## Tag Archives: tensor product

### When the tensor product of finite field extensions is a field

Let $K_1$ and $K_2$ be finite extensions of a field $F$ contained in a field $K$. Prove that the $F$-algebra $K_1 \otimes_F K_2$ is a field if and only if $[K_1K_2 : F] = [K_1:F][K_2:F]$.

First, define $\psi : K_1 \times K_2 \rightarrow K_1K_2$ by $(a,b) \mapsto ab$. Clearly $\psi$ is $F$-bilinear, and so induces an $F$-module homomorphism $\Psi : K_1 \otimes_F K_2 \rightarrow K_1K_2$. In fact $\Psi$ is an $F$-algebra homomorphism. Using Proposition 21 in D&F, if $A = \{\alpha_i\}$ and $B = \{\beta_j\}$ are bases of $K_1$ and $K_2$ over $F$, then $AB = \{\alpha_i\beta_j\}$ spans $K_1K_2$ over $F$. In particular, $\Psi$ is surjective.

Suppose $K_1 \otimes_F K_2$ is a field. Now $\mathsf{ker}\ \Psi$ is an ideal of $K_1 \otimes_F K_2$, and so must be trivial- so $\Psi$ is an isomorphism of $F$-algebras, and thus an isomorphism of fields. Using Proposition 21 on page 421 of D&F, $K_1 \otimes_F K_2$ has dimension $[K_1:F][K_2:F]$, and so $[K_1K_2 : F] = [K_1:F][K_2:F]$ as desired.

Conversely, suppose $[K_1K_2 : F] = [K_1:F][K_2:F]$. That is, $K_1 \otimes_F K_2$ and $K_1K_2$ have the same dimension as $F$-algebras. By Corollary 9 on page 413 of D&F, $\Psi$ is injective, and so $K_1 \otimes_F K_2$ and $K_1K_2$ are isomorphic as $F$-algebras, hence as rings, and so $K_1 \otimes_F K_2$ is a field.

### A fact about the rank of a module over an integral domain

Let $R$ be an integral domain with quotient field $F$ and let $M$ be any (left, unital) $R$-module. Prove that the rank of $M$ equals the dimension of $F \otimes_R M$ over $F$.

Recall that the rank of a module over a domain is the maximal number of $R$-linearly independent elements.

Suppose $B \subseteq M$ is an $R$-linearly independent set, and consider $B^\prime \subseteq F \otimes_R M$ consisting of the tensors $1 \otimes b$ for each $b \in B$. Suppose $\sum \alpha_i (1 \otimes b_i) = 0$. For some $\alpha_i \in F$. Clearing denominators, we have $\sum r_i ( 1 \otimes b_i) = 0$ for some $r_i \in R$. Now $\sum 1 \otimes r_ib_i = 0$, and we have $1 \otimes \sum r_ib_i = 0$. By this previous exercise, there exists a nonzero $r \in R$ such that $\sum rr_ib_i = 0$. Since $B$ is linearly independent, we have $rr_i = 0$ for each $i$, and since $r \neq 0$ and $R$ is a domain, $r_i = 0$ for each $i$. Thus $\alpha_i = 0$ (since the denominators of each $\alpha_i$ are nonzero). So $B^\prime$ is $F$-linearly independent in $F \otimes_R M$. In particular, we have $\mathsf{rank}_R(M) \leq \mathsf{dim}_F(F \otimes_R M)$.

Now note that if $b_i^\prime = \sum \alpha_{i,j} \otimes m_{i,j} \in F \otimes_R M$, is a linearly independent set, with $\alpha_{i,j} = r_{i,j}/s_{i,j}$, then ‘clearing denominators’ by multiplying $b_i^\prime$ by $\prod_j s_{i,j}$, we have a second linearly independent set with the same cardinality whose elements are of the form $1 \otimes m_i$ for some $m \in M$. Suppose there exist $r_i \in R$ such that $\sum r_im_i = 0$; then $\sum r_i(1 \otimes b_i) = 0$, and (since the $1 \otimes b_i$ form a basis) we have $r_i = 0$. So the $m_i$ are $R$-linearly independent in $M$. Thus $\mathsf{rank}_R(M) \geq \mathsf{dim}_F(R \otimes_R M)$.

### A fact about the Alt map on a tensor power

Let $F$ be a field in which $k!$ is a unit and let $V$ be an $F$-vector space. Recall that $S_k$ acts on the tensor power $\mathcal{T}^k(V)$ by permuting the components, and that $\mathsf{Alt}_k$ is defined on $\mathcal{T}^k(V)$ by $\mathsf{Alt}_k(z) = \sum_{\sigma \in S_k} \epsilon(\sigma) \sigma z$. Prove that $\mathsf{im}\ \mathsf{Alt}_k$ is the unique largest subspace of $\mathcal{T}^k(V)$ on which each permutation $\sigma \in S_k$ acts as multiplication by $\epsilon(\sigma)$.

Suppose $z \in \mathcal{T}^k(V)$ such that $\sigma z = \epsilon(\sigma) z$ for all $\sigma \in S_k$. Then $\mathsf{Alt}_k(z) = \frac{1}{k!} \sum_{\sigma \in S_k} \epsilon(\sigma) \sigma z$ $= \frac{1}{k!} \sum_{\sigma \in S_k} \epsilon(\sigma) \epsilon(\sigma) z$ $= \frac{1}{k!} \sum_{\sigma \in S_k} z$ $= z$, so that $z \in \mathsf{im}\ \mathsf{Alt}_k$.

In particular, any subspace of $\mathcal{T}^k(V)$ upon which every permutation $\sigma \in S_k$ acts as scalar multiplication by $\epsilon(\sigma)$ is contained in $\mathsf{im}\ \mathsf{Alt}_k$.

Note also that $\sigma \in S_k$ acts on $\mathsf{im}\ \mathsf{Alt}_k$ as multiplication by $\epsilon(\sigma)$ since $\mathsf{im}\ \mathsf{Alt}_k \cong_F \bigwedge^k(V)$.

### Exterior powers of subdomains of a field of fractions

Let $R$ be an integral domain, and let $F$ be its field of fractions.

1. Considering $F$ as an $R$-module in the usual way, prove that $\bigwedge^2(F) = 0$.
2. Let $I \subseteq F$ be a submodule. Prove that $\bigwedge^k(I)$ is torsion for all $k \geq 2$.
3. Exhibit an integral domain $R$ and an $R$-submodule $I \subseteq F$ such that $\bigwedge^k(I) \neq 0$ for all $k \geq 0$.

Let $\frac{a}{b} \otimes \frac{c}{d} \in \mathcal{T}^2(F)$ be a simple tensor. Note that $\frac{a}{b} \otimes \frac{c}{d} = \frac{ad}{bd} \otimes \frac{cb}{bd}$ $= abcd(\frac{1}{bd} \otimes \frac{1}{bd}) \in \mathcal{A}^2(F)$. In particular, we have $\frac{a}{b} \wedge \frac{c}{d} = 0$ in $\bigwedge^2(I)$.

Suppose $\frac{a_1}{b_1} \wedge \frac{a_2}{b_2} \wedge \cdots \wedge \frac{a_k}{b_k} \in \bigwedge^k(I)$ be nonzero; then the $a_i$ are nonzero, and certainly the $b_i$ are nonzero. Now $a_1a_2b_1b_2 \neq 0$ in $R$, and evidently $a_1a_2b_1b_2(\frac{a_1}{b_1} \wedge \frac{a_2}{b_2} \wedge \cdots \wedge \frac{a_k}{b_k}) = \frac{a_1a_2}{1} \wedge \frac{a_1a_2}{1} \wedge \cdots \wedge \frac{a_k}{b_k} = 0$. So every element of $\bigwedge^k(I)$ is torsion, and thus $\bigwedge^k(I)$ is torsion as an $R$-module.

Now consider $R = \mathbb{Z}[x_1,\ldots,x_n]$, and let $I = (x_1,\ldots,x_n)$. We claim that if $\sum \alpha_ix_i = \sum \beta_ix_i \in I$, then there exist $h_i \in I$ such that $\alpha_i - \beta_i = h_i$. To see this, choose some $j$ and consider $\alpha_jx_j - \beta_jx_j = \sum_{i \neq j} (\beta_i - \alpha_i)x_i$. Since $R$ is a domain, $x_j$ divides the right hand side of this equation; say $\sum_{i \neq j} (\beta_i - \alpha_i)x_i = x_jh_j$. Note that every term on the left hand side is divisible by some $x_i$ other than $x_j$. Hence every term in $h_j$ is divisible by some $x_i$ other than $x_j$. In particular, $h_j \in I$, and $\alpha_i - \beta_i = h_i$.

Now consider the elements of $\prod^k I$ as “column vectors”- that is, write $\sum \alpha_i x_i$ as $[\alpha_1\ \alpha_2\ \ldots\ \alpha_n]^\mathsf{T}$. Note that this does not give a unique representation of the elements of $I$. However, if two column vectors $A$ and $B$ represent the same element in $I$, then there exists a third column vector $H$ such that $A = B+H$, and moreover the entries of $H$ are in $I$.

Now write the elements of $R^k$ as (square) matrices. Let us consider the determinant of such a matrix $A$, as an element of $R$, reduced mod $I$. This is an alternating bilinear map on the set of matrices over $R$; we claim that this is also well-defined up to our nonunique identification of matrices over $R$ with elements of $I^k$. To that end, suppose matrices $A$ and $B$ represent the same element of $I^k$; then we have a matrix $H$ such that $A = B+H$. Consider the determinant of both sides mod $I$. Using the combinatorial formula for computing determinants, we have $\mathsf{det}(B+H) = \sum_{\sigma \in S_n} \epsilon(\sigma) \prod (\beta_{\sigma(i),i} + h_{\sigma(i),i})$. Note that each $h_{i,j}$ is divisible by some $x_i$, and so goes to zero in the quotient $R/I$. So in fact $\mathsf{det}(A) = \mathsf{det}(B+H) \equiv \mathsf{det}(B) \mod\ I$; thus the map $\mathsf{det} : I^k \rightarrow R/I$ is a well-defined alternating bilinear map. Since $\mathsf{det}(x_1 \otimes \cdots \otimes x_n) = 1$ is nonzero, this map is nontrivial. Thus $\bigwedge^k(I) \neq 0$ for all $k$.

### Exhibit a nontrivial alternating bilinear map from a tensor square

Let $R = \mathbb{Z}[x,y]$ and $I = (x,y)$.

1. Prove that if $ax + by = a^\prime x + b^\prime y$ in $R$, then there exists $h \in R$ such that $a^\prime = a + yh$ and $b^\prime = b - xh$.
2. Prove that the map $\varphi(ax+by,cx+dy) = ad-bc \mod (x,y)$ is a well-defined, alternating, bilinear map $I \times I \rightarrow R/I$.

Note that $(a-a^\prime)x = (b^\prime - b)y$. Since $x$ and $y$ are irreducible (hence prime) in the UFD $R$, we have that $y$ divides $a - a^\prime$ and $x$ divides $b^\prime - b$. Say $a - a^\prime = yh$ and $b^\prime - b = xh^\prime$. Since $R$ is a domain, we see that $h = h^\prime$, and thus $a^\prime = a - yh$ and $b^\prime = b + xh$.

Suppose now that $a_1x + b_1y = a_2x + b_2y$ and $c_1x + d_1y = c_2x + d_2y$. Then there exist $h,g \in R$ such that $a_2 = a_1 - yh$, $b_2 = b_1 + xh$, $c_2 = c_1 - yg$, and $d_2 = d_1 + xg$. Evidently, $a_2d_2 - b_2c_2 \equiv a_1d_1 - b_1c_1 \mod (x,y)$. Thus $\varphi$ is well defined. Note that $\varphi((a_1+a_2)x + (b_1+b_2)y, cx+dy) = (a_1+a_2)d - (b_1+b_2)c$ $= (a_1d - b_1c) + (a_2d - b_2c)$ $= \varphi(a_1x+b_1y,cx+dy) + \varphi(a_2x+b_2y,c_x+d_y)$, so that $\varphi$ is linear in the first argument; similarly it is linear in the second argument. Moreover, $\varphi((ax+by)r,cx+dy) = \varphi(arx+bry,cx+dy)$ $= ard-brc$ $= \varphi(ax+by,r(cx+dy))$. Thus $\varphi$ is $R$-bilinear. Since $\varphi(ax+by,ax+by) = ab-ab = 0$, $\varphi$ is alternating. Finally, note that $\varphi(1,0) = 1$. Thus there exists a nontrivial alternating $R$-bilinear map on $I \times I$.

### A fact about 2-tensors in the tensor algebra of the abelian group ZZ

Compute the image of the map $\mathsf{Sym}_2$ on $\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z}$. Conclude that the symmetric tensor $1 \otimes 1$ is not in the image of $\mathsf{Sym}_2$.

Recall that $\mathsf{Sym}_2 : \mathcal{T}^2(M) \rightarrow \mathcal{T}^2(M)$ is given by $\mathcal{Sym}_2(z) = \sum_{\sigma \in S_2} \sigma \cdot z$, where the action of $S_n$ on $\mathcal{T}^n(M)$ permutes the entries of a tensor.

Let $a \otimes b \in \mathcal{T}^2(M)$. Now $\mathsf{Sym}_2(a \otimes b) = a \otimes b + b \otimes a$ $= 2 (a \otimes b)$, using the fact that $\mathbb{Z}$ has a 1. In particular, $\mathsf{im}\ \mathsf{Sym}_2 \subseteq \{k(a \otimes b) \ |\ a, b \in \mathbb{Z}, k \in 2\mathbb{Z} \}$. The reverse inclusion is clear as well, so that these sets are in fact equal.

Suppose now that $1 \otimes 1 = \mathsf{Sym}_2(a \otimes b) = 2(a \otimes b)$. Consider the map $\psi : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}/(2)$ given by $(a,b) \mapsto \overline{ab}$. This map is clearly $\mathbb{Z}$-bilinear, and so induces an abelian group homomorphism $\overline{\psi} : \mathcal{T}^2(\mathbb{Z}) \rightarrow \mathbb{Z}/(2)$ by the universal property of tensor products. Note that, under this map, $\overline{\psi}(1 \otimes 1) = \overline{1}$, while $\overline{\psi}(2(a \otimes b)) = \overline{0}$. Thus $1 \otimes 1$ is not in the image of $\mathsf{Sym}_2$.

### The matrix of an extension of scalars

Let $K$ be a field and let $F \subseteq K$ be a subfield. Let $\psi : V \rightarrow W$ be a linear transformation of finite dimensional vector spaces over $F$.

1. Prove that $1 \otimes \psi : K \otimes_F V \rightarrow K \otimes_F W$ is a $K$-linear transformation.
2. Fix bases $B$ and $E$ of $V$ and $W$, respectively. Compare the matrices of $\psi$ and $1 \otimes \psi$ with respect to these bases.

Note that since $K$ is an $(K,F)$-bimodule, $1 \otimes \psi$ is a $K$-module homomorphism- that is, a linear transformation.

Let $B = \{v_i\}$ and $E = \{w_i\}$. We claim that $B^\prime = \{ 1 \otimes v_i \}$ and $E^\prime = \{1 \otimes w_i\}$ are bases of $K \otimes_F V$ and $K \otimes_F W$, respectively, as $K$ vector spaces. In fact, we showed this in a previous exercise. (See part (1).)

Suppose $\psi(v_j) = \sum a_{i,j} w_i$. Then $(1 \otimes \psi)(1 \otimes v_j) = \sum a_{i,j} (1 \otimes w_j)$. Thus the columns of $M^E_B(\psi)$ and $M^{E^\prime}_{B^\prime}(\psi)$ agree, and we have $M^E_B(\psi) = M^{E^\prime}_{B^\prime}(\psi)$.

### A module is flat if and only if the tensor product of 1 with any injective homomorphism from a finitely generated module is injective

Let $R$ be a ring and let $A$ be a unital right $R$-module. Prove that $A$ is flat if and only if for all injective left $R$-module homomorphisms $\psi : L \rightarrow M$ where $L$ is finitely generated, the map $1 \otimes \psi : A \otimes_R L \rightarrow A \otimes_R M$ is injective.

Certainly if $A$ is flat, then every such map $1 \otimes \psi$ is injective.

Suppose now that $A$ is a right $R$-module such that for all injective left $R$-module homomorphisms $\psi : L \rightarrow M$ with $L$ finitely generated, $1 \otimes \psi$ is injective. Let $\psi : L \rightarrow M$ be an arbitrary injective module homomorphism. Suppose $(1 \otimes \psi)(\sum a_i \otimes \ell_i) = 0$. Now there exists a finitely generated submodule $L^\prime \subseteq L$ containing all of the $\ell_i$. (For example, the submodule generated by the $\ell_i$.) The restriction $\psi^\prime$ of $\psi$ to $L^\prime$ is injective by our hypothesis, and certainly $(1 \otimes \psi)(\sum a_i \otimes \ell_i) = (1 \otimes \psi^\prime)(\sum a_i \otimes \ell_i)$. Thus $\sum a_i \otimes \ell_i = 0$. In particular, $\mathsf{ker}\ 1 \otimes \psi = 0$, so that $1 \otimes \psi$ is injective.

Thus $A$ is flat.

### The change of base of a flat module is flat

Let $R$ and $S$ be rings with 1 and let $\theta : R \rightarrow S$ be a ring homomorphism such that $\theta(1) = 1$. Make $S$ into a left $R$-module by $r \cdot s = \varphi(r)s$. In fact, $S$ is an $(R,S)$-bimodule. Let $M$ be a right, unital $R$-module. Show that if $M$ is flat as an $R$-module, then $M \otimes_R S$ is flat as a right $S$-module.

Let $\psi : L \rightarrow N$ be an injective homomorphism of left $S$-modules. Note that $S$ is free as a right $S$-module, hence flat, so that $1 \otimes \psi : S \otimes_S L \rightarrow S \otimes_S N$ is injective. Now we claim that $1 \otimes \psi$ is a homomorphism of left $R$-modules. To see this, let $r \in R$ and let $s \otimes \ell$ be a simple tensor in $S \otimes_S L$. Now $(1 \otimes \psi)(r \cdot (s \otimes \ell)) = (1 \otimes \psi)(rs \otimes \ell)$ $= rs \otimes \psi(\ell)$ $= r \cdot (s \otimes \psi(\ell))$ $= r \cdot (1 \otimes \psi)(s \otimes \ell)$. Since $M$ is flat as a right $R$-module, $1 \otimes (1 \otimes \psi) : M \otimes_R (S \otimes_S L) \rightarrow M \otimes_R (S \otimes_S N)$ is injective. Via the canonical isomorphisms, $1 \otimes \psi : (M \otimes_R S) \otimes_S L \rightarrow (M \otimes_R S) \otimes_S N$ is injective. Thus $M \otimes_R S$ is flat.

### Over a commutative ring, the tensor product of two flat modules is flat

Let $R$ be a commutative ring and let $M$ and $N$ be $R$-modules. Show that if $M$ and $N$ are flat over $R$, then $M \otimes_R N$ is flat over $R$.

This is a special case of the previous exercise, since both $M$ and $N$ can naturally be considered $(R,R)$-bimodules.