# Algebra Problems Oct 31

Reminders: 

- **RRT**: $f(x) = a_nx^n + \cdots + a_1x + a_0, f(r/s) = 0$ in lowest terms then $r\mid a_0$ and $s\mid a_n$. 

  Proof: expand $f(r/s)$, move $a_0s^n$ to the RHS, factor out $r$.

- **Gauss**: $f\in \mathbf Z[x]$ factors in $\mathbf Z[x] \iff f$ factors in $\mathbf Q[x]$.

- If $f\in \mathbf{Q}[x]$ has no rational root, it can only have at least a quadratic irreducible factor.

- **Undetermined coefficients**: e.g. $x^4-x^3 +2 = (x^2+ax+b)(x^2+cx + d)$ and solve for integers $a,b,c,d$.

- **Reduction**: if $f\in \mathbf Z[x]$ is reducible then it is reducible $\mod p$ for every $p$; so if $\bar f\in \mathbf F_p[x]$ is irreducible for any $p$ then $f\in \mathbf Q[x]$ is irreducible.

- **Eisenstein**: $f(x) = a_nx^n + \cdots + a_1 x + a_0$, $a_j \equiv 0\mod p$ for all $0\leq j\leq n-1$ with $a_n\not\equiv 0\mod p$ and $a_0\not\equiv 0 \mod p^2$ implies $f\in \mathbf Q[x]$ is irreducible.

  Proof: reduce mod $p$ to get $\bar f(x) = \bar a_n x^n = \bar g(x) \bar h(x)$, each a monomial, and show that if $g(x) = b_kx^k +\cdots + b_0$ then $\bar b_k \neq 0, \bar b_0\neq 0$, forcing $g$ to be constant.

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- 3.2.2a
  ![image-20221031132137506](/home/zack/.config/Typora/typora-user-images/image-20221031132137506.png)
  - $\subseteq$: STS that $\sqrt 2$ and $i$ are in the latter.
    Let $s = \sqrt{2} + i$, then $s^2 = 2i\sqrt{2}+1$ and $s^3 = -\sqrt{2} + 5i$, so $s^3 + s =i + 5i = 6i\in K := \mathbf Q[\sqrt 2 + i] \implies (1/6)6i = i \in K \implies s-i = \sqrt{2}\in K$. 
  - $\supseteq$: Clear.
  - For $\mathbf Q[i\sqrt 2]$: fields are closed under products, so $\sqrt 2, i\in \mathbf Q [\sqrt 2, i]\implies \sqrt 2 \cdot i \in \mathbf Q [\sqrt 2, i]$. For strict containment, it STS there is an element in the latter and not the former. 
    Could check minimal polynomials: $t\:= i\sqrt 2 \implies t^2 = -2 \implies t^2+2 = 0$, so this is a quadratic extension, but $\mathbf Q[\sqrt 2, i]$ is a degree 4 extension.
    
    More directly, probably $i\not\in K := \mathbf Q[i\sqrt 2]$ -- see Ex. 3.2.2 in the chapter. Write $t=i\sqrt 2$, then $t^2 = -2$ so $K = \left\{a + bt \mid a,b,\in \mathbf Q\right\}$. Set $i= a+b\sqrt 2i$, this forces $b\neq 0$, subtract $b\sqrt 2 i$ to get $a = i-b\sqrt{2} i = i(1- b\sqrt 2)$, square both sides to get $a^2 = -(1-b\sqrt 2)^2 = 1 -2b\sqrt 2 + 2b^2$ and rearrange to get $\sqrt{2} = {1-a^2+2b^2\over 2b}$, contradicting $\sqrt{2}\not\in \mathbf Q$.
    $$
    i = a+bi\sqrt 2 \implies a = i-bi\sqrt 2 = i(1-b\sqrt 2)\implies a^2= -(1-b\sqrt 2)^2 = 1 - 2b\sqrt 2 + 2b^2 \implies \sqrt{2} = {\cdots\}
    $$
    
- 3.3.2abcdfi (Eisenstein, RRT, reduction mod $p$, undetermined coefs)
  ![image-20221031142254564](/home/zack/.config/Typora/typora-user-images/image-20221031142254564.png)
  - a: Irreducible. Reduce to $\mathbf{F}_2$.
  - b: Irreducible. Eisentstein with $p=3$.
  - c: Reducible. RRT implies roots are $\pm 1$, check $f(-1) = -1 + 1 -1 + 1 = 0$.
  - d: Irreducible. Eisenstein with $p=2,3,5$.
  - f: Irreducible. Reduce to $\mathbf F_2$ and check that $\bar f(\bar 1) = \bar 1, \bar f (\bar 0) = \bar 1$.
  - i: Reducible: RRT implies roots are $\pm 1$, so check $f(-1) = -1 + -1 + 1 + 1 = 0$.
- 3.3.3ce
  ![image-20221031142314010](/home/zack/.config/Typora/typora-user-images/image-20221031142314010.png)
  - c: For a root $r/s$, we have $s\mid 2, r\mid 6$, so $\pm s=1,2$ and $\pm r=1,2,3,6$ and thus $x= {\pm 1,2,3,6\over \pm 1,2}$. Guess and check yields $f(-3/2) = 0$. Long division yields $f(x) = (x^3+2)(2x + 3)$, so that's the only one.
  - e: Need integer polynomial, so $f(x) = 0 \iff 5x^3 - x^2 - 20 x + 4 = 0$ so $x = {\pm 1,2,4 \over \pm 1,5}$. Guess $f(2) = 0$ and $f(-2) = 0$, long division by $(x-2)(x+2) = x^2 + 4$ yields $f(x) = (x^2+4)(x-{1\over 5})$, or just guess $f(1/5)=0$.
- 3.3.4c
  - Roots are of the form $x = {\pm 1,2,3,4,6,12 \over \pm 1}$.
    - $k$ even: 
      - $f(1) = 1 + 3-12 \neq 9$
      - $f(-1) = 1 -3 - 12 \neq 0$
      - $f(2) = 4^k + 6 \cdot2^k - 12$, for $k\geq 1$ the first part is larger than 12
      - Factor $f(x) = x^k(x^k + 3x) - 12$, then $f(-2) = -2^k(-2^k-6) -12 = 2^k(2^k+6)-12$ and the first term is bigger than 12 for $k\geq 1$
      - Similarly $f(3) = 3^k(3^k + 9) - 12$, same issue with first term.
    - $k$ odd: 
      - Similar analysis.
- 3.3.6
  ![image-20221031143830981](/home/zack/.config/Typora/typora-user-images/image-20221031143830981.png)
  - a. Check $f(-1) = f(1) = \sum_{i=0}^n a_i = 0 \iff \#\left\{ i \mid a_i\neq 0 \right\} \equiv_2 0$.
  - b. All polynomials. Degree $n\implies 2^n$ possibilities. Count in binary.
    - Degree 2: 
      - $000\leadsto 0$
      - $001\leadsto 1$
      - $010 \leadsto x$, irreducible.
      - $011\leadsto x+1$, reducible by above.
      - $100\leadsto x^2$, reducible.
      - $101\leadsto x^2+1$ , reducible by above.
      - $110\leadsto x^2 + x$, reducible by above.
      - $111\leadsto x^2 + x + 1$, irreducible.