User:Struggle Town Mayor/MATH5735

PS1

 * 1) $A \mapsto A^T$
 * 2) Yes, scalar, no (?)
 * 3) asdf
 * 4) asdf
 * 5) Multiply a lot of junk
 * 6) Ideal of $k[x]/\langle x^3 \rangle$ has form $I/\langle x^3 \rangle$ where $I \triangleleft k[x]$ and $\langle x^3 \rangle \subseteq I$. $k[x]$ a PID implies $I$ is principal. But $\langle x,y \rangle / \langle x^2,xy,y^2 \rangle$ is an ideal of $k[x,y]/\langle x^2,xy,y^2 \rangle$, not principal.
 * 7) $m0 = m(r-r) = mr-mr = 0$, $m(-1) + m(1) = m(-1+1) = m0 = 0$.
 * 8) Is there just a direct isomorphism you can see by writing any element of $k[x]/k[x^2]$? I feel I've missed something. (?)
 * 9) Don't see what there is to prove here
 * 10) (a) Write $r=\sum \alpha_i \sigma^i$, multiply junk to see $r \in \text{ker}(\phi)$ iff $\alpha_0+\alpha_2+\alpha_4 = 0 = \alpha_1+\alpha_3+\alpha_5$. (b) $\text{span}\{1+\sigma^2+\sigma^4,\sigma+\sigma^3+\sigma^5\}$
 * 11) Look for homomorphisms $\phi : k[x] \to k[x]/\langle x^3 \rangle$ with $\phi(\langle x^2 \rangle) = 0$. nontrivial ones correspond to polynomials with positive degree (?)

PS2

 * 1) universal property twice, $\text{Hom}_R(R,R) \cong R$.
 * 2) CRT
 * 3) yes (CRT), no get contradiction. Also: (??) $\langle x^2 +x +1 \rangle$ is gen. by an irreducible and $\mathbb{R}[x]$ is PID, so this ideal is maximal, so $M$ is simple and hence indecomposable. (??) Jason can you elaborate on your method? I'm not sure if the example in lecs works here. [what lecture method? you can assume that it's true and get a contradiction.] - Pete [I don't remember editing this question haha] - Jason
 * 4) $3$ and $5$ coprime ??? profit (CRT..?)
 * 5) show 1 in or not in the ideal. Scalar mult on right: $\partial(x) = 1$ is in module generated. Scalr mult on left: always have $\partial$ right-most, so this isn't everything.
 * 6) mult one elt by $2^{1/4}$ (havent checked this totally but seems legit?). Have $\mathbb{Z}[^4\sqrt{2}] \cong (\mathbb{Z}[\sqrt{2}] \oplus (^4\sqrt{2})(\mathbb{Z}[\sqrt{2}])$.
 * 7) Tute Week 7
 * 8) Isom given by map $\phi: kH \oplus kH \to kG$ defined as $(1, \tau) : (\sum \lambda_i \sigma^i, \sum \lambda_j \sigma^j) \mapsto \sum \lambda_i \sigma^i +  \tau \sum \lambda_j \sigma^j$. With basis $\{1, \tau\}$, the matrix is $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$.
 * 9) Tute Week 11
 * 10) ?? $R^2/M \cong \frac{k[x]}{\langle x+1 \rangle} \oplus \frac{k[x]}{\langle x-1 \rangle} \oplus \frac{k[x]}{\langle x-1 \rangle} \oplus \frac{k[x]}{\langle x^2 + x +1 \rangle}$ ?? Got a different answer - maybe one of us messed the calcs - Pete
 * 11) Assignment 2

PS3

 * 1) Need only show that the image of mult by 2  = ker of mult by 2 = $\{0,2\}$.
 * 2) $1=e+e'$, so $e=1-e'$. Now, $e=(e+e')e = e^2 + e'e$ and $e'e \in M \cap M' = 0$, so $e = e^2$.
 * 3) ?? Not exactly sure what needs to be checked. Vaguely, equal as sets from completeness, and orthogonality gives $e_i B \cap e_j = 0$ if $i \neq j$.
 * 4) Noetherian follows from Prop. 13(a). ($M$ is a fin-gen $A$-module and $A$ is noeth.) (?? artinian??) $M$ can be broken up into simples $M_i$. Each simple $M_i \cong R/ I_i$ for some maximal ideal $I_i$. As $R_R$ is noeth and art, we have $R/I_i$ is noet and art, so $M_i$ and hence $M$ is noeth and art...
 * 5) To see it is not noetherian, take the submodules of the form $\frac{1}{2^m}\mathbb{Z}$ for increasing $m$, will form an increasing chain without limit. To see it is artinian, SH $\Rightarrow$ every module of the form $\frac{1}{2^m}\mathbb{Z}$, then argue only have containment if $m>m'$ etc. (think the SH follows from hint). Also, see http://planetmath.org/exampleofanartinianmodulewhichisnotnoetherian
 * 6) Parts done in Tute Week 11. These questions make me sad.
 * 7) Part done in Tute Week 9.
 * 8) Composition series: $0 < \frac{\langle x^2 \rangle}{\langle  x^2(x-1) \rangle} <  \frac{\langle  x\rangle}{\langle x^2(x-1)  \rangle} < \frac{\langle k[x] \rangle}{\langle  x^2(x-1) \rangle}.$ Composition factors: $\frac{\langle x^2\rangle}{\langle x^2(x-1)  \rangle}, \frac{\langle x\rangle}{\langle x^2  \rangle}, \frac{\langle k[x]\rangle}{\langle  x \rangle}.$ First one is simple from 3IT. They are all isomorphic to $k$ (??).
 * 9) All composition factors are $k$. Series for second one: $0 <  \frac{\langle x, xy, y^2 \rangle}{\langle x^2, xy, y^2  \rangle} < \frac{\langle x,y\rangle}{\langle  x^2, xy, y^2  \rangle} <  \frac{\langle k[x,y] \rangle}{\langle x^2, xy, y^2   \rangle}.$ Not isomorphic since e.g. $x$ acts as 0 on $M$ and nonzero on $N$.
 * 10) SH $\Rightarrow$ composition factors are $\frac{\langle 2x \rangle}{\langle 6,x^2 \rangle}$, $\frac{\langle 2 \rangle}{\langle 6,x^2 \rangle}$... Direct sum of simples? NFI.

PS4

 * 1) Umm...
 * 2) Tute Week 13
 * 3) Tute Week 9
 * 4) SH $\Rightarrow$ the direct sum is R-linear and so if the image of one is zero, that component is zero?
 * 5) (i) semi-simple; $M_n(\mathbb{R})$ is semi-simple from P16.1 and $\mathbb{C}$ is simple, so product is semi-simple. (ii) ???? (iii) Is basically the same as $\mathbb{Q}^n$, so it is semi-simple. (iv) Use Maschke's theorem. (v) ????
 * 6) Maschke + Wedderburn, and can't have matrices since they're non-commutative.
 * 7) Tute Week 9
 * 8) SH $\implies A_4$. $S_4/A_4 \cong \mathbb{Z}/2\mathbb{Z}$, so it's abelian and $[G,G] \leq A_4$, which implies $|[G,G]|\ \mid\ |A_4| = 12$. But $(1\ 2)^{-1}(1\ 3\ 2)^{-1}(1\ 2)(1\ 3\ 2) = (1\ 3\ 2)(1\ 3\ 2) = (1\ 2\ 3)$, so all 3-cycles are in $[G,G]$, but there are $\frac{4!}{3} = 8$ of them. So $|[G,G]| \geq 8$ but divides 12, which means $|[G,G]| = 12$ and $[G,G] = A_4$.
 * 9) Tute Week 11
 * 10) zzzz
 * 11) from memory: 4 1-dim'l groups, rest 2 dim'l.

PS5

 * 1) checking defs. main thing is for $\phi' : N \to N'$ and $\phi : N' \to N'{}'$, and any $\psi : M \to N$, we have \[ [F(\phi\phi')](\psi) = (\phi\phi')\circ\psi = \phi \circ (\phi' \circ \psi) = \phi \circ [F(\phi')(\psi)] = [F(\phi)F(\phi')](\psi)\].
 * 2) checking defs, and that $f^G$ is well defined.
 * 3) checking defs.
 * 4) ???? Want to construct an inverse to map given by the FIT using universal property of tensor products?
 * 5) Mult is well defined by P23.4 and $\dim (V \otimes W) = \dim V \times \dim W$. Assoc. of multiplication ok. Identity is trivial representation, also have $k \otimes_k V \cong V$. Inverse: Given a $kG$-module $V$ with corresponding representation $\rho_V$, def $\rho_{V^{-1}}$ by $\rho_{V^{-1}}(g) v:= \rho_V(g^{-1})v$. ??? This is a representation?? Then $(\rho_V(g) \otimes \rho_{V^{-1}}(g))(v \otimes w) = g v \otimes g^{-1} w = v \otimes w$. This is naturally identified with group of representations :\ Q seems strange since we define the tensor product of $kG$ modules in terms of representations anyway. \begin{align*}&\text{Vague unformed thought from GT: I think this ^ is a legitimate representation if we take $V^{-1}$ to be the $kG$-module $V^*$, which is well-defined}\\&\text{for any $kG$-mod $V$. Then $\rho_{V^*}(g) = \rho_{V}(g^{-1})$ so all happy, I think.? --- (Mayor edit to keep numbering)}\end{align*}
 * 6) Tute Week 13
 * 7) Want to consider the eigenvalues of $\rho_V(g^{-1})$. If $\mu$ is an eigenvalue of $\rho_V(g)$ with e'vec $v$, then $v = \rho_V(g^{-1}g) = \rho_V(g^{-1})\rho_V(g) v$ implies $\mu^{-1}$ is eval of $\rho_V(g^{-1})$. Since $g$ has finite order, $\mu$ is root of unity, so $\mu^{-1}= \overline{\mu}$. (Sorry, brain fail but how does this help find the character identity that we want?)
 * 8) SH $\Rightarrow$ follows from fact that the traces of the dihedral rep's are real valued along with the previous statement & T25.3.2?
 * 9) Tute Week 13 (Wednesday)
 * 10) By computation, $H=\{1,(12)(34),(13)(24),(14)(23)\}\leq[G,G]$. So $|G_{ab}|=\frac{|G|}{|[G,G]|}\leq\frac{12}{4}=3$. So we have at most 3 1-dim representations. But then the only way to write 12 as sum of squares is $12=1+1+1+3^2$ so $|G_{ab}|=3\Longrightarrow|[G,G]|=4\Longrightarrow [G,G]=H$. So we have 4 conjugacy classes, since we have 4 irreducible representations. 1 and {(12)(34),(13)(24),(14)(23)} are 2 conjugacy classes and the other 2 classes have 4 elements each, by (not sure about this...) symmetry. After 13 hours of computation by hand the classes are $\{(123) (142) (134) (243)\}$, $\{\text{the inverses of the ones before}\}$.

The 3 one-dimensional representations are given by p47 in joels notes. Let the 1-dim characters be $\chi_1,\chi_2,\chi_3$ and the 3-dim be $\chi_4$. So our character table is

$C_i$      1  (12)(34)  (123)  (132) $|C_i|$    1       3      4      4

$\chi_1$       1       1      1      1 $\chi_2$       1       1      w      $w^{-1}$ $\chi_3$       1       1   $w^{-1}$   w      $\chi_4$        3       a      b       c

where $w=\exp{\frac{2\pi i}{3}}$. So now use fourier stuff to solve for a,b,c. The equations are \begin{align*} 0&=3+3a+4b+4c\\ 0&=3+3a+4bw^{-1}+4cw\\ 0&=3+3a+4bw+4cw^{-1} \end{align*}

we then get a=-1, b=c=0. Now we need to find the 4-dim representation $\rho_W$ explicitly. The matrices are 4x4 and

\begin{align*} \rho_W(1)&=I_4\\ \rho_W(12)(34)&=(a_{ij}), a_{12}=a_{21}=a_{43}=a_{34}=1 \text{ and everywhere zero}\\ \rho_W(123)&=(b_{ij}), b_{13}=b_{21}=b_{32}=b_{44}=1 \text{ and everywhere zero}\\ \rho_W(132)&=(c_{ij}), c_{12}=c_{23}=c_{31}=c_{44}=1 \text{ and everywhere zero} \end{align*}

So we have

$\chi_W$       4       0      1       1

Now,

\begin{align*} <\chi_W,\chi_1>&=1/12(4\times1+4\times1+4\times1)=1\\ <\chi_W,\chi_2>&=1/12(4\times1+4\times w+4\times w^{-1})=0\\ <\chi_W,\chi_3>&=1/12(4\times1+4\times w^{-1}+4\times w)=0\\ <\chi_W,\chi_4>&=1/12(4\times3)=1 \end{align*}

Hence, $W=V_1\oplus V_4$, corresponding to the trivial representation and the 3-dim representation.

In Joel's notes we are asked to find the 3-dim repr explicitly, would be nice if someone could figure it out.

...Show isotypic component corresponding to $k$ is...
Let $M = M_0 \oplus M_1 \oplus \dots \oplus M_n$ be the decomposition into isotypic components; want to show $M_0 = M^G$. Note $M^G$ is clearly a $kG$-submodule of $M$.

Decompose into isotypic components: \[ M^G = M_0^G \oplus M_1^G \oplus \dots \oplus M_n^G. \] Note that $M_i^G \subseteq M_i$, by (4) (from the bit before earlier in the lecture). Suffices to show \begin{align*} (1) M_0^G &= M_0\\ (2) M_i^G &= 0 \quad \forall i \neq 0. \end{align*} For $(1)$, we show that every element of $M_0$ is fixed by $G$. But $M_0 \cong \bigoplus_{i \in I} k$ so consider $m = (m_i)_{i \in I} \in \bigoplus_{i \in I} k$. Check \[ gm = g(m_i) = (gm_i) = (m_i) (???) = m. \] some wizardry going on here, or maybe I'm just dumb. For $(2)$, let $M_i^G \cong \bigoplus_{j \in J} D_j^{n_j}$. If some $m \in M_i^G \neq 0$, consider left module $kGm = km \cong k$, the trivial module, which by semisimplicity is a direct summand of $M_i^G$, a contradiction as $M_i^G$ is a direct sum of $D_i^{n_i}$ (???).