Factoids #0

I) unit polynomials with non-vanishing degree

$2t+1\in\mathbb{Z}_4[t]$ is its own multiplicative inverse, showing that $R[t]^*=R^*$ does not hold in a general commutative Ring with one.

This phenomenon is uniquely characterized by the following equivalence:

$R[t]^* = R^* \iff \nexists \,0\neq a,b\in R:a\cdot b=0=a+b$

Proof. Negated replication.

Let $R\not\owns f=\sum_{i=0}^n\alpha_it^i\in R[t]^*,\alpha_n\neq0$ be a unit polynomial of non-vanishing degree $n\geq 1$ . Let $g=\sum_{j=0}^m\beta_jt^j\in R[t]^*,\beta_m\neq0$ denote its multiplicative inverse, i.e. $f\cdot g=1$ .

Claim. The polynomial $g$ has non-vanishing degree $m\geq 1$ .
Proof. Suppose $g \in R$ . Since $f\cdot g=\sum_{i=0}^n(\alpha_i\cdot g)t^i$ , it follows from $\alpha_n\cdot g=0$ that $g$ is a zero divisor. However, at the same time $a_0\cdot g=1$ implies that $g$ is a unit, arriving at a contradiction.

Since both $n,m\geq 1$ , one concludes $\exists 1\leq k\leq m$ as well as $\alpha_n\cdot\beta_m=0$ .

Existence of the desired ring elements $a,b$ is assured by the following construction.

Let $k=1 \nearrow m$ rise discretely.

If $a:=\alpha_n\beta_{m-k}\neq 0$ , implying $b:=\sum_{i=1}^k\alpha_{n-i}\beta_{m-k+i}\neq 0$ , holds, since the construction arrived at this point, one finds

$a\cdot b=\alpha_n\beta_{m-k}\cdot \sum_{i=1}^k\alpha_{n-i}\beta_{m-k+i}=\sum_{i=1}^k \underbrace{\alpha_n\beta_{m-k+i}}_{=0}\cdot \alpha_{n-i}\beta_{m-k}=0.$

The above condition is met for at least one $1\leq k\leq m$ , since otherwise $k=m$ would imply $\alpha_n\beta_{m-m} = 0$ , which is impossible since $\alpha_n\neq 0$ and $\beta_0$ is a unit element.

By construction, $0\neq a,b$ as well as $a+b=0$ are given.

Negated implication.

Setting $f:=at+1, g:=bt+1$ , one calculates

$f\cdot g=(at+1)\cdot (bt+1)=abt^2+(a+b)\cdot t+1=0t^2+0t+1=1$ ,

showing $R\not\owns f,g\in R[t]^*$ .

q.e.d.

As a corollary, the property $R[t]^*=R^*$ follows for any integral domain.

Furthermore, looking at $\mathbb{Z}/6\mathbb{Z}$ , this ring’s zero divisors are $\{0,2,3,4\}$ , with no mutual zero divisors summing to zero. Using the above, $\mathbb{Z}/6\mathbb{Z}[t]^*=\mathbb{Z}/6\mathbb{Z}^*$ follows.

II) A closing bijection

It defines

$\varphi:(0, 1)\to(0,1],\alpha\mapsto\begin{cases}1/n, & \exists n\in\mathbb{N}:\alpha=1/(n+1)\\\alpha, & \text{otherwise}\end{cases}$

an isomorphism in the category Set.

III) A ring full of zero divisors

It defines

$\left\{ \begin{pmatrix}0&0\\0&0\end{pmatrix}, \Lambda := \begin{pmatrix}0&1\\0&0\end{pmatrix}, \begin{pmatrix}1&0\\0&0\end{pmatrix}, \begin{pmatrix}1&1\\0&0\end{pmatrix} \right\}\leq \mathrm{Mat}_2(\mathbb{F}_2)$

a non-commutative ring without one of cardinality four in which every element is a zero divisor with left-annihilating element Λ:

Thanks to Nathan Tiggemann for finding this marvelous algebraic structure.

Generalizing, any commutative ring with one R induces a non-commutative ring without one on which Λ acts as an omni-right-annihilator, namely

$\tilde R:=\left\{\begin{pmatrix}a&b\\0&0\end{pmatrix}\Biggm| a,b\in R\right\}$ .

As a corollary, by constructing the above ring using the reals, one obtains a ring with a (left-factored) polynomial ring housing a polynomial of degree one having uncountably many roots:

$\begin{pmatrix}0&1\\0&0\end{pmatrix}\cdot t\in \tilde{\mathbb{R}}[t]$ .

15th of Jun, 201914th of Jun, 2019 by jfrech Categories: MathematicsTags: 2019, algebra, rings Leave a comment

I) unit polynomials with non-vanishing degree

II) A closing bijection

III) A ring full of zero divisors

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