## May 26, 2017

### A conference poster of BTABM

Here is a conference poster of the ABM project that I have been working on with Paul Mitchell.

(linked to the original file)

Bacillus thuringiensis (Bt) corn is a transgenic seed variety, used for controlling pest damage, whose adoption involves many stakeholders. We think that it is appropriate to model the situation as a complex system. Like many others, I like Python and so use Mesa as a platform. We try to get it done during the summer!

## May 15, 2017

### The irrationality of √2

When we start to learn analysis or in particular real numbers, a proof for the irrationality of $\sqrt{2}$ typically goes like this.

Assume, for contradiction, there exists $p, q \in \mathbb{Z}$ satisfying $$\left( \frac{p}{q} \right)^2 = 2$$ where $p$ and $q$ are relatively prime and $q \neq 0$. Then, rearranging, we have $p^2 = 2q^2$ and so $p$ is even or $p = 2r$ where $r\in \mathbb{Z}$. But, substituting it above, we also have $q^2 = 2r^2$ and $q$ is even as well. This is a contradiction to being relatively prime.

In the lecture on Field automorphism (at 6:35), Professor Macauley gives another proof based on the following proposition. If $\phi$ is an automorphism of $\mathbb{Q}(\sqrt{2})$ ($\mathbb{Q}$ adjoining $\sqrt{2}$), then, $$\phi(q) = q, \quad \forall q\in \mathbb{Q} .$$ One of such automorphisms is $\phi(a+b\sqrt{2}) = a-b\sqrt{2}$. But, now, $\phi(\sqrt{2}) = -\sqrt{2} \neq \sqrt{2}$. Thus, $\sqrt{2}$ is not rational. Constructing the extension field $\mathbb{Q}(\sqrt{2})$, it is tautological but nonetheless cute.

## May 4, 2017

### RR[x]/(x²+1) isomorphic to CC

It is a very basic problem in ring theory, but I keep forgetting the key ideas involved. The trick is to identify the following evaluation homomorphism: $$\varphi_i: \mathbb{R}[x] \to \mathbb{C},\quad \varphi_i(p(x)) = p(i) .$$ To me, this tells a lot. First, it maps all the real coefficients of $\mathbb{R}[x]$ into $\mathbb{C}$, reducing $\mathbb{R}^n$ to $\mathbb{R}^2$. Second, it maps $x$ into $i$ by essentially evaluating each polynomial $p(x)\in\mathbb{R}[x]$ at $i\in\mathbb{C}$. Although $i \notin \mathbb{R}$, it seems okay because the codomain $\mathbb{C}$ is an extension field of $\mathbb{R}$. Finally, it is surjective because for any $a+bi \in \mathbb{C}$ there exists $a+bx \in \mathbb{R}[x]$ such that $\varphi(a+bx) = a+bi$. The kernel captures the very motive for going to this trouble, i.e. no roots of $x^2+1$ in $\mathbb{R}$. $$\ker(\varphi) = \{p(x)\in\mathbb{R}[x] \mid p(i)=0\} = \{(x^2+1)q(x) \mid q(x)\in\mathbb{R}[x]\} = \langle x^2+1 \rangle .$$ Thus, by the first isomorphism theorem, $\mathbb{R}[x]/\langle x^2+1 \rangle \cong \mathbb{C}$.