Jonathan. Frech’s WebBlog

Factoids #0 (#216)

Jonathan Frech

I) Unit polynomials with non-vanishing degree

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

This phenomenon is uniquely characterized by the following equivalence:

$$R[t]^*=R^*\iff\nexists\,0\neq a,b\in R:a\cdot b=0=a+b$$$$R[t]^*=R^*\iff\nexists\,0\neq a,b\in R:a\cdot b=0=a+b$$
Proof. Negated replication. Let $R\not\owns f=\sum_{i=0}^n\alpha_it^i\in R[t]^*,\alpha_n\neq 0$$R\not\owns f=\sum_{i=0}^n\alpha_it^i\in R[t]^*,\alpha_n\neq 0$ be a unit polynomial of non-vanishing degree $n\geq 1$$n\geq 1$. Let $g=\sum_{j=0}^m\beta_jt^j\in R[t]^*,\beta_m\neq 0$$g=\sum_{j=0}^m\beta_jt^j\in R[t]^*,\beta_m\neq 0$ denote its multiplicative inverse, i. e. $f\cdot g=1$$f\cdot g=1$.
Claim. The polynomial $g$$g$ has non-vanishing degree $m\geq 1$$m\geq 1$.
Proof. Suppose $g\in R$$g\in R$. Since $f\cdot g=\sum_{i=0}^n(\alpha_i\cdot g)t^i$$f\cdot g=\sum_{i=0}^n(\alpha_i\cdot g)t^i$, it follows from $\alpha_n\cdot g=0$$\alpha_n\cdot g=0$ that $g$$g$ is a zero divisor. However, at the same time $a_0\cdot g=1$$a_0\cdot g=1$ implies that $g$$g$ is a unit, arriving at a contradiction.
Since both $n,m\geq 1$$n,m\geq 1$, one concludes $\exists 1\leq k\leq m$$\exists 1\leq k\leq m$ as well as $\alpha_n\cdot\beta_m=0$$\alpha_n\cdot\beta_m=0$.
Existence of the desired ring elements $a,b$$a,b$ is assured by the following construction.
  • Let $k=1\nearrow m$$k=1\nearrow m$ rise discretely.
  • If $a:=\alpha_n\beta_{m-k}\neq 0$$a:=\alpha_n\beta_{m-k}\neq 0$, implying $b:=\sum_{i=1}^k\alpha_{n-i}\beta_{m-k+i}\neq 0$$b:=\sum_{i=1}^k\alpha_{n-i}\beta_{m-k+i}\neq 0$, holds, since the construction arrived at this point, one finds
    $$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.$$$$a\cdot b=\alpha_n\beta_{m-k}\cdot \sum_{i=1}^k\alpha_{n-i}\beta_{m-k+i}=\sum_{i=1}^k \underbrace{\alpha_n\beta_{m-k+i}}_{=0}\cdot \alpha_{n-i}\beta_{m-k}=0.$$
  • The above condition is met for at least one $1\leq k\leq m$$1\leq k\leq m$, since otherwise $k=m$$k=m$ would imply $\alpha_n\beta_{m-m}=0$$\alpha_n\beta_{m-m}=0$, which is im­pos­si­ble since $\alpha_n\neq 0$$\alpha_n\neq 0$ and $\beta_0$$\beta_0$ is a unit element.
By construction, $0\neq a,b$$0\neq a,b$ as well as $a+b=0$$a+b=0$ are given.
Negated implication. Setting $f:=at+1$$f:=at+1$, $g:=bt+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,$$$$f\cdot g=(at+1)\cdot (bt+1)=abt^2+(a+b)\cdot t+1=0t^2+0t+1=1,$$
showing $R\not\owns f,g\in R[t]^*$$R\not\owns f,g\in R[t]^*$.
-=-

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

Furthermore, looking at $\mathbb{Z}/6\mathbb{Z}$$\mathbb{Z}/6\mathbb{Z}$, this ring’s zero divisors are $\{0,2,3,4\}$$\{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}^*$$\mathbb{Z}/6\mathbb{Z}[t]^*=\mathbb{Z}/6\mathbb{Z}^*$ follows.

II) A closing bijection

It defines

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

an isomorphism in the category Set.

III) A ring full of zero divisors

It defines

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

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

Addition and multiplication table.

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

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

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:

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