Can Boolean ring without unit be embedded into a boolean ring?

Solution 1:

Part of modern, abstract algebra is turning imagination into reality by fiat. This is a rather general technique: say you have some concrete object $A$ and you want an element with property $P$: then just adjoin a formal element to $A$ and quotient the result by the collection of all relations that must be satisfied by that element in order for that element to have property $P$. This approach can be used for multiple different elements, and in multiple different contexts (rings, groups, fields, etc.).

Now say $A$ is a Boolean ring (so it is commutative and has characteristic two, remember). We want a new element $\epsilon$ in it that acts as the identity. What relations does the identity need to satisfy; can you turn these relations into an ideal to quotient $A[\epsilon]$ by? Now you simply need to check that the quotient $A[\epsilon]/I$ is Boolean and unital and contains a copy of $A$ (i.e. $A\to A[\epsilon]/I$ is injective).

Solution 2:

The most elementary proof is to consider $A'=\mathbf{2}\times A$, where $\mathbf{2}=\{0,1\}$ is the two element boolean ring. The operations are defined by

\begin{gather} (\alpha,a)+(\beta,b)=(\alpha+\beta,a+b)\\ (\alpha,a)(\beta,b)=(\alpha\beta,\alpha b+a\beta+ab) \end{gather} and $0a=0$, $1a=a$ by definition.

Proving that the axioms of rings are satisfied is easy. The map $A\to A'$ defined by $a\mapsto(0,a)$ is a ring homomorphism. Moreover $$ (\alpha,a)^2=(\alpha^2,\alpha a+a\alpha+a^2)=(\alpha,a) $$ so it is a Boolean ring. It has a unit, because $$ (\alpha,a)(1,0)=(\alpha,a). $$

This is the same as the usual Dorroh extension for generic rings (or $\mathbb{Z}$-algebras), only we use the fact that $A$ is a $\mathbf{2}$-algebra.

We are basically adding a complement to each element: indeed $(1,0)+(0,a)=(1,a)$, so $(1,a)$ is the complement of $(0,a)$.

Solution 3:

edit:

TLDR

This answer assumes that for any ring $R$ (with or without unit) a polynomial ring $R[x]$ can be constructed containing some element $x \in R[x]$. This is a valid assumption and so the answer stands.

On $R[x]$ for non-unital ring $R$

Mattia F. brought to my attention that $R$ didn't have a unit from start and so something like $x + I$ (with $I$ an ideal of $R[x]$) wouldn't make sense. I accepted the criticism since to refer to $x$ would require some element $1 \in R$, as polynomials over $R$ would have to be like:

$$r_0 + r_1 x + r_2 x^2 + \cdots + r_n x^n$$

Afterwards, from Martin Brandenburg's comment, I've learned that the required polynomial ring indeed makes sense. To construct $R[x]$ a formal element $x$ is adjoined to $R$ resulting, assuming coefficients $r_i \in R$ and $z_j \in \mathbb{Z}$, in elements of the form:

$$r_0 + r_1 x + \cdots + r_n x^n + z_1 x + \cdots + z_m x^m$$

For this to make sense the multiplication: $$(\cdot) : \mathbb{Z} \times R \to R$$ must be defined, as usual, as:

$$ n \cdot r = \begin{cases} 0, & \text{if $n = 0$} \\ \sum_{i=1}^{n} r, & \text{if $n > 0$} \\ (-n) \cdot (-r), & \text{if $n < 0$} \end{cases} $$

original answer:

I'll expand on the accepted answer from @anon, so maybe people should better read (and try) it first.

Let $A$ be a boolean ring without unit (that is, a ring without unit in which every element is idempotent). As a consequence of this idempotence $A$ is commutative and so consider $A[x]$, the ring of polynomials in one indeterminate over $A$. It's straightfoward that $A[x]$ has characteristic $2$, as does $A$.

Let's find a suitable ideal $I \subseteq A[x]$ such that the quotient $A[x]/I$ is a boolean ring with unit in which the original ring $A$ can be easily embedded. Furthermore let's consider only quotients in which $x$ is the desired unit. From this, for all $p = p_n x^n + \cdots + p_0 \in A[x]$ it should be that: $$ p x \equiv p \pmod{I}$$ or equivalently, as $p=-p$: $$ px + p \in I \text{, that is:}$$ $$p_nx^{n+1} + (p_n + p_{n-1})x^n + \cdots + (p_1 + p_0)x + p_0 \in I$$

Therefore, let $I$ be the set of all elements in this form together with $0$. This definition indeed yields an ideal, but to see this clearly, let's find an equivalent representation. Let $p = c_n x^n + \cdots + c_0 \in I$ be of degree $n > 0$. By definition of $I$, there are $n$ elements $p_0, \ldots, p_{n-1} \in A$ such that:

$$ \begin{cases} c_n = p_{n-1} \\ c_{n-1} = p_{n-1} + p_{n-2}\\ \cdots \\ c_1 = p_1 + p_0\\ c_0 = p_0 \end{cases} \iff \begin{cases} c_0 + c_1 + \cdots + c_{n-1} + c_n = 0 \\ c_0 + c_1 + \cdots + c_{n-1} = p_{n-1}\\ \cdots \\ c_0 + c_1 = p_1\\ c_0 = p_0 \end{cases} $$

So $I$ can be expressed by: $$ I = \Big\{ c_n x^n + \cdots + c_0 \in A[x] ~\|~ c_n + \cdots + c_0 = 0 \text{, with } n \geq 0\Big\}$$

The sum of elements of $I$ results in polynomials whose coefficients summed up also give $0$. Let $a = a_m x^n + \cdots + a_0$ be any polynomial in $A[x]$ and $p = p_n x^n + \cdots + p_0 \in I$ and consider their product $q = a p$. It's not difficult to see that the sum of the coefficients of $q$ gives: $$(a_m + \cdots + a_0)(p_n + \cdots + p_0) = 0$$ and therefore $q \in I$. It remains to prove that all elements in $A[x]/I$ are idempotent.

Observe that, as $A$ has characteristic $2$ and all of its elements are idempotent, for any $a_n x^n + \cdots + a_0 \in A[x]$: $$ (a_n x^n + \cdots + a_0)^2 = a_n^2 x^{2n} + \cdots + a_1^2x^2 + a_0^2 = a_n x^{2n} + \cdots + a_1x^2 + a_0 $$

From this, it follows that: $a^2 + a \in I$, that is: $$a^2 \equiv a \pmod{I}$$

The element $x + I$ is an unit of $A[x]/I$ by construction, but just to make sure, observe that $ax + a \in I$, for in the sum of all coefficients of $ax + a$, the coefficients of $a$ are summed exactly two times.

At last, if $a = b \pmod{I}$ for $a, b \in A$, then the sum of the coefficients of $a + b$ equals $0$, wich means that $a = b$. So the homomorphism $i: A \to A[x]/I$ given by $a \mapsto a + I$ is injective.