(parentheses)

Polynomial $A = x^n + a_{n-1}x^{n-1} + .. + a_1 x + a_0$ , where each $a_i$ is divisible by some number $p$ , is irreducible in $\mathbb Q$ when $p^2$ does not divide $a_0$ .
If A is not irreducible then there are some $F$ and $G$ such as $A = FG$ .
But that means that $a_0 = f_0 g_0$ , which means that $(p | f_0)$ XOR $(p | g_0)$ (since $p^2$ does not divide $a_0$ ).
Let’s assume that $p | f_0$ . Now let’s take a map $Z[x] \rightarrow (Z/pZ)[x]$ (I believe this method is called homomorphism-reduction).
$[A] = x^n$ , but $[G] = \cdots + [g_0]$ , where $[g_0] \neq 0$ .
That is possible only when $[G]$ is constant and $[F] = [A]/[G]$ . Which means that $A$ is irreducible.

Note: here $[a]$ denotes $f(a)$ where $f$ is a homomorphism $Z[x] \rightarrow (Z/pZ)[x]$ .

Example of using Eisenstein criterion..

Written by oh

September 29, 2011 at 12:42 pm

Posted in Math

Tagged with algebra, math, ring theory

Union of countable sets is countable.

with 2 comments

Let’s say we have a possibly infinite collection of countable sets $A_1, A_2, \cdots, A_n, \cdots$ , prove that $\bigcup A_i$ is countable too. We are going to do this using Axiom of Choice.

Since every set $A_i$ is countable, there is a set of injections from $N$ to $A_i$ , let’s call it $M_i$ . Using the axiom of choice we can create a set of maps $\{g_1, g_2, \cdots\}$ from $N$ to $A_i$ . Now, we are going to create an injection from $(N, N)$ to $\bigcup A_i$ . If $x_k$ is an element of $\bigcup A_i$ then it’s also an element of some set $A_m$ . Therefore we can construct $f(z, m) = x_k$ where $x_k = g_m(z)$ . And since we know that $(N, N)$ is countable, we can map $N$ to $\bigcup A_i$ . QED.