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Factoids #0

2019-06-15, post № 216

mathematics, #algebra, #rings

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\neq 0 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\neq 0 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]^*.

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:

factoids-0_a-ring-full-of-zero-divisors_addition-and-multiplication-table-of-mat2-f2.png
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\}.

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].