Wikipedia:Reference desk/Archives/Mathematics/2015 August 4

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August 4

Finding a basis for a free abelian group

Suppose $G$ is an abelian group, which I'm going to think of as a $\mathbb {Z}$-module. For a set $X\subset G$, say that $X$ has an unexpected division if there are coefficients $c_{1},\dots ,c_{n}\in \mathbb {Z}$, distinct $x_{1},\dots ,x_{n}\in X$ and a prime $p$ such that $p|(c_{1}x_{1}+\cdots +c_{n}x_{n})$ but $p\not |c_{i}$ for some $i$.

If $G$ is a free abelian group and $X$ is maximal with the property that it has no unexpected divisions, does this make $X$ a basis for $G$? My plan to prove this was to extend $G$ to a $\mathbb {Q}$-space, show that $X$ spans the resulting space, and conclude that $X$ is a basis for $G$. The third step is no problem, but I'm stuck on showing that $X$ spans the space.

If it's not true, is there any property P which is closed under subset and such that we can define a basis as a set which is maximal with property P?--203.97.79.114 (talk) 09:39, 4 August 2015 (UTC)

An equivalent way of phrasing the problem is as follows. Suppose that G is an abelian group and X such that X is a basis of the vector space $G\otimes _{\mathbb {Z} }\mathbb {Z} _{p}$ for all p. Is X a basis of G? The answer is "yes". Here is a proof that X spans G. Let A denote the integral span of X in G and B=G/A. By right exactness of the tensor product, the exact sequence

$0\to A\to G\to B\to 0$

goes over to an exact sequence

$A\otimes \mathbb {Z} _{p}\to G\otimes \mathbb {Z} _{p}\to B\otimes \mathbb {Z} _{p}\to 0$.

By hypothesis, $A\otimes \mathbb {Z} _{p}=G\otimes \mathbb {Z} _{p}$, so we have $B\otimes \mathbb {Z} _{p}=0$. Hence B is a torsion group, no element of which has order divisible by p. Since this is true for all p, B=0. Sławomir
Biały 13:57, 4 August 2015 (UTC)

B does not have to be a torsion group. For an abelian group B, $B\otimes \mathbb {Z} _{p}=0$ iff the scalar multiplication by p map (thinking of B as a Z-module) is surjective. GeoffreyT2000 (talk) 15:46, 4 August 2015 (UTC)

Good point. I had implicitly in mind the case where B is finitely generated. But the proof no longer works without that hypothesis. I thought possibly this hypothesis was not really necessary, e.g., that we could show $B'\otimes \mathbb {Z} _{p}=0$ for any finitely generated submodule B' of B. Sławomir
Biały 18:07, 4 August 2015 (UTC)

Maybe I'm missing something simple, but I don't understand why your rephrasing is correct. My maximality requirement says that $X$ has the property that any superset is linearly dependent in some $G\otimes _{\mathbb {Z} }\mathbb {Z} _{p}$. Why would that imply that any superset is linearly dependent in every $G\otimes _{\mathbb {Z} }\mathbb {Z} _{p}$?--203.97.79.114 (talk) 19:26, 4 August 2015 (UTC)

In the statement of the problem, it wasn't clear to me that p was meant to be fixed. In that case, doesn't $G=\mathbb {Z}$ and $X=\{2\}$ give a counterexample, with p any odd prime? Sławomir
Biały 21:24, 4 August 2015 (UTC)

It's not meant to be fixed.--130.195.253.4 (talk) 22:07, 4 August 2015 (UTC)

(This is 203...) To give more detail: As you observed, what I called being unexpectedly divisible by $p$ is equivalent to being linearly dependent in $G\otimes _{\mathbb {Z} }\mathbb {Z} _{p}$. So I'm interested in sets $X$ which are linearly independent in every $G\otimes _{\mathbb {Z} }\mathbb {Z} _{p}$, and are maximal with this property. So for every proper superset, there exists a $p$ such that the superset is linearly dependent in $G\otimes _{\mathbb {Z} }\mathbb {Z} _{p}$.

Saying that $X$ spans $G\otimes _{\mathbb {Z} }\mathbb {Z} _{p}$ is equivalent to saying that any proper superset of $X$ is linearly dependent in $G\otimes _{\mathbb {Z} }\mathbb {Z} _{p}$. So saying that $X$ is a basis for every $p$ is saying that for every proper superset, for every $p$, the superset is linearly dependent in $G\otimes _{\mathbb {Z} }\mathbb {Z} _{p}$. You're flipping a quantifier from "there exists a $p$" to "for every $p$", and I don't see why that's justified.--130.195.253.4 (talk) 22:16, 4 August 2015 (UTC)

I have a no answer, sadly.

Consider the free abelian group on countably many generators, and let $a_{1}=(1,p_{1},0,0,\dots ),a_{2}=(1,0,p_{2},0,0,\dots ),a_{3}=(1,0,0,p_{3},0,0,\dots )$, etc. If a finite sum $\sum m_{i}a_{i}$ is divisible for $p_{j}$, then by considering the $i+1$ coordinate for $i\neq j$ we see that $p_{j}|m_{i}$. Then by considering the first coordinate we see that $p_{j}|m_{j}$. So the set $X=\{a_{1},a_{2},\dots \}$ has the desired property.

On the other hand, suppose $\{b_{1},b_{2},\dots \}$ were any basis extending $X$. Let $e_{1}=(1,0,0,\dots )$. Then $e_{1}=\sum k_{i}b_{i}$ for some finite sum. Fix $a_{j}$ not appearing in this sum. Then $a_{j}-\sum k_{i}b_{i}$ is divisible by $p_{j}$, but the coefficient in front of $a_{j}$ is 1, which is not divisible by $p$. So there can be no basis extending $X$, meaning a superset of $X$ which is maximal with the desired property is a counterexample.--130.195.253.4 (talk) 05:36, 5 August 2015 (UTC)

Chances of getting a 6 throwing a die

If you throw a die once, you have 1/6 of getting a 6. If you throw it twice, you have 2/6. Do you have 7/6 chances when you throw the die 7 times?--Jubilujj 2015 (talk) 21:56, 4 August 2015 (UTC)

No, it doesn't work like that. It's easiest to look at the chances of NOT getting a six, and then subtract that from one:

After one roll:

1 - (5/6) = 1/6 ≈ 0.167

After 2 rolls:

1 - (5/6)2 = 1 - 25/36 = 11/36 ≈ 0.306

After 7 rolls:

1 - (5/6)7 = 1 - 78125/279936 ≈ 1 - 0.279 = 0.721

To verify that this is correct, let's look at all the possible rolls in the 2 roll case:

1-1 1-2 1-3 1-4 1-5 1-6
2-1 2-2 2-3 2-4 2-5 2-6
3-1 3-2 3-3 3-4 3-5 3-6
4-1 4-2 4-3 4-4 4-5 4-6
5-1 5-2 5-3 5-4 5-5 5-6
6-1 6-2 6-3 6-4 6-5 6-6

As you can see, there are 36 possible combos, 11 of which (bolded) contain one or more six (one contains the double six). If you have some spare time, you can list all 279936 possible combos for 7 rolls. :-) StuRat (talk) 22:24, 4 August 2015 (UTC)

Just making a point more explicit:

There are not 2 chances in 6 of getting a 6 if you throw two dice (or one twice, which is the same). There are 11 possibilities 6,6; 6,1-5; 1-5,6. That's 11/36 ≈ 0.306. You don't just add two probabilities like that if you throw two dice. The same applies to the case where you throw 7 times. --Scicurious (talk) 00:28, 5 August 2015 (UTC)

Use the inclusion-exclusion principle: sum up the probabilities of getting a 6 on each die, then subtract the probabilities of getting two 6s on each pair of dice, then add the probabilities of getting three 6s on each triple of dice, and so on. GeoffreyT2000 (talk) 01:58, 5 August 2015 (UTC)