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  • i) $\mathbb{Z}/2\mathbb{Z}$ has no $\mathbb{Z}$ basis.

  • ii) $\mathbb{Z}/2\mathbb{Z}$ has no $\mathbb{Z/2\mathbb{Z}}$ basis.

  • iii) Suppose $R= \mathbb{Z}[t]$. Then $2R+tR$ has no R-basis.

i) Let $R= \mathbb{Z}$ an $L_{module}=\mathbb{F}_{2}=\mathbb{Z}/2\mathbb{Z}$. Then $L\subset \mathbb{Z} 1 (1 \in \mathbb{F}_{2})$. And $L=\mathbb{Z}1$. But $\mathbb{Z}1$ is not linearly independent, because $2\in \mathbb{Z}, 2 \ne 0$, but $21 = 0$ So $\mathbb{Z}/2\mathbb{Z}$ does not have a $\mathbb{Z}$ basis.

ii) Let $R= \mathbb{Z}/2\mathbb{Z}$ an $L_{module}=\mathbb{F}_{2}=\mathbb{Z}/2\mathbb{Z}$ Then $L= R = \mathbb{Z}/2\mathbb{Z}$. So it is a basis of itself, and thats why the assumption is not correct.

iii) It is given that $R= \mathbb{Z}[t]$ an $I=2R+Rt$. (I believe) that it is a principal ideal domain. It is a R module, with generators 2,t an N=2. It can not have a basis because a basis (w) (n=1) would mean $I=Rw$, but that is not possible (I believe). Also (2,t) can not be a basis since $2,t$ are dependent : $-t, 2 \in R $ not all 0 , $(-t)2+2t=0$.

Everywhere where I believe something, I am not able to prove it. Can somebody help me prove my beliefs? Thanks.

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What is your definition of "$R$-basis"? In some instance, I've seen things like "if $r_1m_1+\cdots+r_km_k=0$ then $r_im_i=0$ for each $i$", as opposed to "if $r_1m_1+\cdots+r_km_k = 0$ then $r_i=0$ for all $i$". –  Arturo Magidin Dec 1 '11 at 16:56
    
definition: $x_{1},...,x_{n})$ are a R-basis for M if $x_{1},...,x_{n}$ are R-linear and independend and : $M=Rx_{1}+...+Rx_{n} = \{ r_{1}x_{1} + ... + r_{n}x_{n} ; r_{1},...,r_{n} \in R $ –  Tashi Dec 1 '11 at 17:08
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Your response does not address the point that I specifically mentioned. Which is your definition of "independent"? Is it that the only way to add up to zero is that each summand be zero, or that the only way to add up to zero is that each coefficient be zero? –  Arturo Magidin Dec 1 '11 at 17:09
    
definition: Let M b a R-module, an let $x_{1},...x_{n}$ be in M. $x_{1},...,x_{n}$ are R linear independent if : $r_{1},..,r_{n}\in R, r_{1}x_{1}+...+r_{n}x_{n}=0 \Rightarrow r_{1}=...=r_{n}=0$ –  Tashi Dec 1 '11 at 17:17
    
Okay, the coefficients. Thank you. –  Arturo Magidin Dec 1 '11 at 17:18
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2 Answers

up vote 3 down vote accepted

Aside: Why introduce new letters and new symbols to clutter up things? What's the point of introducing $L$ if the only thing we are going to do with $L$ is say that it is equal to $\mathbb{F}_2$, and why introduce $\mathbb{F}_2$ if it's just $\mathbb{Z}/2\mathbb{Z}$?

Also, "$L_{module}$" does not really clarify things; it makes them worse. It's clunky notation, doesn't seem to mean anything (what does it mean when you write "module" under a structure? To specify that we are considering a certain abelian group $M$ as an $R$ module, we usually write $M_R$ or ${}_RM$, not $M_{module}$, and we certainly don't write $R_{module}$).


(i) The reason $\mathbb{Z}/2\mathbb{Z}$ has no $\mathbb{Z}$-basis is that every element of $\mathbb{Z}/2\mathbb{Z}$ satisfies $2x = 0$; this means that no nonempty set is linearly independent, and since $\mathbb{Z}/2\mathbb{Z}\neq\{0\}$, then the empty set does not generate the module. So no set can be both linearly independent and span.

(ii) On the other hand, $\mathbb{Z}/2\mathbb{Z}$ does have a $\mathbb{Z}/2\mathbb{Z}$ basis. The singleton $\{\overline{1}\}$ is a basis (we are dealing with vector spaces here, and vector spaces always have bases).

(iii) $\mathbb{Z}[t]$ is not a principal ideal domain: if $\mathbb{Z}[t]$ were a principal ideal domain, then $(2,t)$ would be generated by a $\gcd$ of $2$ and $t$. But $2$ and $t$ have no common divisors other than $1$ and $-1$ ($\mathbb{Z}[t]$ is a UFD). Yet $(2,t)\neq (1)$.

The reason $(2,t)$ does not have a basis: as noted above, any generating set must have more than one element (the ideal is principal). But in $\mathbb{Z}[t]$, any two elements are linearly dependent over $\mathbb{Z}[t]$: given $p(t)$ and $q(t)$, not both zero, we have $q(t)p(t) + (-p(t))q(t) = 0$. If they are both zero, then $1p(t)+1q(t)=0$. So no set with more than one element can be independent, and no set with one element or fewer can generate $(2,t)$, so $(2,t)$ does not have a basis.

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Thank you Arturo Magidin! What does singleton mean? By $\{\overline{1}\}$ do you mean the equivalence class? –  Tashi Dec 1 '11 at 17:41
    
@Tashi: "Singleton" = set with only one element. $\overline{1}$ is the image of $1$ in $\mathbb{Z}/2\mathbb{Z}$. $\{\overline{1}\}$ is the set whose only element is the element $1+2\mathbb{Z}$ of $\mathbb{Z}/2\mathbb{Z}$. –  Arturo Magidin Dec 1 '11 at 18:01
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Ummmm...given any commutative ring with identity $R$, there is a basis for $R$ as an $R$-module, namely $\{1\}$. In fact, an $R$-module $M$ has a basis (of elements in $R$) precisely when $M$ is a free $R$-module. Since $\mathbb{Z}/2\mathbb{Z}$ is finite and any free $\mathbb{Z}$ module (except $0$) is infinite, it is clear that $\mathbb{Z}/2\mathbb{Z}$ is not a free $\mathbb{Z}$ module and hence has no basis in $\mathbb{Z}$.

Of course, $R$ has a basis as an $R$-module, namely $\{1\}$. Therefore ii) is incorrect.

Also, $\mathbb{Z}[t]$ (where, presumably, $t$ is an indeterminate) is not a principal ideal domain, since $I=(2,t)=2R+tR$ is not principal (if you're not sure why this is the case, it would be worth writing down a proof).

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Thank you!...... –  Tashi Dec 1 '11 at 17:42
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