tensor products of modules

**hatsoff** · March 6th 2011, 07:01 AM

Hey guys.

I'm struggling with the notion of a tensor product of modules. My textbook starts off with some easy exercises, but since I'm having trouble with the definition, even the easy exercises are tough for me.

In particular, we have

Originally Posted by Dummit & Foote, Abstract Algebra, 3rd Ed, p375

Show that the element $2\otimes 1$ is $0$ in $\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}/2\mathbb{Z}$ but is nonzero in $2\mathbb{Z}\otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$ .

I can prove the first part, albeit not easily. But the second part is giving me more trouble. I can prove that if $2\otimes 1$ is zero then the whole tensor product is zero, i.e. $2\mathbb{Z}\otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}=\{0\}$ . But how do I derive a contradiction from this? Or is there an altogether easier way of showing it?

Any help would be much appreciated. Thanks!

**tonio** · March 6th 2011, 10:53 AM

Originally Posted by hatsoff

Hey guys.

I'm struggling with the notion of a tensor product of modules. My textbook starts off with some easy exercises, but since I'm having trouble with the definition, even the easy exercises are tough for me.

In particular, we have

I can prove the first part, albeit not easily. But the second part is giving me more trouble. I can prove that if $2\otimes 1$ is zero then the whole tensor product is zero, i.e. $2\mathbb{Z}\otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}=\{0\}$ . But how do I derive a contradiction from this? Or is there an altogether easier way of showing it?

Any help would be much appreciated. Thanks!

The tensor product of modules is the quotient of a free module over a module N generated by some elements.

Just show that $2\otimes 1$ doesn't belong to N by checking what characterizes those elements in N...

Tonio

**hatsoff** · March 6th 2011, 01:22 PM

Unfortunately I can't check that $2\otimes 1$ belongs to the denominator, so to speak, of the underlying quotient group, because I don't know how to perform operations on that quotient group. In particular, I don't know how to perform operations on the numerator of that quotient group, which is perhaps the SOURCE of my problem in understanding the definition of a tensor product.

According to my definition, we construct $S\otimes_R N$ by considering the free $\mathbb{Z}$ -module over $S\times N$ . But $S\times N$ is supposed to be an abelian group in its own right, and I don't understand how. For example, one idea would be to say that since $S$ and $N$ are both abelian groups individually, then we can just carry over the addition operations from each of them so that $(s_1,n_1)+(s_2,n_2)=(s_1+s_2,n_1+n_2)$ . Yet clearly this is not the case. But if not that, then how do we perform operations inside the group $S\times N$ ?

Here's what my textbook has to say on the subject: link

**NonCommAlg** · March 6th 2011, 03:21 PM

Originally Posted by hatsoff

Hey guys.

I'm struggling with the notion of a tensor product of modules. My textbook starts off with some easy exercises, but since I'm having trouble with the definition, even the easy exercises are tough for me.

In particular, we have

I can prove the first part, albeit not easily. But the second part is giving me more trouble. I can prove that if $2\otimes 1$ is zero then the whole tensor product is zero, i.e. $2\mathbb{Z}\otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}=\{0\}$ . But how do I derive a contradiction from this? Or is there an altogether easier way of showing it?

Any help would be much appreciated. Thanks!

if $R$ is a commutative ring with unity, $I$ is an ideal of $R$ and $M$ is an $R$ -module, then $M \otimes_R R/I \cong M/IM.$ this is very easy to see:

define the map $f : M \times R/I \longrightarrow M/IM$ by $f(x,r+I)=rx + IM$ and the map $g: M/IM \longrightarrow M \otimes_R R/I$ by $g(x+IM)=x \otimes_R (1+I).$ obviously $f$ is $R$ -bilinear and thus it induces an $R$ -module homomorphism

$\overline{f} : M \otimes_R R/I \longrightarrow M/IM$

defined by $\overline{f}(x \otimes_R (r+I))=rx+IM.$ It is easy to see that $g$ is well-defined (prove it!). finally, both $\overline{f}g$ and $g \overline{f}$ are identity maps.
so $\overline{f}$ is an $R$ -module isomorphism. so $x \otimes_R (r+I)=0$ if and only if $rx \in IM$ . now, apply this to your question.

**NonCommAlg** · March 6th 2011, 04:05 PM

I don't know what the heck is going on with MHF. It's been extremely slow and annoying. Anyway, I just wanted to add to my previous post that you don't even need the map $g$ for your problem. Just the homomorphism $\overline{f}$ is good enough to solve your problem.

**tonio** · March 6th 2011, 05:49 PM

Originally Posted by hatsoff

Unfortunately I can't check that $2\otimes 1$ belongs to the denominator, so to speak, of the underlying quotient group, because I don't know how to perform operations on that quotient group. In particular, I don't know how to perform operations on the numerator of that quotient group, which is perhaps the SOURCE of my problem in understanding the definition of a tensor product.

According to my definition, we construct $S\otimes_R N$ by considering the free $\mathbb{Z}$ -module over $S\times N$ . But $S\times N$ is supposed to be an abelian group in its own right, and I don't understand how.

ANY $\mathbb{Z}-$ module, whether free or not, is an abelian group.

The example in page 339 already dealt with this important case

For example, one idea would be to say that since $S$ and $N$ are both abelian groups individually, then we can just carry over the addition operations from each of them so that $(s_1,n_1)+(s_2,n_2)=(s_1+s_2,n_1+n_2)$ . Yet clearly this is not the case. But if not that, then how do we perform operations inside the group $S\times N$ ?

Coordinatewise, of course. This is dealt with in page 353.

Here's what my textbook has to say on the subject: link

As you can see, your book (a very good one, by the way) says relevant stuff many pages before 10.4.

In particular, in this case, $2\mathbb{Z}\otimes \mathbb{Z}/2\mathbb{Z}=(2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z})/N$ , with

$N:=\langle\, (s_1+s_2,m)-(s_1,m)-(s_2,m)\,,\,(s,m_1+m_2)-(s,m_1)-(s,m_2)\,,\,(sr,m)-(s,rm)\,\rangle$ ,

$s,s_1,s_2\in 2\mathbb{Z}\,,\,m,m_1,m_2\in \mathbb{Z}/2\mathbb{Z}\,,\,r\in\mathbb{Z}$ , and thus

$w:=2\otimes 1=0$ in our tensor product iff $w\in N$ , so if we can find a characteristic in

the elements of $N$ that isn't present in w we're done.

Hint: i) you can't "take away" the 2 and write [tex] $2\otimes 1 = 1\otimes 2$ (why?)

ii) In the generating elements of N, m is always 0 or 1 (mod 2)

Tonio

**TheArtofSymmetry** · March 7th 2011, 01:01 AM

Originally Posted by hatsoff

Hey guys.

I'm struggling with the notion of a tensor product of modules. My textbook starts off with some easy exercises, but since I'm having trouble with the definition, even the easy exercises are tough for me.

In particular, we have

I can prove the first part, albeit not easily. But the second part is giving me more trouble. I can prove that if $2\otimes 1$ is zero then the whole tensor product is zero, i.e. $2\mathbb{Z}\otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}=\{0\}$ . But how do I derive a contradiction from this? Or is there an altogether easier way of showing it?

Any help would be much appreciated. Thanks!

This might give you some intuitive idea.
Since $\mathbb{Z}$ is a right $\mathbb{Z}$ -module and $\mathbb{Z}/2\mathbb{Z}$ is a left $\mathbb{Z}$ -module, it satisfies $ar \otimes b=a \otimes rb$ for $a \in \mathbb{Z}$ , $b \in \mathbb{Z}/2\mathbb{Z}$ , and $r \in \mathbb{Z}$ .
Thus, $2 \otimes 1 = (1\cdot 2) \otimes 1=1 \otimes (2\cdot 1) = 1 \otimes 0=0$ in $\mathbb{Z}\otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$ . However, this rule does not applies to $2\mathbb{Z}\otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$ , since 1 is not an element of $2\mathbb{Z}$ . Further verfications show that $2\otimes 1$ is a non-zero element in $2\mathbb{Z}\otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$ .

Unfortunately I can't check that $2\otimes 1$ belongs to the denominator, so to speak, of the underlying quotient group, because I don't know how to perform operations on that quotient group. In particular, I don't know how to perform operations on the numerator of that quotient group, which is perhaps the SOURCE of my problem in understanding the definition of a tensor product.

According to my definition, we construct $S\otimes_R N$ by considering the free $\mathbb{Z}$ -module over $S\times N$ . But $S\times N$ is supposed to be an abelian group in its own right, and I don't understand how. For example, one idea would be to say that since $S$ and $N$ are both abelian groups individually, then we can just carry over the addition operations from each of them so that $(s_1,n_1)+(s_2,n_2)=(s_1+s_2,n_1+n_2)$ . Yet clearly this is not the case. But if not that, then how do we perform operations inside the group $S\times N$ ?

Here's what my textbook has to say on the subject: link

Let S be an R-module and N be a left R-module. You may already know that $S\otimes_R N$ is a left R-module such that $r(s \otimes n)=rs \otimes n$ for all $r \in R, s\in S,$ and $n \in N$ .

It is abelian (by the definition of R-module) in the sense that $m_1 \otimes n_1 + m_2 \otimes n_2 = m_2 \otimes n_2 + m_1 \otimes n_1$ , where $m_1 \otimes n_1$ and $m_2 \otimes n_2$ are elements in R-module $S\otimes_R N$ .

**topspin1617** · March 7th 2011, 07:56 AM

The previous post gives a good explanation.

For an actual PROOF that $2\otimes1\neq 0$ , try this:

Define the map $\varphi :2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\rightarrow \mathbb{Z}/2\mathbb{Z}$ by $(a,b)\mapsto \frac{ab}{2}(\mathrm{mod}\,2)$ .

You can prove (unless I'm wrong about this, which is possible..) that $\varphi$ is a linear map; thus, it corresponds to a homomorphism $\Phi :2\mathbb{Z}\otimes \mathbb{Z}/2\mathbb{Z}\rightarrow \mathbb{Z}/2\mathbb{Z}$ . Also notice that $\varphi (2\otimes 1)=1$ , so $\Phi (2\otimes 1)=1$ , which implies $2\otimes 1\neq 0$ .

EDIT: Could also do this: use the result of example 8 to see that $2\mathbb{Z}\otimes \mathbb{Z}/2\mathbb{Z}\cong 2\mathbb{Z}/(2\mathbb{Z}\cdot 2\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}\neq 0$ and, as you noted earlier, $2\otimes 1=0\Rightarrow 2\mathbb{Z}\otimes \mathbb{Z}/2\mathbb{Z}=0$ .

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