Group Generation Through Exterior Algebras (11.5.9)

Dummit and Foote Abstract Algebra, section 11.5, exercise 9:

Let $R=\mathbb{Z}G$ be the group ring over the group $G = \{1,σ\}$. Letting $M = \mathbb{Z} × \mathbb{Z}$ and defining $σ(e_1)=e_1+2e_2, σ(e_2)=-e_2$, show $M$ is an $R$-module and $\bigwedge^2 M$ is a group of order 2 generated by $e_1 \wedge e_2$.

Proof: Since there is a unique representation for $r∈R$ and $m∈M$ we may define the action of $R$ on $M$ from their mutual bases' actions to obtain $rm = (z_1σ+z_2)(a,b) = (z_1a+z_2a,2z_1a-z_1b+z_2b)$. Note that $σ(σ(a,b)) = (a,b)$ so that it satisfies action associativity.

Observe$$(a,b) \wedge (c,d) = ( (a,0)+(0,b) ) \wedge ( (c,0) + (0,d) ) = (ad+bc)(e_1 \wedge e_2)$$so that $\bigwedge^2 M$ is generated by $e_1 \wedge e_2$ as an $R$-module. As well, we have $σ(e_1 \wedge e_2) = (σe_1) \wedge e_2 = (e_1+2e_2) \wedge e_2 = e_1 \wedge e_2$ and $σ(e_1 \wedge e_2) = e_1 \wedge (σe_2) = e_1 \wedge -e_2 = -(e_1 \wedge e_2)$ so that $(z_1σ+z_2)(e_1 \wedge e_2) = (\overline{z_1+z_2})(e_1 \wedge e_2)$ where the overline denotes reduction modulo 2, implying $\bigwedge^2 M$ is either of order 1 or 2. To prove the latter, we shall construct a bilinear alternating form $φ$ on $M$ such that $φ(e_1,e_2) \neq 0$.

First, define $ψ : R → \mathbb{Z}/2\mathbb{Z}$ by $ψ(z_1σ+z_2)=\overline{z_1+z_2}$ and prove it is a ring homomorphism$$ψ(z_1σ+z_2+z_1'σ+z_2')=\overline{z_1+z_2+z_1'+z_2'}=ψ(z_1σ+z_2)+ψ(z_1'σ+z_2')$$$$ψ((z_1σ+z_2)(z_1'σ+z_2'))=ψ((z_1z_2'+z_2z_1')σ+(z_1z_1'+z_2z_2'))=$$$$\overline{z_1z_2'+z_2z_1'+z_1z_1'+z_2z_2'}=\overline{(z_1+z_2)(z_1'+z_2')}=ψ(z_1σ+z_2)ψ(z_1'σ+z_2')$$Now define $φ : M × M → \mathbb{Z}/2\mathbb{Z}$ by $φ((a,b),(c,d))=ψ(ad-bc)$. It alternates and is 1 on $(e_1,e_2)$ so all that remains to show is that it is bilinear by being additive in its components$$φ((a,b)+(a',b'),(c,d))=ψ((a+a')d-(b+b')c))=$$$$ψ(ad-bc)+ψ(a'd-b'c)=φ((a,b),(c,d))+φ((a',b'),(c,d))$$and linear over $R$$$φ((z_1σ+z_2)(a,b),(c,d))=φ((z_1a+z_2a,2z_1a+z_2b-z_1b),(c,d))=$$$$\overline{z_1ad+z_1bc+z_2ad-z_2bc}=\overline{z_1ad-z_1bc+z_2ad-z_2bc}=$$$$\overline{z_1(ad-bc)+z_2(ad-bc)}=(z_1σ+z_2)φ((a,b),(c,d))~\square$$

Exterior Algebras and Fraction Fields (11.5.8c)

Dummit and Foote Abstract Algebra, section 11.5, exercise 8(c):

(c) Give an example of an integral domain $R$ with fraction field $F$ and ideal $I⊆F$ considered as an $R$-module such that $\bigwedge^n I \neq 0$ for all $n$.

Proof: Let $R = \mathbb{Z}[x_1,...]$ and $I = (x_1,...)$. It suffices to find an alternating $n$-multilinear map $φ_n : I × ... × I → \mathbb{Z}$ ($n$ factors) such that $φ_n(x_1,...,x_n) = 1$ for all $n$. To that end, define $φ_n$ as follows,$$φ_n: I × ... × I → \mathbb{Z}$$$$φ_n(∑a_{1,i}x_i,...,∑a_{n,i}x_i)=\text{det }(a_{ij}')_{1≤i,j≤n}$$where $a'_{ij}$ is the constant term of $a_{ij}$. Note that $∑a_{i}x_i=∑b_{i}x_i$ implies that $a_i'=b_i'$ for all $i$, so that $(a_{ij}')_{1≤i,j≤n}$ is uniquely determined and $φ_n$ is well defined. This map is multilinear alternating on components of $I × ... × I$ just as $\text{det}$ is multilinear alternating on matrix rows. Here $(x_1,...,x_n)$ represents the identity matrix, and as such $φ_n(x_1,...,x_n) = 1$.$\square$

Common Annihilator Through Matrices (11.4.3)

Dummit and Foote Abstract Algebra, section 11.4, exercise 3:

Let $R$ be a commutative ring with $1$ and let $V$ be an $R$-module with $x_1,...,x_n∈V$. Letting $W$ be the column matrix of these elements, assume that for some $A∈M_{n × n}(R)$,$$AW=0$$Prove $(\text{det }A)x_i=0$ for $i∈\{1,...,n\}$.

Proof: This implies a system$$α_{11}x_1+α_{12}x_2+...+α_{1n}x_n=0$$$$α_{21}x_1+α_{22}x_2+...+α_{2n}x_n=0$$$$...$$$$α_{n1}x_1+α_{n2}x_2+...+α_{nn}x_n=0$$In fashion of constructing the cofactor formula for the determinant along the first column, multiply the first row by $\text{det }A_{11}$ and for $k∈\{2,...,n\}$ add $(-1)^{k+1}\text{det }A_{k1}$ times the $k^{th}$ row to the first row to obtain$$(\text{det }A)x_1+\sum_{j=2}^n (α_{1j}\text{det }A_{11}+\sum_{k=2}^n(-1)^{k+1}α_{kj}\text{det }A_{k1})x_j=0$$$$(\text{det }A)x_1+\sum_{j=2}^n (\sum_{k=1}^n(-1)^{k+1}α_{kj}\text{det }A_{k1})x_j=0~~~~~(*)$$For each $j$, let $B_j$ be the matrix $A$ with the first column replaced by the $j^{th}$ column. We see$$0 = \text{det }B_j = \sum_{k=1}^n(-1)^{k+1}β_{k1}\text{det }B_{k1} = \sum_{k=1}^n(-1)^{k+1}α_{kj}\text{det }A_{k1}$$so that $(*)$ collapses to$$(\text{det }A)x_1 = 0$$By interchanging arbitrary $x_i$ with $x_1$ and letting $A_i$ be the matrix $A$ with its first and $i^{th}$ column interchanged, since this operation negates the determinant and by the argument above we have$$-(\text{det }A)x_i = 0 = (\text{det }A)x_i~~~\square$$

Dual Annihilators (11.3.3)

Dummit and Foote Abstract Algebra, section 11.3, exercise 3:

Let $S \subseteq V$ for $V$ some finite dimensional space. Define $\text{Ann}(S)=\{v∈V~|~f(v)=0\text{ for all }f∈S\}$.

(a) Show $\text{Ann}(S)$ is a subspace of $V$.
(b) Let $W_1,W_2 \subseteq V^*$ be subspaces. Show $\text{Ann}(W_1+W_2)=\text{Ann}(W_1)∩\text{Ann}(W_2)$ and $\text{Ann}(W_1∩W_2)=\text{Ann}(W_1)+\text{Ann}(W_2)$.
(c) Prove $W_1=W_2 ⇔ \text{Ann}(W_1)=\text{Ann}(W_2)$.
(d) Prove $\text{Ann}(S)=\text{Ann}(\text{span }S)$.
(e) Assume $V$ has for basis $\{v_1,...,v_n\}$. Prove that if $S=\{v_1^*,...,v_k^*\}$ then $\text{Ann}(S)$ has basis $\{v_{k+1},...,v_n\}$.
(f) Assume $V$ is finite dimensional. Prove that if $W^* \subseteq V^*$ is a subspace then $\text{dim }\text{Ann}(W^*) = \text{dim }V - \text{dim }W^*$.

Proof: We shall approach this problem from another angle to obtain the final result first, and then recover the rest as implications. But first, (a):$$v_1,v_2∈\text{Ann}(S)⇒f(v_1+αv_2)=f(v_1)+αf(v_2)=0~~~~~∀f∈S$$ Let $S$ be a finite set of linearly independent linear functionals $v_1^*,...,v_k^*$, perhaps a basis for an arbitrary subspace of $V^*$. Define a linear transformation$$F : V → F^k$$$$F(v)=(v_1^*(v),...,v_k^*(v))$$Under this terminology, we can see $\text{Ann}(S)=\text{ker }F$. Letting $A$ be the $k × n$ reduced row echelon matrix of $F$ from basis $e_1,...,e_n$ (of $V$) to $e_1,...,e_k$ (of $F^k$) we observe the linear functionals manifest as linearly independent rows, so that the kernel is of dimension $n-k=\text{dim }V-\text{dim span }S$. This is (f), since (d) is immediately evident. (e) then follows easily, as the $⊇$ argument for the basis is clear and dimensional restrictions show it spans.

When $S ⊆ V$, define the subspace $\text{Ann}(S)=\{v^*∈V^*~|~v^*(s)=0~\text{for all }s∈S\}$ to be the dual notion of the $\text{Ann}$ defined above. A parallel argument by evaluations at $v_i$ gives $\text{dim Ann}(S)=\text{dim }V^*-\text{dim span }S$. When $S$ is a subspace of $V$ or of $V^*$ we show $\text{Ann}(\text{Ann}(S))=S$ since clearly $⊇$ holds, and dimension restrictions show $⊆$, which now easily gives (c).

We now show $\text{Ann}(W_1+W_2)=\text{Ann}(W_1)∩\text{Ann}(W_2)$ when $W_1,W_2$ are subspaces of $V^*$ (or $V$). ($⊇$) Let $v∈\text{Ann}(W_1)∩\text{Ann}(W_2)$. We have $(w_1+w_2)(v)=w_1(v)+w_2(v)=0$ (respectively, $v(w_1+w_2)=v(w_1)+v(w_2)=0$). ($⊆$) Let $v∈\text{Ann}(W_1+W_2)$. We have $w_1(v)=(w_1+0)(v)=0$ and similarly $w_2(v)=0$ (respectively, $v(w_1)=v(w_1+0)=0$ and $v(w_2)=0$). Now, to finish, we have$$\text{Ann}(W_1∩W_2)=\text{Ann}(W_1)+\text{Ann}(W_2)⇔$$$$\text{Ann}(\text{Ann}(W_1∩W_2))=\text{Ann}(\text{Ann}(W_1)+\text{Ann}(W_2))⇔$$$$W_1∩W_2=\text{Ann}(\text{Ann}(W_1))∩\text{Ann}(\text{Ann}(W_2))=W_1∩W_2~\square$$

Dual Endomorphism Ring (11.3.1)

Dummit and Foote Abstract Algebra, section 11.3, exercise 1:

Let $V$ be a vector space of finite dimension $n$. Prove that the map$$\psi : \text{End}(V) \rightarrow \text{End}(V^*)$$$$\psi(\varphi) = \varphi^*$$is an isomorphism of vector spaces. Show $\psi$ is not a ring isomorphism when $n \geq 2$. Exhibit an $F$-algebra isomorphism from $\text{End}(V)$ to $\text{End}(V^*)$.

Proof: Recall that $\text{dim } V = \text{dim } V^*$ implying $\text{dim End}(V) = \text{dim End}(V^*)$, so that it suffices to show $\psi$ is a nonsingular linear transformation. To show linearity, we must show$$\psi(\varphi_1 + \alpha \varphi_2) = \psi(\varphi_1)+\alpha \psi(\varphi_2)$$$$(\varphi_1 + \alpha \varphi_2)^*(f) = \varphi_1^*(f) + (\alpha \varphi_2^*)(f)~~~~~\forall f \in V^*$$$$(\varphi_1 + \alpha \varphi_2)^*(f)(v) = \varphi_1^*(f)(v) + (\alpha \varphi_2^*)(f)(v)~~~~~\forall v \in V$$$$f \circ (\varphi_1 + \alpha \varphi_2)(v) = f \circ \varphi_1^*(v) + f \circ \alpha \varphi_2^*(v)$$which follows from the linearity of $f$. To show nonsingularity, suppose $\psi(\varphi)=0$, which is to say $\varphi^*(f) = 0$ for all $f \in V^*$, which is to say $\varphi^*(f) (v) = f \circ \varphi (v) = 0$ for all $v \in V$. Assuming $\varphi$ is nonzero implies $\varphi(v) = \sum a_ie_i$ where some $a_k$ is nonzero. Letting $f$ be the element that sends $e_k$ to $1$ and all other $e_j$ to zero, we obtain a contradiction for $f \circ \varphi (v) = f(\sum a_ie_i) = a_k$.

Suppose $n \geq 2$ and $\psi$ is a ring isomorphism$$\psi(\varphi_1 \circ \varphi_2) = \psi(\varphi_1) \circ \psi(\varphi_2)$$$$\psi(\varphi_1 \circ \varphi_2)(f) = \psi(\varphi_1) \circ \psi(\varphi_2)(f)~~~~~\forall f \in V^*$$$$f \circ (\varphi_1 \circ \varphi_2)= f \circ (\varphi_2 \circ \varphi_1)$$Choosing $\varphi_1,\varphi_2$ as $n \times n$ matrices such that $\varphi_2 \circ \varphi_1 = 0$ but $\varphi_1 \circ \varphi_2$ is nonzero, we can choose $f$ that maps a single basis element involved in the nonzero image of an element $v$ under the latter to $1$ so that the equality is violated and $\psi$ is not a ring isomorphism. Note that this approach does prove $\psi$ is a ring isomorphism when $n=1$ as $1 \times 1$ matrices commute.

Let $V$ have basis $e_1,...,e_n$, so that $V^*$ has basis $e_1^*,...,e_1^*$, $\text{End}(V)$ has basis $e_{ab}$ (considered as the $n \times n$ matrix with $1$ in position $a,b$ and zeros elsewhere) and $\text{End}(V^*)$ has basis $e_{ab}^*$ for $1≤a,b≤n$. Define $ψ : \text{End}(V) → \text{End}(V^*)$ by its action on the basis of $\text{End}(V)$ via $ψ(e_{ab})=e_{ab}^*$. This extends to a linear transformation that sends basis to basis and as such is nonsingular. All that remains is to demonstrate multiplicativity. We may observe this$$ψ(φ_1 \circ φ_2)=ψ(φ_1) \circ ψ(φ_2)$$as the composition seen as matrix multiplication remains the same through the transformation.$~\square$

Span and Linear Dependence Computations (11.2.27)

Dummit and Foote Abstract Algebra, section 11.2, exercise 27:

Let $V$ be an $m$-dimensional vector space with basis $e_1,...,e_m$ and let $v_1,...,v_n$ be vectors in $V$. Let $A$ be the $m \times n$ matrix translating the vectors into this basis and let $A'$ be the reduced row echelon form of $A$.

(a) Let $B$ be any matrix row equivalent to $A$. Let $w_1,...,w_n$ be the vectors described by the columns of $B$. Prove that any linear relation$$x_1v_1+...+x_nv_n = 0$$implies$$x_1w_1+...+x_nw_n = 0$$(b) Prove that the vectors given by the pivotal columns of $A'$ are linearly independent and the rest are linearly dependent on these.
(c) Prove $v_1,...,v_n$ are linearly independent if and only if $A'$ has $n$ nonzero rows.
(d) By (c), the vectors $v_1,...,v_n$ are linearly dependent if and only if $A'$ has nonpivotal columns. The solutions to the linear dependence relations among $v_1,...,v_n$ are given by the linear equations defined by $A'$. Show that the variables $x_1,...,x_n$ corresponding to nonpivotal columns can be prescribed arbitrarily and the remaining variables are then uniquely defined to give the linear dependence relation.
(e) Prove that the subspace $W$ spanned by $v_1,...,v_n$ has dimension $r$ where $r$ is the number of nonzero rows of $A'$ and that a basis for $W$ is given by the original vectors corresponding to the pivotal columns of $A'$.

Proof: (a) Since the other two row operations can be replicated by addition of scalar-multiplied rows to one another, it suffices to prove the relations are preserved for one operation. Viewing the equation in the form of the original basis, we can see that the other unmodified rows (i.e. basis vectors) result in zero in the sum, and since the scalar-multiplied row to be added when multiplied and summed with the $x_i$ is already known to be zero, its addition to any of the rows does not change the sum.

(b) These pivotal columns must be of the form $e_i$, so are clearly independent. Every other vector is composed of a column whose nonzero entries have already been preceded by a pivotal element, and are thus generatable by these.

(c) Viewing $A'$ as $A$ under a nonsingular matrix multiplication, we can see that $v_1,...,v_n$ are linearly independent if and only if $w_1,...,w_n$ are linearly independent if and only if $A'$ has $n$ nonzero rows.

(d) After prescribing these variables arbitrarily, the partial sum represents a vector in basis vectors $e_1,...,e_n$ which are zero in coordinates not reached by a pivotal vector, so that there are unique scalars to the pivotal vectors to make this sum zero.

(e) Again viewing $A'$ as $A$ under a nonsingular linear transformation, we can see that the vectors mapping to the pivotal column vectors of $A'$ precisely form the basis for $W$, of which pivotal columns there are exactly $r$.

Stable Subspaces of Linear Transformations (11.2.9)

Dummit and Foote Abstract Algebra, section 11.2, exercise 9:

Let $φ∈\text{End}(V)$, and let $W \subseteq V$ be a $\varphi$-stable subspace. Show that $\varphi$ induces linear tranformations $\varphi_{|W}$ and $\overline{\varphi}$ on the spaces $W$ and $V/W$. Show that if $\varphi_{|W}$ and $\overline{\varphi}$ are nonsingular then $\varphi$ is nonsingular. Show the converse holds when $V$ is finite dimensional, but not necessarily when $V$ is infinite dimensional.

Proof: $\varphi_{|W}$ is clearly a linear transformation by the stability of $W$, and define $\overline{\varphi}(\overline{v})=\overline{\varphi(v)}$. For well definedness, suppose $\overline{v_1}=\overline{v_2}$; then $v_1-v_2 \in W$ so $v_1=v_2+w$ and $\overline{\varphi(v_1)}=\overline{\varphi(v_2+w)}=\overline{\varphi(v_2)}$.

Assume these two are nonsingular, and now assume $φ(v)=0$. If $v∈W$, then since $φ_{|W}$ is nonsingular we have $v=0$. If $v∉W$, then $\overline{v}≠0$ and since $\overline{φ}$ is nonsingular we have $φ(v)∉W$ so a fortiori $φ(v)≠0$, a contradiction.

For the converse, since $φ$ is nonsingular we naturally have $φ_{|W}$ is nonsingular. Assume $V$ is finite dimensional to show $\overline{φ}$ is nonsingular; now that we may assume $W$ is finite dimensional, assume $φ(v)∈W$. Then since $φ_{|W}$ is nonsingular and thus surjective, we may obtain $w∈W$ such that $φ(w)=φ(v)$, implying $v=w$ so that in particular $v∈W$ and $\overline{v}=0$.

Now assume $V$ is the direct sum of the countably infinite number of copies of $\mathbb{R}$. Letting $φ$ be the right shift operator with $W$ the subspace consisting of vectors whose first coordinate is zero, we see $φ$ is nonsingular yet $\overline{φ}$ is manifestly not nonsingular by $\overline{φ}(\overline{e_1})=0$.

Bases and Cardinality (11.1.12-14a)

Dummit and Foote Abstract Algebra, section 11.1, exercises 12-14a:

12. If $F$ is a countable field and $V$ is an infinite dimensional vector space with basis $\mathcal{B}$, prove $|\mathcal{B}|=|V|$.
13. Prove $\mathbb{R}^n≅\mathbb{R}$ as vector spaces over $\mathbb{Q}$.
14a. Let $\mathcal{A}$ be a basis for the infinite dimensional vector space $V$ over $F$. Prove $V ≅ \oplus_{a∈\mathcal{A}}F$.

Proof: Throughout these exercises we shall assume the fact that countable unions and finite direct products of an infinite set $S$ fix the cardinality.

(12) By the inclusion mapping we clearly see $|\mathcal{B}| ≤ |V|$. Letting $V_i$ for $i∈\mathbb{N}$ be the set of elements of $V$ whose basis sum includes exactly $i$ nonzero vectors, we have $V$ is the countable union of the $V_i$, so it suffices to show $|V_i| = |\mathcal{B}|$. We can observe$$|\mathcal{B}| = |\mathcal{B} \times \mathcal{B}| \geq |F \times \mathcal{B}| = |(F \times \mathcal{B})_1 \times ... \times (F \times \mathcal{B})_i| \geq |V_i|$$so that$$|\mathcal{B}| = |\bigsqcup_{i∈\mathbb{N}} \mathcal{B}| \geq |\bigsqcup_{i∈\mathbb{N}} V_i| = |V|$$and now $|\mathcal{B}|=|V|$.

(13) First we prove $\mathbb{R}^2≅\mathbb{R}$, so that by induction the proposition easily follows. First, assume the basis $\mathcal{A}$ of $\mathbb{R}$ is finite. Then $\mathbb{R}$ is isomorphic to the direct sum of a finite number of copies of $\mathbb{Q}$ by the next exercise, which would imply $\mathbb{R}$ is countable, a contradiction. Now, there is a basis $\mathcal{B}$ for $\mathbb{R}^2$ given by the basis $\mathcal{A}$ in each component whose cardinality is equal to $|\mathcal{A} \sqcup \mathcal{A}|$ so that $|\mathcal{B}|=|\mathcal{A}|$. Linearly extend the homomorphism induced by this bijection to obtain an isomorphism.

(14a) By mapping the coefficients of a vector sum of the basis to the proper coordinates in the direct sum we obtain the evident isomorphism. Now, the direct product is clearly a vector space by closure with componentwise multiplication by scalars of $F$.

Basis Calculation (11.1.1)

Dummit and Foote Abstract Algebra, section 11.1, exercise 1:

Let $V=\mathbb{R}^n$ and let $(a_1,...,a_n)∈V$ be fixed. Let $W \subseteq V$ be the set of vectors $(x_1,...,x_n)$ such that $x_1a_1+...+x_na_n=0$. Prove $W$ is a subspace and find a basis for $W$.

If all $a_i=0$ then clearly $W=V$ and $e_1,...,e_n$ suffice as a basis. Otherwise, let $a_m≠0$. For $(x_1,...,x_n),(y_1,...,y_n)∈W$ we see$$(x_1,...,x_n)-r(y_1,...,y_n)=(x_1-ry_1,...,x_n-ry_n)$$as well as$$(x_1-ry_1)a_1+...+(x_n-ry_n)a_n=$$$$(x_1a_1+...+x_na_n)-r(y_1a_1+...+y_na_n)=0$$so that $W$ is a subspace. We claim $e_i-(a_i/a_m)e_m$ for all $i∈\{1,...,m-1,m+1,...,n\}$ is a basis for $W$. Foremost, all of these are seen to be vectors of $W$. Linear independence:$$b_1(e_1-(a_1/a_m)e_m)+...+b_{m-1}(e_{m-1}-(a_{m-1}/a_m)e_m)+$$$$b_{m+1}(e_{m+1}-(a_{m+1}/a_m)e_m)+...+b_n(e_n-(a_n/a_m)e_m)=0⇒$$$$b_1e_1+...+b_{m-1}e_{m-1}+b_{m+1}e_{m+1}+...+b_ne_n-(\sum_{i≠m} b_i a_i/a_m)e_m=0⇒$$$$b_1,...b_{m-1},b_{m+1},...,b_n=0$$This implies $W$ is either of dimension $n-1$ or $n$. If it were the latter, then we would have $W=V$ despite $e_m∉W$ so that necessarily $W$ is of dimension $n-1$ and now this set must be a basis.$~\square$

Flat Tensor Products (10.5.23)

Dummit and Foote Abstract Algebra, section 10.5, exercise 23:

When $M$ is a right flat $R$-module and $S$ is a ring considered as a left $R$-module by some identity-fixing homomorphism $R → S$, prove that $M \otimes_R S$ is a right flat $S$-module.

Proof: Let $0 → A → B$ be an exact sequence of $S$ modules by $ψ$. Since $S$ is a free right $S$-module of rank 1, it is flat, and therefore $1 \otimes_S ψ : S \otimes_S A → S \otimes_S B$ is injective. Moreover, this produces an exact sequence of $R$-modules, and since $M$ is right flat, $1 \otimes_R (1 \otimes_S) : M \otimes_R S \otimes_S A → M \otimes_R S \otimes_S B$ is injective, which is the associated homomorphism induced by the functor $M \otimes_R S \otimes_S \_$, which is to say $M \otimes_R S$ is a right flat $S$-module.$~\square$

Rings Inducing Projective and Injective Modules (10.5.6)

Dummit and Foote Abstract Algebra, section 10.5, exercise 6:

Prove every $R$-module is projective if and only if every $R$-module is injective.

Proof: Every $R$-module being projective implies every short exact sequence $0 → D → M → N → 0$ splits since $N$ is an $R$-module and thus projective, therefore arbitrary $D$ is also injective. Likewise, if every $R$-module is injective, then for an arbitrary $R$-module $D$ we have $0 → L → M → D → 0$ splits as $L$ is injective, so that $D$ is projective.$~\square$

Direct Sums of Special Modules (10.5.3-5)

Dummit and Foote Abstract Algebra, section 10.5, exercise 4:

3. Prove $Q_1 \oplus Q_2$ is a projective $R$-module if and only if $Q_1$ and $Q_2$ are projective $R$-modules.
4. Prove $Q_1 \oplus Q_2$ is an injective $R$-module if and only if $Q_1$ and $Q_2$ are injective $R$-modules.
5. Prove $Q_1 \oplus Q_2$ is a flat $R$-module if and only if $Q_1$ and $Q_2$ are flat $R$-modules. Prove $\sum A_i$ is a flat $R$-module if and only if each $A_i$ is a flat module.

Proof: Lemma: Let $\mathcal{F}_i$ be functors of $R$-modules. $\bigoplus \mathcal{F}_i$ is exact if and only if every $\mathcal{F}_i$ is exact. Proof: ($\Leftarrow$) Letting $0 → L → M → N → 0$ be exact by $ψ$ and $φ$, since $\mathcal{F}_i$ are exact functors, then $(\bigoplus \mathcal{F}_i)(ψ)$ is injective by observation of components, likewise $(\bigoplus \mathcal{F}_i)(φ)$ is surjective, and $(\bigoplus \mathcal{F}_i)(φ)$ is zero on and only on elements whose individual coordinates belong to the images of their respective functored $ψ$ homomorphisms, i.e. $\text{ker }(\bigoplus \mathcal{F}_i)(φ)=\text{img }(\bigoplus \mathcal{F}_i)(ψ)$. ($⇒$) For some inexact $\mathcal{F}_n$, we can observe inexactness in $(\bigoplus \mathcal{F}_i)$ in a natural fashion. For example, if $\text{ker }(\mathcal{F}_n)(φ)≠\text{img }(\mathcal{F}_n)(ψ)$, then the kernel and images of $(\bigoplus \mathcal{F}_i)(φ)$ and $(\bigoplus \mathcal{F}_i)(ψ)$ don't coincide by adducing the kernel-image inexactness in the coordinate in question.$~\square$

Proof: (3-5) $M_i$ are all projective (or injective or flat [here $I$ possibly infinite]) if and only if $Hom_R(M_i,\_)$ (or $Hom_R(\_,M_i)$ or $M_i \otimes_R \_$) are all exact if and only if $\bigoplus Hom_R(M_i,\_)≅Hom_R(\bigoplus M_i,\_)$ (or $\bigoplus Hom_R(\_,M_i)≅Hom_R(\_,\bigoplus M_i)$ or $\bigoplus (M_i \otimes \_) ≅ (\bigoplus M_i) \otimes \_)$) if and only if $\bigoplus M_i$ is projective (or injective or flat).$~\square$