Abstract Nonsense

Crushing one theorem at a time

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<m 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<m 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<a,b<ab 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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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 - Posted by | Algebra, Ring Theory | , , , , , ,


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