(All of these answers are my own and my not be correct)

For each of the following statments, determine whether it is true or false. No proofs are required.
1. A subring of a commutative ring is commutative
  
  True by definition
2. If $R,S$ are rings and there exists a ring homomorphism $f:R\to S$ which is not injective, then $R$ and $S$ are not isomorphic.
  
  False, not necessarily. There can exist another ring homomorphism that is bijective. $R$ and $S$ can be isomorphic
  
  Example:
  
  $R\cong S=\Bbb Z \oplus \Bbb Z$
  
  $f: \Z\oplus\Z \to \Z\oplus Z\\ f(a,b)=(a,a)$
3. The ring $\Z_2[x]$ is a PID
  
  $F \text{ is a field}\implies F[x] \text{ is a PID}$
  
  $\Z_2$ is a field $\implies \Z_2[x]$ is a PID
  
  True, $p=2$ is prime so $\Z_2$ is prime
4. If $f(x) \in F[x]$ is a reducible polynomial over a field and $\deg (f)=3$ then $f(x)$ has a root in $F$
  
  True, if $\deg f \leq 3,$ and $f$ has a root, then $f$ is reducible
5. Every polynomioal of degree $4$ over $\R$ is reducible
  
  True: Possible degrees of polynomials is (1)(2)
  
  Irreducibles over Reals have odd degree
  
  Theorem: Irreducibles polynomials over reals have only degree 1 or 2
6. If $F$ is a field of cardinality $49$ and $x\in F$ satisfies $x^6=x,$ then $x=0$ or $x=1$
  
  $|F|=49,\\ x^6=x\xRightarrow{?}x=0\text{ or } x=1$
  
  $x\neq 0\implies x^5=1\implies \text{ord}_{F^x}(x) \mid 5$
  
  $\text{ord}_{F^x}(x)\in \{\ 1,5\ \}$
  
  Using Legrange theorem,
  
  $|F^x|=49-1=48$
  
  $\text{ord}_{F^x}(x)=1\implies x=1$
  
  $|F^x|=|F|-1$
  
  True statement but unrelated, do not get confused: $|\Z_n^x|\implies n$ is prime
  
  True
Let $\Z_p \subseteq \mathbb{K}$ be a field extension. How many roots does the polynomial $x^{p^2}-x^p$ have in $K$?

$f(x)\in \Z_p[x]$

$f(x)^p=f(x^p)$

$(a+b)^p=a^p+b^p,$ if $\text{char}F=p$

$f(x)=a_nx^n+\cdots + a_nx+a_0$

$f(x)^p=(a_nx^n+\cdots + a_1x+a_0)^p=(a_nx^n)^p+\cdots +(a_nx)^p+a_0^p=a_n^p(x^p)^n+\cdots + a_1^p(x^p)+a_0^p=a_n(x^p)^n+\cdots + a_1(x^p)+a_0$

Full credit solution

$x^{p^2}-x^p=(x^p-x)^p$ ← Because char = p

$\alpha \in K$ is a root of $(x^p-x)^p\iff\alpha$ is a root of $x^p-x\iff \alpha \in \{\ 0,1,\cdots, p-1\ \}$

$\implies p$ roots

Variations

What if $x^{p^2}-x$ ?

At most $p^2$ roots

$K = \Z_p\implies$ only $p$ roots

If field is $|K|=p^2$

We proved in lectures that $\{\ \alpha \in K\mid \alpha ^{p^n}\ \}\subseteq K$ is a subring and a field

If $|K|=p^{even}$ $\implies p^2$ roots

$|K|=p^{odd} \implies p$ roots (Homework #6)

$\Z_p\subseteq K$ $st.\ K<\infin$

$x^{p^2}-1$ in $K$

$x^{p^2}=1$

$\text{ord}{K^x}(x)\mid p^2\implies \text{ord}{K^x}(x)\in \{1,p,p^2\}$

$|K^x|=p^n-1\implies$ The only divisor of $p^2-1$ is 1 $\implies x=1$
Let $F$ be a finite field of characteristic $2$. Prove that there exists $\alpha \in F$ such that $F[x]/\braket{\ x^2+x+\alpha \ }$ is a field.

$F[x]/\braket{\ x^2+x+\alpha\ }\iff x^2+x+\alpha$ is irreducible $\xLeftrightarrow{deg=2} x^2+x+\alpha$ has no roots in $F$

$\exists\ \alpha \in F\ st. \ x^2+x+\alpha \neq 0$

$x^2+x\neq -\alpha$

$g:F\to F$ function not a homomorphism

$g(x)=x^2+x$

So, we want to show: $g$ is not surjective

(Then, $\exists\ \beta\ \in F:\forall \ x\in F, \ x^2+x\neq \beta,\ \text{take } \alpha =-\beta$ and we are done. $x^2+x+\alpha$ would not have a root)

$g: \underbrace{F}_{\text{is finite}}\to F$ ←Pigeon Hole principle

Equivalently, let us show that $g$ is not injective

$g(0_F)=0_F^2+0_F=0_F$

$g(1_F)=1_F^2+1_F=1_F+1_F=0_F$

$x^2+x=0$

If $\text{char }\neq 2$ just $g(0)=g(-1)$

So, G is not surjective which implies that $x^2+x+\alpha$ has no roots in $F$
Let $f(x)=x^5 +x+1$. Is it irreducible over $\Z_2$? Over $\Bbb R$? Over $\Z_7$? Prove your answers
1. $\Z_2$ - We check whether there are roots of this polynomial over $\Z_2$ by checking $f(\alpha)=0\ \forall\ \alpha \in \Z_2$
  
  $\Z_2=\{\ 0,1\ \}, \implies f(0)=1\neq 0,\ f(1)=3\neq 0$
  
  So, there are no roots.
  
  Now we know that it cannot split into (1)(4), (4)(1)
  
  Only possibilites are (2)(3), (3)(2)
  
  $f(x)=(x^2+b_1x+c_1)(x^3+a_2x^2+b_2x+c_2)$
  
  $=x^5+a_2x^4+b_2x^3+b_1x^4+a_2b_1x^3+b_1b_2x^2+b_1c_2x+c_1x^3+c_1a_2x^2+c_1b_2x+c_1c_2$
  
  $=x^5+(a_2+b_1)x^4+(b_2+a_2b_1+c_1)x^3+(c_2+b_1b_2+c_1a_2)x^2+(b_1c_2+c_1b_2)x+1$
  
  $\implies a_2+b_1=0\implies a_1=-b_1$
  
  $\implies b_2+a_2b_1+c_1=0$
  
  $\implies c_2+b_1b_2+c_1a_2=0$
  
  $\implies b_1c_2+c_1b_2=1\implies c_2=\frac{1}{c_1}$
  
  $=(x^2+b_1x+c_1)(x^3-b_1x^2+b_2x+\frac{1}{c_1})$
  
  $=x^5+(b_2-b_1^2+c_1)x^3+(\frac{1}{c_1}+b_1b_2-b_1c_1)x^2+(\frac{b_1}{c_1}+c_1b_2)x+1$
  
  $\implies b_2=b_1^2-c_1$
  
  $\implies 1+b_1b_2c_1-b_1c_1^2=0$
  
  $\implies b_1+c_1^2b_2=0$
  
  Degree 5, so we check that it has no roots. The only possibilites are $(2)(3)$
  
  If 2 is reducible then we would have a root and there would be a contradiction
  
  So, just check that $x^5+x+1$ is divisible by $x^2+x+1$
  
  Reducible
2. $\R \$ Reducible because degree is greater than 2 Look up this theorem in the textbook
3. $\Z_7 \$ Reducible, $x=2$ is a root $\implies$ reducible

Notes from last lecture

Format of the exam

First question is a gift
Second question is related to frobenius $x\mapsto x^p$
Modulo certain polynomial field extension “In the spirit of the 3rd problem on the practice final”
Reducibility, one polynomial and asked if it is irreducible over certain fields
Bonus problem

Office Hours (scattered notes)

Homework #6

Question: 4

First step, multiply the polynomial by $x-1$

$x^6+x^5+x^4+x^3+x^2+x+1=0$ over $F$, $|F|=2^6=64$

$x^7 -1$

Full credit solution

Variations

Notes from last lecture

Format of the exam

Office Hours (scattered notes)

Homework #6

Question: 4