Plane Curve Multiplicities Under Polynomial Mappings (3.1.8)

William Fulton Algebraic Curves, chapter 3, section 1, exercise 8:

Let $k$ be an algebraically closed field, $F∈k[x,y]$, and $T : \mathbb{A}^2 → \mathbb{A}^2$ be a polynomial map with $T(P)=Q$. Show that $m_P F^T ≥ m_Q F$, with equality achieved if the Jacobian of $T$, defined as the matrix $(dT_i / dx_j (Q))$ where $T = (T_1,T_2)$, is invertible. Show that the converse is false in considering $T = (x^2,y)$ and $F=y-x^2$, $P=Q=(0,0)$.

Proof: Suppose $P = Q = (0,0)$. Then necessarily $T_1$ and $T_2$ are of the form $$T_1 = ... + ax+by$$ $$T_2 = ... + cx+dy$$ with no constant terms. Write $F = F_m + ... + F_n$ in terms of its form decomposition; then we see $F^T = F_m^T + ... + F_n^T$, and for each $i$, $F_i^T$ contributes to the sum forms of degree $≥i$. This shows the nonzero forms in $F^T$ are all of degree $≥m$, so that $m_{(0,0)} F^T ≥ m_{(0,0)} F$. Now consider the general case for points $P,Q$. Let $S_1, S_2$ be the translations mapping the origin to $P,Q$ respectively. Then $$m_Q F^T = m_{(0,0)} F^{T∘S_2} = m_{(0,0)} F^{S_1∘S_1^{-1}∘T∘S_2}$$ $$= m_{(0,0)} (F^{S_1})^{S_1^{-1}∘T∘S_2} ≥ m_{(0,0)} F^{S_1} = m_P F$$ Now to consider the properties of the Jacobian: Again suppose $P = Q = (0,0)$, and factor $F_m$ into a product of linear factors. Since $F_i^T$ for $i > m$ only contribute forms of degree $> m$, it will suffice to show $F_m^T$ contributes a nonzero form of degree $m$ when the Jacobian is invertible. When the Jacobian is invertible, that is to say its rows are linearly independent, so that if $ex+fy$ is a nonzero linear term, then $T(ex+fy) = eT_1(x) + fT_2(y)$ will have a nonzero form of degree $1$ in its decomposition. Since the smallest nonzero form of a product is the product of the smallest nonzero forms, we see that $m_{(0,0)} F^T = m_{(0,0)} F$. The argument for the general case of $P, Q$ is exactly as before, together with the observations that (1) the Jacobian of a composition of polynomial maps is the same as the matrix product of the Jacobian of polynomial maps at the appropriate points, and (2) that the Jacobians of translations are the identity. As for the demonstration that the converse is not in general true, we note that in the example given, $m_{(0,0)} F = 1 = m_{(0,0)} y - x^4 = m_{(0,0)} F^T$ despite the fact that the Jacobian of $T$ at $(0,0)$ is the noninvertible $$J_{(0,0)}T = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}~~\square$$

Finite Dimensional Rings Containing Algebraically Closed Fields (2.9.47)

William Fulton Algebraic Curves, chapter 2, section 9, exercise 47:

Let $R$ be a commutative ring sharing an identity with $k⊆R$, an algebraically closed field. Show that $R$ is isomorphic to a finite direct product of local rings.

Proof: Let $r_1,...,r_n$ be a basis for $R$ over $k$. Then $k[x_1,...,x_n] / I ≅ R$ for some ideal $I$ via the natural identification $x_i ↦ r_i$. We see that distinct (surjective) morphisms $R → k$ preserving $k$ thus correspond to maximal ideals in $k[x_1,...,x_n]$ containing $I$, which themselves correspond to points in the locus of $I$. It follows that if we can prove there are only finitely many such morphisms, then Proposition 6 of this same section will imply the result. To this end, we show that the image of $r_i$ in $k$ under a morphism $φ$ may be among only finitely many elements; indeed, this follows by considering that $1,r_i,r_i^2,...$ are linearly dependent over $k$, and thus $φ(r_i)$ must be a root of the polynomial of $r_i$ over $k$ induced by this dependence.$~\square$

Multiple Representations are Required to Identify the Pole Set of a Rational Function (2.4.20)

William Fulton Algebraic Curves, chapter 2, section 4, exercise 20:

Let $k$ be algebraically closed, and consider the rational function $f = x/y = z/w$ over the algebraic variety $V = \mathcal{Z}(xw-yz)$ in $\mathbb{A}^4(k)$. Show that the pole set of $f$ is precisely $\{(t_1,0,t_2,0)~|~t_1,t_2∈k\}$, and that it is impossible to write $f = a/b$ for $b(p)≠0$ for all $p∈V$ outside the pole set of $f$.

Proof: First of all, $y$ and $w$ are nonzero on points of $V$, so they are nonzero in $\Gamma(V)$. As well, $\mathcal{I}(\mathcal{Z}(xw-yz)) = (xw-yz)$ since $xw-yz$ is irreducible. Let $J_f = \{G ∈ \Gamma(V)~|~Gf ∈ \Gamma(V)\}$. It is clear that $J_f$ is an ideal in $\Gamma(V)$, and that $y,w∈ J_f$. Suppose $gf = u ∈ \Gamma(V)$ for some $g∈k[x,z]$; then we may write $$gx - uy = 0 \mod (xw-yz)$$ $$gx = 0 \mod (xw,y)$$ from which we conclude $g = 0$, so that $J_f = (y,w)$. This allows us to conclude that the denominator of $f$ in any way that we write it will always have $\{(t_1,0,t_2,0)\}$ as zeros, so that this is precisely the pole set of $f$. Suppose we could write $f = a / b$ with $b$ having zeros precisely at the pole set of $f$; this would be to say $$\mathcal{Z}(b,xw-yz) = \mathcal{Z}(y,w,xw-yz) = \mathcal(Z)(y,w)$$ $$\mathcal{Z}(b,xw-yz) ∩ \mathcal{Z}(x,z) = \mathcal{Z}(x,y,w,z)$$ $$\mathcal{Z}(b,x,z) = \{(0,0,0,0)\}$$ Since $b∈(y,w)⊆\Gamma(V)$, it will thus suffice to show that given $b ∈ (y,w) ⊆ k[y,w]$, we may find a zero besides $(0,0)$. Indeed, this is equivalent to showing $$\mathcal{Z}(b) = \mathcal{Z}(y,w)$$ is impossible. Indeed, every nonconstant polynomial in multiple variables over an algebraically closed field has infinitely many zeros (1.2.14), while $\mathcal{Z}(y,w)$ is just a single point.$~\square$

Wedderburn Decomposition of Finite Group Rings Over Algebraically Closed Fields

Dummit and Foote Abstract Algebra, section 18.2:

Let $G$ be a finite group, $F$ a field, and $FG ≅ R = R_1 × ... × R_r$ be the Wedderburn decomposition of the group ring $FG$. Each $R_i$ is a ring of $n_i × n_i$ matrices with entries from a division ring $Δ_i$. We show in this case that in fact $Δ_i ≅ F$.

We have seen already that $Z(R_i)$ is the set of scalar $n_i × n_i$ matrices with entries from $Z(Δ_i)$. Since $F⊆Z(FG)$, this means (after composition with the projection map $π_i : R → R_i$) $F$ embeds into $Z(Δ_i)$ for each $i$. Now, consider the left ideal $I_i$ of $R$ consisting of elements with $0$ in every coordinate but the $i^\text{th}$, where the value may be any $n_i × 1$ column matrix. This left ideal $I_i$ is a left $R$-module, with an action that restricts to turn $I_i$ into a $Δ_i$-module (multiplication with scalar matrices in $R_i$), which in turns restricts to turn $I_i$ into a vector space over $F$. As a left $FG$-module (inherited from isomorphism of $FG$ with $R$), we may decompose $I_i$ into a finite direct sum of cyclic $FG$-modules (of rank $≤n_i$), hence $I_i$ is a finite vector space over $F$ (of dimension $≤n_i|G|$). $I_i$ is also a ring, with a subring $J_i$ (of $1×1$ matrices) isomorphic to $Δ_i$. Since $J_i$ is an $F$-subspace of $I_i$, it too is finite dimensional, with a vector space action agreeing with multiplication from $F⊆J_i$. That is to say, ultimately, that $Δ_i$ is a finite dimensional vector space over $F$ such that $F⊆Z(Δ_i)$.

If $F$ is algebraically closed, then necessarily $F=Δ_i$, since each ring extension $F[α]⊆Δ_i$ of $F$ by an element $α∈Δ_i$ will in fact be a field extension by $F⊆Z(Δ_i)$, hence be of degree one as a vector space over $F$ by algebraic closure, implying $α∈F$.

Basic Group Representations (18.1.4,18-19,21-22,24)

Dummit and Foote Abstract Algebra, section 18.1, exercises 4, 18-19, 22, 24:

In the exercises that follow, $G$ is a finite group, and $F$ a field.

4. Prove that if $G$ is nontrivial, then every irreducible $FG$-module has degree $< |G|$.

18. Prove that if $φ : G → GL_n(ℂ)$ is an irreducible matrix representation and $A$ is an $n×n$ matrix commuting with $φ(g)$ for all $g∈G$, then $A$ is a scalar matrix. Deduce that if $φ$ is a faithful, irreducible, complex representation then the center of $G$ is cyclic and $φ(g)$ is a scalar matrix for all elements $z$ in the center of $G$.

19. Prove that if $G$ is an abelian group then any finite dimensional complex representation of $G$ is equivalent to a representation into diagonal matrices.

21. Let $G$ be a group with noncyclic center acting on a finite-dimensional vector space $V$ over $F$, where $F$ is a field of characteristic $p$ not dividing the order of $G$.
(a) Prove that if $W$ is an irreducible $FG$-submodule of $V$ then there is some nonidentity $g∈G$ such that $W⊆C_V(g)$, where $C_V(g)$ is the set of elements of $V$ fixed by $g$.
(b) Prove that $V$ is generated as an additive group by the sets $C_V(g)$ as $g$ runs over all nonidentity elements of $G$.

22. Let $p$ be a prime, let $P$ be a $p$-group and let $F$ be a field of characteristic $p$. Prove that the only irreducible representation of $P$ over $F$ is the trivial representation.

24. Let $p$ be a prime, let $P$ be a nontrivial $p$-group and let $F$ be a field of characteristic $p$. Prove that the regular representation is indecomposable.

Proof: (4) Let $V$ be an $FG$-module, and let $v∈V$ be nonzero. Then either $\{gv~|~g∈G\}$ are linearly dependent over $F$, in which case they generate a $G$-stable subspace of dimension strictly less than $|G|$, or they are linearly independent, in which case the nonzero element $\sum_{g∈G} gv$ generates a one-dimensional line that is fixed by each $g∈G$. As such, we see that any $FG$-module not containing any proper nonzero $FG$-submodule must be of dimension $< |G|$.

(18) Let $λ$ be an eigenvalue of $A$, i.e. $A-λ$ has nontrivial kernel as a linear transformation $ℂ^n → ℂ^n$. Since $A$ (and $λ$) commute with every element of $φ(G)$, we see the kernel of $A-λ$ is a $G$-submodule. Since the representation is irreducible, this is to say the kernel is all of $ℂ^n$, so that $A-λ = 0$ and $A$ is a scalar matrix. When $φ$ is faithful, this implies the center of $G$ embeds into the group of nonzero scalar matrices, i.e. $ℂ^×$, so that the center of $G$ is cyclic.

(19) By the above exercise, we find that the action of $G$ on any irreducible $G$-stable subspace of $ℂ^n$ is that of scalar multiplication. If we decompose $ℂ^n$ into a direct sum of irreducible $ℂG$-submodules and observe the matrix representation of $G$ with respect to the $ℂ$-basis of this decomposition, we find that they are diagonalized matrices.

(21)(a) This is a consequence fact that for any irreducible representation $φ$ of a group $G$, we find $Z(G/\text{ker }φ)$ is cyclic (cf. exercise 14d). For if $W$ is an irreducible $FG$-submodule, this offers an irreducible representation of $G$ (by the action on $W$). It follows that if $Z(G)$ is not cyclic, then $\text{ker }φ$ is nontrivial, i.e. there exists a nonidentity element of $G$ that fixes all of $W$.

(b) Decompose $V$ into a finite direct sum of irreducible $FG$-submodules by Maschke's theorem. Each of these summands $W_i$ are contained in $C_V(g_i)$ for some nonidentity $g_i∈G$, by part (a). It follows $V$ is generated as an abelian group by the $C_V(g_i)$.

(22) We may assume a representation $φ$ of $P$ over $F$ is faithful by passing to the quotient $P / \text{ker }φ$ if necessary. But if $P$ is a nontrivial $p$-group, it has a nontrivial center, and there is thus an element $x∈P$ of order $p$ commuting with every element in $P$. If $FP$ is acting irreducibly on a vector space $F$ over $V$ via the representation $φ$, then $V$ is finite dimensional, and as finite-dimensional matrices we have $φ(x)^p-1 = (φ(x)-1)^p = 0$, thus $\text{det }φ(x)-1 = 0$. As $x$ commutes with every element of $P$, we have $\text{ker }φ(x)-1$ determines a $P$-stable subspace of $V$, hence is all of $V$, hence $x=1$, a contradiction since $φ$ is faithful.

(24) Lemma: Let $G$ be a group (not necessarily finite), let $F$ be a field, and let $V$ be an $FG$-module. If $N \unlhd G$ is a normal subgroup, then the set of elements $W ⊆ V$ that are fixed by $N$ is an $FG$-submodule, and there is a natural $F(G/N)$-module action on $W$. The $FG$-submodules and $F(G/N)$-submodules of $W$ are the same. Furthermore, if the module $V$ is itself $FG$ affording the regular representation on $G$, then the induced module is isomorphic to the regular representation of $G/N$.

Proof: It is clear $W$ is a subspace, and if $n ∈ N$, $g ∈G$, and $w ∈ W$, then $ng·w = gg^{-1}ng·w = g(g^{-1}ng)·w = g·w$, so that $W$ is $G$-stable. As well, note that the module structure of $FG$ on $V$ is given by the action of $G$ on $V$; since $W$ is an $G$-stable, this restricts to an action of $G$ on $W$; since $N$ fixes each element of $W$, this induces a natural action of $G/N$ on $W$, thereby turning $W$ into a $F(G/N)$-module under this action. We see an $F$-subspace is $G$-stable if and only if it is $G/N$-stable, so the $FG$-submodules and $F(G/N)$-submodules are the same.

Suppose $V$ is the module given by the regular representation, i.e. $V=FG$ and the module action is given by multiplication. For each distinct left coset $xN$ define $α_{xN} = \sum_{g∈xN} g ∈ FG$. It is clear, then, that $α_{xN}∈W$, the $α_{xN}$ are linearly independent over $F$, and that any element of $FG$ fixed by each element of $N$ must retain equivalent coefficients on elements in the same left coset of $N$, hence be an $F$-linear combination of the $α_{xN}$, so the $α_{xN}$ form an $F$-basis for $W$. Define an $F$-linear isomorphism $φ : W→F(G/N)$ given by $φ(α_{xN}) = \overline{x}$. This is in fact an $F(G/N)$-module isomorphism, since $φ(\overline{g}α_{xN}) = φ(α_{gxN}) = \overline{gx} = \overline{g}φ(α_{xN})$.$~\square$

We proceed by induction on the order of $P$. Suppose $FP = V_1 \oplus V_2$ as $FP$-modules. If $P$ is nontrivial, let $x∈Z(P)$ be of order $p$. Then on any $FP$-submodule of $FP$, we have $(x-1)^p = x^p - 1 = 0$ as $FP$-module transformations, hence $x-1$ has nontrivial kernel in each of $V_1, V_2$; this is to say $\langle~x~\rangle$ fixes nontrivial elements in each of $V_1,V_2$, say $W_1,W_2$ respectively. Then if $W$ is the $FP$-submodule of elements of $FP$ fixed by $\langle~x~\rangle$, we in fact have $W = W_1 \oplus W_2$. This nontrivial direct sum expression as $FP$-modules translates to the same as $F(P/\langle~x~\rangle)$-modules. But now $W ≅ F(P/\langle~x~\rangle)$ may be written as a nontrivial direct sum, a contradiction.$~\square$