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3.3. Generating Modules, Torsions, Annihilators.

The concepts we have introduced so far are not new. In fact, this is the third time you've probably seen all of these concepts. However, module theory is very deep, and here is where we will start seeing new concepts. \ \ Generating Modules.\ Let \(M\) be an \(R\)-module and suppose \(S \subset M\) where \(S\) is nonempty. Denote the smallest submodule of \(M\) containing \(S\) as \(\left< S \right>\), and observe that

\[ \left< S \right> = \bigcap_{\alpha \in \lambda} S_{\alpha} \]

where \(\{S_\alpha\}_{\alpha \in \lambda}\) is a family of submodules containing \(S\).

Note that this is in fact a submodule, since arbitrary intersections of submodules yield a submodule.

Now consider the set of all finite linear combinations of elements of \(S\) with coefficients in \(R\). That is,

\[ S' = \Big\{\sum_{i = 1}^{n} a_is_i \mid a_i \in R, s_i \in S \text{ for all } i \in N\Big\}. \]

This is of course also an \(R\)-module. We claim that \(\left< S \right> = S'\).

  • \phantom{meow} \(\bf{\left< S \right> \subset S'.}\) To see this, consider any \(s \in \left<S\right>\). Then \(s \in S_{\alpha}\) for all \(\alpha \in \lambda\). Furthermore, since \(S \subset S'\) we see that \(S'\) is one of the members of the families of all submodules containing \(S\). Therefore we must have that \(s \in S'\), and thus \(\displaystyle s = \sum_{i = 1}^{n}a_is_i\) for some \(a_i \in R\) and \(s_i \in S\). Hence, \(\left<S \right> \subset S'\).
  • \phantom{meow} \(\bf{S' \subset \left< S \right>}.\) Simply observe that since \(\left< S \right>\) contains \(S\), and because \(\left< S \right>\) is a submodule, it must be that \(\left< S \right>\) contains all linear combinations of elements of \(S\) with coefficients in \(R\). That is, \(\displaystyle \sum_{i = 1}^{n} a_is_i \in \left< S \right>\) for any \(a_i \in R, s_i \in S\).

Thus what we have shown is the following theorem.

Let \(M\) be an \(R\)-module and suppose \(S \subset M\). Then if \(\left< S \right>\) is the smallest submodule of \(M\) containing \(S\) then

\[ \left< S \right> = \bigcap_{\alpha \in \lambda} S_\alpha \]

where \(\{S_\alpha\}_{\alpha \in \lambda}\) is the family of submodules containing \(S\). More explicilty, we have that

\[ \left< S \right> = \Big\{\sum_{i = 1}^{n} a_is_i \mid a_i \in R, s_i \in S \text{ for all } i \in N\Big\}. \]

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The above theorem then leads a very useful definition that we will work with frequently.

Let \(M\) be an \(R\)-module and \(S \subset M\).

  • 1. The \(R\)-submodule \(\left< S \right>\) of \(M\) is called the submodule of \(M\) generated by \(S\).
  • 2. If \(M = \left< S \right>\) for some \(S \subset M\), then we say that \(M\) is generated by \(S\). Hence, the elements of \(S\) are referred to as the generators of \(M\).
  • 3. If \(M\) is generated by \(S\) and \(S\) is finite, then we say \(M\) is finitely generated by \(S\). In this case we refer to \(|S|\), denoted as \(\mu(M)\), as the rank of \(M\). Furthermore, if \(S = \{x_1, x_2, \dots, x_n\}\) it is convenient to write \(M = \left< x_1, x_2, \dots, x_n \right>\).

In the case of a cyclic group \(G\), we see that \(G\) has rank one and one generator. Thus we see that this concept is generalized in module theory if the generating set for some module is of cardinality one.

Note we can also union modules together to get another module.

Let \(M\) be an \(R\)-module and \(\{S_{\alpha}\}\) a family of submodules of \(R\). Then the **submodule generated by \(\{N_{\alpha**\}_\{\alpha \in \lambda\}\)} is

\[ \left< \bigcup_{\alpha \in \lambda}S_\alpha\right> \]

which we often denote as \(\displaystyle \sum_{\alpha \in \lambda}S_{\alpha}\).

Next we move onto the concept of an annihilator, which we define as follows.

Let \(M\) be an \(R\)-module, and suppose \(X \subset M\). Then we define the set

\[ \ann(X) = \{a \in R \mid ax = 0 \text{ for all }x \in X\} \]

to be the annihilator of \(X\). \textcolor{NavyBlue}{Note that \(\ann(X) \subset R\) and that \(0 \in \ann(X)\) for any \(X \subset M\). If \(\ann(X) = \{0\}\), we of course say that it is trivial.}

The annihilator captures all of the coefficients of \(R\) which annihilate every element in \(X\). Thus one can imagine that the size of \(\ann(X)\) increases as the size of \(X \subset M\) increases (of course, if \(\ann(X)\) is not trivial for all \(X\).)

Let \(M\) be an \(R\)-module and let \(X \subset M\) be nonempty. Then

  • 1. \(\ann(X)\) is a left ideal of \(R\)
  • 2. If \(N\) is a submodule of \(M\), then \(\ann(N)\) is an ideal of \(R\).
  • 3. If \(R\) is commutative and \(N\) is a cyclic submodule of \(M\) generated by \(x \in N\) then \(\ann(N) = \{a \in R \mid ax = 0 \}\).

  • 1. Let \(X \subset M\). In order for \(\ann(X)\) to be a left ideal, it must be a subring of \(R\) which absorbs left multiplication.

    • It's a Subring. We can apply the subring test to prove this. Recall earlier we said that \(0 \in \ann(X)\) for any \(X \subset M\), so \(\ann\) is nonempty.

    Now let \(a, b \in \ann(X)\), so that \(ax = bx = 0\) for all \(x \in X\). Then clearly \(abx = 0\) and \((a - b)x = ax -bx = 0\) so that \(ab \in \ann(X)\) and \(a-b \in \ann(X)\). Thus it is a subring of \(R\). * It's an ideal. For any \(r \in R\) and \(a \in \ann(X)\) we have that \(rax = r(ax) = 0\). Therefore \(ra\in I\) for all \(r \in R\), which proves it is a left ideal. \ \textcolor{Plum}{Why isn't it also a right ideal? Well, observe that we would need \(arx = 0\) whenever \(a \in \ann(X)\) and \(r \in R\). This would require either that \(ar = 0\), which we can't always guarantee, or that \(rx \in X\). But we don't know if \(rx \in X\); we could only guarantee that if \(X\) was an \(R\)-module, which we'll see in the next proof. } * 2. Let \(N\) be a submodule of \(M\). Since we showed in (1.) that \(\ann(X)\) is a left ideal for any \(X \subset M\), we must simply show that \(\ann(N)\) also absorbs right multiplication of \(R\) as well in order to show it is an ideal.

Thus let \(r\in R\) and \(a \in \ann(N)\). Since \(N\) is a submodule of \(N\) we know that \(rx \in N\). Hence \(a(rx) = 0\) as \(an = 0\) for all \(n \in N\). Therefore \(ar \in \ann(N)\) whenever \(r \in R\) and \(a \in \ann(N)\), proving that \(\ann(X)\) absorbs right multiplication and is therefore an ideal. * 3. Consider \(\ann(N)\) where \(N\) is cyclic and generated by \(x\) and let \(|N| = k\). Suppose we have an \(a \in R\) such that \(ax = 0\). Then we see that \(ax^2 = (ax)x = 0\), \(ax^3 = (ax)x^2 = 0\), and that in general \(ax^j = (ax)x^{j-1} = 0\) for any \(j \in \{1, 2, \dots, k\}\). Since every \(n \in N\) is of the form \(x^j\) for some \(j \in \{1, 2, \dots, k\}\) we have \(an = 0\) for all \(n \in N\). Hence,

\[ \ann(X) = \{a \in R \mid an = 0 \text{ for all } n \in N\} = \{a \in R \mid ax = 0\}. \]

\textcolor{red}{Where does commutativity come into play?}


Note that in general we denoted \(\ann(x)\) to be \(\ann(N)\) where \(N\) is cyclic and generated by \(x\). This only really makes sense if \(R\) is commutative.

For an \(R\)-module \(M\) and a family of submodules \(\{N_\alpha\}_{\alpha \in \lambda}\), we have been able to define the arbitrary intersection and addition of submodules. Next we define the product of \(R\)-modules. We now define a product of \(R\)-modules.

If \(M\) is an \(R\)-module and \(I\) is an ideal, then we can define

\[ IM = \Big\{ \sum_{i = 1}^{n}a_im_i \mid a_i \in R, m_i \in M \text{ for } n \in N \Big\} \]

which is a submodule of \(R\). Thus our main properties of submodules are intersection, addition, and products with ideals.

Let \(R\) be an integral domain and \(M\) an \(R\)-module. Suppose \(x \in M\). Then

  • 1. If \(\ann(x) \ne \{0\}\) then we define \(x\) to be torsion element. We define the set of torsion elements of \(M\) to be the torsion submodule and denote this as \(M_{\tau}\).
  • 2. If \(M_\tau = \{0\}\) then we say \(M\) is torsion free, while if \(M_\tau = M\) we say that \(M\) is torsion module.

Let \(R\) be an integral domain and \(M\) an \(R\)-module. Then

  • 1. \(M_\tau\) is a submodule of \(M\).
  • 2. \(M/M_\tau\) is torsion free.

  • 1. We can use the submodule test to show that this is a submodule of \(M\). First, recall that \(M_\tau\) is nonempty as it always contains 0. Let \(a, b \in R\) and suppose \(m_1, m_2 \in M_\tau\). Then observe that $(am_1 + bm_2)x = am_1x
  • bm_2x = 0$ since \(m_1x = 0\) and \(m_2x = 0\). Hence, this is a submodule.
  • 2. Suppose that \((m + M_\tau)x = M_\tau\) where \(m \in M\) for some \(x \in M\). Then this implies that \(mx \in M_\tau\). Therefore there exists a \(n \in R\) such that \(n(mx) = (nm)x = 0\). Since \(R\) is an integral domain, \(nm \ne 0\) so that we must have \(x = 0\). Thus the torsion module is trivial.

Let \(R\) be an integral domain and suppose \(M = \left< x_1,x_2, \dots, x_n \right>\) (that is, \(M\) is finitely generated). Then

\[ \ann(M) = \ann(x_1) \cap \ann(x_2) \cap \dots \cap \ann(x_n). \]

Note that \(0 \in \ann(x_i)\) for all \(i \in \{1, 2, \dots, n\}\). Hence, the above intersection is never empty.

  • \phantom{m} \(\bf{\ann(M) \subset \ann(x_1) \cap \ann(x_2) \cap \dots \cap \ann(x_n)}\). Observe that for any \(a \in \ann(M)\), we see that \(ax = 0\) for all \(x \in M\). In particular, \(ax_i = 0\) for \(i \in \{1, 2, \dots, n\}\). Hence we see that \(a \in \ann(x_i)\) for each \(x_i\), so that \(a \in \ann(x_1) \cap \ann(x_2) \cap \dots \cap \ann(x_n)\).
  • \phantom{m} \(\bf{\ann(x_1) \cap \ann(x_2) \cap \dots \cap \ann(x_n) \subset }\bf{\ann(M)}\). Observe that for $ a \in \ann(x_1) \cap \ann(x_2) \cap \dots \cap \ann(x_n)$ we see that \(ax_i = 0\) for each \(i \in \{1, 2, \dots, n\}\).

By 1.\ref{submodule_intersection_theorem} we know that for any \(m \in M\) we have that \(\displaystyle m = \sum_{i = 1}^{n}a_ix_i\) for some \(a_i \in R\) and \(x_i \in M\). But note that

\[ am = a\sum_{i =1}^{n}a_ix_i = \sum_{i =1}^{n}aa_ix_i = \sum_{i =1}^{n}a_i(ax_i) =0 \]

where in the last step we used the commutativity of \(R\). Therefore \(a \in \ann(X)\), proving that \(\ann(x_1) \cap \ann(x_2) \cap \dots \cap \ann(x_n) \subset \ann(M)\).

With both directions of the proof complete, we can conclude that \(\ann(M) = \ann(x_1) \cap \ann(x_2) \cap \dots \cap \ann(x_n)\) as desired.