# Algebra Review Dec 5



![image-20221205091920001](/home/zack/.config/Typora/typora-user-images/image-20221205091920001.png)



# 4.2.2

I. Sec. $4.2 \# 2,3$ abcef 

(Remember, to prove $R / I \cong S$, you want to find a surjective ring homomorphism $\phi: R \rightarrow S$ with $\operatorname{ker} \phi=I$. In part b, omit the bar over the 6 . In part $\mathrm{f}$, use the fact that $\mathbb{Z}_{12}=\mathbb{Z} /\langle 12\rangle$ and apply FHT.)



![image-20221205094017670](/home/zack/.config/Typora/typora-user-images/image-20221205094017670.png)



Definition of a ring:

![image-20221205172517561](/home/zack/.config/Typora/typora-user-images/image-20221205172517561.png)



Domain: never, since $(0, 1) \cdot (1, 0) = (0, 0)$, so there are nonzero zero divisors.

# 4.2.3.abcef

![image-20221205094141769](/home/zack/.config/Typora/typora-user-images/image-20221205094141769.png)



![image-20221205172757864](/home/zack/.config/Typora/typora-user-images/image-20221205172757864.png)



- $\phi: \R[x] \to \C$ where $\phi(f) = f(\sqrt{-6})$, then $\phi(f) = 0 \iff f(\sqrt -6) = 0 \iff x^2-6 \mid f \iff (x^2-6)\ni f$.
  Surjective: if $\lambda \in \R$, take the constant polynomial $f(x) = \lambda$.

  If $\lambda \in \C$, $bx+a \mapsto a + i\sqrt 6 b$
  

- $\phi: \Z/18\Z \to \Z/6\Z$ where $[n]_{18} \mapsto [n]_6$.
  Then $\phi([n]_{18}) = [0]_6 \iff 6 \mid n \iff n\in (6)$.
  Well-defined: if $[n_1]_{18} = [n_2]_{18}$ then $[n_1-n_2]_{18} = [0]_{18} \iff 18 \mid n_1-n_2$, but if so then $6\mid n_1 -n_2$ and so $[n_1- n_2] = [0]_6 \implies [n_1]_6 = [n_2]_6$.
  Surjective: for $[n]_6$ take $[n]_{18}$ where $n=0,1,\cdots, 5$.

- $\phi: \Q[x] \to \Q[i\sqrt 3]$ where $\phi(f) = f(e^{2\pi i \over 3}) = f(-1/2(1+i\sqrt 3))$.
  Then $\phi(f) = 0 \iff f(e^{2\pi i\over 3}) = 0 \iff x^2+x+1\mid f \iff f\in (x^2+x+1)$.
  Surjectivity: for $c+di\sqrt 3$, take $f(x) = (c+d) + 2dx$.

- $\Phi: F[x] \to F$ where $f\mapsto f(0)$ and $x+a\mapsto a$.
  

- $\Z \to \Z/3 \times \Z/4$ $n\mapsto (n\mod 3, n\mod 4)$, show surjectivity.



# 4.2.5

II. Sec. $4.2$ \# 5 
(Before doing (a), first prove that $R$ is a commutative subring of the (non-commutative!) ring $M_2(\mathbb{R})$. \#4.1.11 might give an idea for (b).)



![image-20221205171704175](/home/zack/.config/Typora/typora-user-images/image-20221205171704175.png)



- $\Phi: R/I\to R$, send to diagonal, show ring morphism (sums and products).

# 4.2.11ab

III. Sec. $4.2$ \# $11 \mathrm{ab}$ (If any of the rings are isomorphic, prove it using FHT if you can.)



![image-20221205171715841](/home/zack/.config/Typora/typora-user-images/image-20221205171715841.png)



- Not $\Z_4$: no element of order 4, and isomorphisms preserve orders of elements.
- Not $\Z_2\times \Z_2$: LHS has nilpotents, this doesn't.

# 4.2.20

IV. Sec. $4.2$ \# 20, 21, 23 ab (Hint: A homomorphism defined on any of the direct products in this problem is completely determined by $\phi(1,0)$ and $\phi(0,1)$-why? Call these something, maybe $(a, b)$ and $(c, d)$. What is the resulting formula for $\phi(x, y)$ ? Now find conditions on $a, b, c, d$ in order that $\phi$ is a homomorphism.)



![image-20221205171725195](/home/zack/.config/Typora/typora-user-images/image-20221205171725195.png)



# 4.2.21



![image-20221205171734529](/home/zack/.config/Typora/typora-user-images/image-20221205171734529.png)

- Irreducible (no roots in $F$).
- See the book: $R[x]/(f)$ for $f$ irreducible is a field (Thm 2.3)



# 4.2.23ab



![image-20221205171744966](/home/zack/.config/Typora/typora-user-images/image-20221205171744966.png)

2 and 1.