp}(u)=\left\{x \in X: r_{x} \neq 0\right\} $$ Given a basis $B$ of $R^{(X)}$ and a finite subset $W \subseteq X$, prove that there are only finitely many elements $b \in B$ with $\operatorname{Supp}(b) \subseteq W .$ Conclude that $|B|=\operatorname{Fin}(X)$, where Fin $(X)$ is the family of all the finite subsets of $X$. Finally, using the fact that $|\operatorname{Fin}(X)|=|X|$ when $X$ is infinite, conclude that $R^{(X)} \cong R^{(Y)}$ implies $|X|=|Y| .$

Let $R$ be a ring with IBN.
(i) If $R^{\infty}$ is a free left $R$-module having an infinite basis, prove that $R \oplus R^{\infty} \cong R^{\infty}$.
(ii) Prove that $R^{\infty} \not R^{n}$ for any $n \in \mathbb{N}$.
(iii) If $X$ is a set, denote the free left $R$-module $\bigoplus_{x \in X} R x$ by $R^{(X)}$. Let $X$ and $Y$ be sets, and let $R^{(X)} \cong R^{(Y)}$. If $X$ is infinite, prove that $Y$ is infinite and that $|X|=|Y|$; that is, $X$ and $Y$ have the same cardinal.
Hint. Since $X$ is a basis of $R^{(X)}$, each $u \in R^{(X)}$ has a unique expression $u=\sum_{x \in X} r_{X} x ;$ define
$$
\operatorname{Supp}(u)=\left\{x \in X: r_{x} \neq 0\right\}
$$
Given a basis $B$ of $R^{(X)}$ and a finite subset $W \subseteq X$, prove that there are only finitely many elements $b \in B$ with $\operatorname{Supp}(b) \subseteq W .$ Conclude that $|B|=\operatorname{Fin}(X)$, where Fin $(X)$ is the family of all the finite subsets of $X$. Finally, using the fact that $|\operatorname{Fin}(X)|=|X|$ when $X$ is infinite, conclude that $R^{(X)} \cong R^{(Y)}$ implies $|X|=|Y| .$

Key Concepts

Invariant Basis Number (IBN)

The IBN property ensures that if two free modules over a ring have bases of different cardinalities, they are not isomorphic. This means any two bases of a free module must have the same size, preventing anomalies in dimension when dealing with module isomorphisms and is essential for understanding when two free modules can be equivalent.

Free Modules

A free module is a module that has a basis, meaning each element can be uniquely expressed as a finite linear combination of basis elements. This unique representation is crucial when analyzing the structure of modules and is fundamental in proving properties about direct sums and module isomorphisms.

Direct Sum of Modules

The direct sum of modules is a construction where a module is expressed as a sum of submodules such that every element of the module can be uniquely written as a sum of elements from each submodule, with only finitely many nonzero components. This concept is key in demonstrations of isomorphisms like showing that the direct sum of a module with an infinite free module is isomorphic to the infinite free module.

Module Isomorphism

Module isomorphism is the idea that two modules are structurally identical if there exists a bijective linear map between them that respects the module operations. This concept underpins many proofs in module theory, including those involving free modules, direct sums, and dimension or cardinality arguments.

Cardinality Arguments in Module Theory

When dealing with free modules, especially those with infinite bases, cardinality arguments are used to compare the sizes of bases or sets indexing the direct sum components. This often involves showing that the number of finite subsets of an infinite set has the same cardinality as the set itself, which is critical when proving that isomorphic free modules have bases of the same size.

Support of an Element in a Free Module

The support of an element in a free module is the set of basis elements that have nonzero coefficients in its unique representation. This concept helps in tracking where the 'action' of an element lies within the module and is used in proofs that involve controlling the number of nonzero contributions from certain subsets of the basis, aiding in cardinality and finiteness arguments.

Transcript

00:01 So if you remember the definition of linearly independence is that we take the equation.

00:09 So we'll do c0 times, we'll do cosine x first, sorry, c0, and then we'll actually do e to the ax, and then cosine bx plus c1 x, x, x, e to the ax, cosine x, and then x, remember, if we take a linear combination of these of all the functions and set it equal to 0, so that linear combination is going to be this plus c1 or plus c m, cm, x to the m, e to the a x, cosine bx, and then plus and then we're going to have d0, e to the a x, cosine, a x, cosine, sorry, not cosine, but sign bx plus d1 x e to the a x, b, or sorry, sign, sine bx, plus and plus d m, x to the m, e to the a x, sine bx, is equal to zero.

01:38 So remember, again, the definition of linearly independent, or linear independence is that there exists a non -trivial solution, sees or non -trial coefficients, all of these here, such that, or the only solution to this, or the not is the trivial solution if it's linearly independent.

02:03 If it's linearly independent, it means that all of these coefficients here have to be equal to zero.

02:09 So let's assume for now that it is dependent, okay? so that means that there exists a non -trivial solution to this equation here.

02:21 So we can factor out, first off, an e to the ax.

02:27 Again, since e to the ax is not equal to zero, it can never be equal to zero, we can take out e to the ax from all of the terms.

02:36 So what we're left with is then after we factor out the cosine bx, so our left is oops, yeah, i'll do that.

02:47 So c0 plus c1x plus dot dot, dot, plus cm, x to the m, times cosine bx, plus then you have d0, plus d1x plus d1, x plus dot dot, dot, plus d m, x to the m, sign of x, or sorry, sign of b of x, is equal to 0.

03:23 So now this we're going to call p of x.

03:33 So p of x, cosine b x, and then plus, and then this is going to be q of x, sine bx, sine b of x, just like the hint says, is equal to zero.

03:47 So this must be equal to zero for all x in our, um, in, in negative infinity of infinity.

03:57 So this px, px cosine bx, we're going to first consider when sine of bx is equal to 0.

04:11 So what are all the x is when sine of bx is equal to 0? that's going to be, for example, just any n pi.

04:21 So whenever we have n pi here, or bx is equal to n pi, so that's whenever we have x is equal to n pi over b.

04:34 So again, whenever this part here is equal to 0, which is when bx is equal to n pi.

04:49 Again, we're just choosing any x that we want.

04:51 So we're going to choose x such that this part is equal to zero.

04:55 So we have x is going to be equal to n pi over b.

05:01 Okay.

05:02 So then now we have x is equal to n pi over b.

05:10 So we can also plug that into here.

05:13 So that's going to be zero.

05:15 So we have p n pi over b and then cosine of n pi.

05:24 So cosine of n pi, this is equal to negative 1 to the n plus 1.

05:31 So that means, remember, where we have a unit circle here, if it's zero, it's going to be 1.

05:40 If it's pi, it's going to be negative 1.

05:43 Sorry, it's actually cosine n pi is equal to negative 1 to the n.

05:50 Okay? so we have negative 1 to the n, and then again, this is just all 0 here, and that's equal to 0.

05:58 So now we can divide by a negative 1 to the n...

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