# Abstract Nonsense

## Characteristic of a Ring (Pt. II)

Point of Post: This is a continuation of this post.

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With this characterization we have the following fascinating theorem:

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Theorem: Let $R$ and $S$ be unital rings with unities $1_R$ and $1_S$. Then, every unital homomorphism $f:R\to S$ induces a unital epimomorphism $f_\cap:R_\cap\twoheadrightarrow S_\cap$.

Proof: It’s evident that we may restrict $f$ to a unital morphism $R_\cap\to S$, thus it suffices to prove that $f(R_\cap)=S_\cap$. To do this we merely note that $f\circ u_R:\mathbb{Z}\to S$ is a unital homomorphism and thus we may conclude from prior theorem that $f\circ u_R=u_S$. Thus,

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$S_\cap=u_S(\mathbb{Z})=(f\circ u_R)(\mathbb{Z})=f(u_R(\mathbb{Z}))=f(R_\cap)$

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and so the conclusion follows. $\blacksquare$

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From this we get the following result which is useful in proving there doesn’t exist unital homomorphisms from a ring $R$ to a ring $S$:

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Theorem: Let $R$ and $S$ be unital rings. If there exists a unital homomorphism $f:R\to S$ then $\text{char}(S)\mid\text{char}(R)$.

Proof: Assume first that $\text{char}(S)=0$, then we have by definition that $S_\cap$ is infinite, and since $\#(R_\cap)\geqslant\#(S_\cap)$ by the previous theorem we may conclude that $R_\cap$ is infinite, and thus $\text{char}(R)=0$, and since $0\mid 0$ this case holds. Suppose then that $\text{char}(S)=m<\infty$. If $\text{char}(R)=0$ we are done since $m\mid 0$, so assume not, so that $\text{char}(R)=n<\infty$. We know then from basic group theory that  since $f_\cap:R_\cap\twoheadrightarrow S_\cap$ is a group epimorphism and $R_\cap,S_\cap$ are finite that $|S_\cap|\mid |R_\cap|$ or $m\mid n$. The conclusion follows. $\blacksquare$

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This allows one to conclude many interesting things about the existence of unital morphisms between two unital rings, things like there is no unital homomorphism between $\mathbb{Z}_{25}[x]\to\mathbb{Z}_{3}[x]$.

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Characteristics for Non-Unital Rings

We now seek to define the notion of characteristic for non-unital rings which generalizes the notion of the characteristic for unital rings. To motivate this consider the following:

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Theorem: Let $R$ be a unital ring with unity $1_R$. Then, $\text{char}(R)=m<\infty$ if and only if $mr=0$ for all $r\in R$ and no smaller such number has this property. Moreover, $\text{char}(R)=0$ if and only if there exists no $m\in\mathbb{N}$ such that $mr=0$ for all $r\in R$.

Proof: Suppose first that that $\text{char}(R)=m<\infty$. Then, for any $r\in R$ one has that $rm=r(1_Rm)=(r1_R)m=0m=0$ and if $k one has that $k1_R\ne 0$ and so clearly $kr=0$ does not hold for every $r\in R$. Conversely, suppose that $rm=0$ for every $r\in R$. Evidently then we have that $m1_R=0$. Note then that for each $k there exists some $r_0\in R$ such that $kr_0\ne 0$ and thus $k1_R\ne 0$ otherwise $kr_0=(k1_R)r_0=0$. Thus, $m=\text{char}(R)$.

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Suppose then that $\text{char}(R)=0$, then for every $m\in\mathbb{N}$ we have that $m1_R\ne 0$ and so $mr=0$ does not hold for every $r\in R$. Conversely, suppose that then that for every $m\in\mathbb{N}$ there exists some $r_0\in R$ such that $mr_0=0$. Clearly then $m1_R\ne 0$ since otherwise $mr_0=(m1_R)r_0=0$. $\blacksquare$

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Thus, it makes sense to generalize the notion of characteristic to non-unital rings by defining the characteristic of a (possibly non-unital) ring $R$, denoted $\text{char}(R)$,  to be $m$ if $mr=0$ for all $r\in R$ and no smaller natural number has that property and $0$ if there does not exists a natural number $m$ such that $mr=0$ for all $r\in R$. This last theorem tells us that this notion of characteristic and the one previously defined agree on unital rings and thus it is unambiguous to write $\text{char}(R)$.

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Restrictions of Characteristic on Integral Domains and Division Rings

What we now show is that the characteristic of some classes of rings are greatly restricted. In particular we shall show that for integral domains and division rings the characteristic is either $0$ or a prime. This makes a lot of sense since,well, it’s so intuitive that the intuition is the proof:

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Theorem: Let $R$ be a integral domain or a division ring, then $\text{char}(R)$ is either zero or a prime.

Proof: Suppose that $\text{char}(R)=ab$, with $a,b>1$. We then have that $ab1_R=(a1_R)(1_R)=0$. That said, since $0 we have by definition that $a1_R,b1_R\ne 0$, but this is impossible if $R$ is either a division ring or an integral domain. The conclusion follows. $\blacksquare$

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as a corollary we obviously have that

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Theorem: Let $k$ be a field, then $\text{char}(k)$ is either zero or prime.

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

June 20, 2011 -