Abstract Nonsense

Crushing one theorem at a time

Polynomial Rings (Pt. III)

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

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Isomorphism Considerations

Just as the case for the product of rings and matrix rings there is a certain ‘functorial’ property for matrix rings. In particular given two commutative unital rings R,S and a morphism f:R\to S there is a natural induced map F(f):R[x]\to S[x] given by a_0+\cdots+a_nx^n\mapsto f(a_0)+\cdots+f(a_n)x^n. Indeed:

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Theorem: Let R and S be commutative unital rings and f:R\to S be a ring morphism. Then, the map F(f):R[x]\to S[x] given by a_0+\cdots+a_n x^n\mapsto f(a_0)+\cdots+f(a_n)x^n is a ring morphism which is a unital morphism if f is. Furthermore, F(\text{id}_R)=\text{id}_{R[x]} and F(g\circ f)=F(f)\circ F(f). Moreover, F carries monomorphisms to monomorphisms and epimorphisms to epimorphisms, and thus isomorphisms to isomorphisms.

Proof: The fact that F(f) is a morphsims which is unital if f is (it’s just a calculation) as well as the fact that F(\text{id}_R)=\text{id}_{R[x]} is obvious. To prove that F(g\circ f)=F(g)\circ F(f) we note that for any a_0+\cdots+a_nx^n\in R[x] one has

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\begin{aligned}\left(F(g\circ f)\right)(a_0+\cdots+a_n x^n) &=g(f(a_0))+\cdots+g(f(a_n))x^n\\ &=(F(g))\left(f(a_0)+\cdots+f(a_n)x^n\right)\\ &=\left(F(g)\circ F(f)\right)(a_0+\cdots+a_n x^n)\end{aligned}

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The fact that F carries monomorphisms to monomorphisms, epimorphisms to epimorphisms, and isomorphisms to isomorphisms is obvious. \blacksquare

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Remark: In other words, for people who like to phrase it in this way, the map \bold{CRing}\to\bold{CRing} by taking R\mapsto R[x] and f\mapsto F(f) is an endofunctor.

From this we get the two useful corollaries:

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Corollary: Let R and S be commutative unital rings, then R\cong S implies R[x]\cong S[x]

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Corollary: Let R be a commutative unital ring and \mathfrak{a} an ideal of R. Then, \mathfrak{a}[x] is an ideal of R[x] and R[x]/\mathfrak{a}[x]\cong (R/\mathfrak{a})[x].

Proof: Consider the epimorphism R[x]\to (R/\mathfrak{a})[x] induced from the canonical epimorphism \pi:R\to R/\mathfrak{a}. Note then that the kernel of this map is evidently \mathfrak{a}[x] and so the result follows immediately from the first isomorphism theorem. \blacksquare


Polynomial Rings in Finitely Many Indeterminates

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Let R be a commutative integral, we define the polynomial ring over R in the indeterminates x_1,\cdots,x_n, denoted R[x_1,\cdots,x_n], inductively by R[x_1,\cdots,x_{n}]=R[x_1,\cdots,x_{n-1}][x_n]. In other words, R[x_1,\cdots,x_n] is the set of all polynomials in the indeterminate x_n with coefficients in R[x_1,\cdots,x_{n-1}]. So, the elements of R[x_1,\cdots,x_m] are of the form

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\displaystyle \sum_{\bold{d}\in \left(\mathbb{N}\cup\{0\}\right)^m}a_{\bold{d}}x^{\bold{d}}

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where x^\bold{d} denotes x_1^{d_1}\times\cdots x_m^{d_m} if \bold{d}=(d_1,\cdots,d_m) and a_{\bold{d}}\in R. In this context \bold{d} is known as the multidegree of the monomial x_1^{d_1}\times x_m^{d_m} and the degree of the monomial is defined by d_1+\cdots+d_m. For an arbitrary polynomial p(x_1,\cdots,x_m)\in R[x_1,\cdots,x_m] we define the degree of p to be the maximum of the degrees of all its monomial terms.

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The only other thing we note about polynomial rings in finitely many indeterminates, besides the definition and degree, is that by applying the two previous results concerning R[x] one can prove that R[x_1,\cdots,x_n] is an integral domain if and only if R is, and that the units of R[x_1,\cdots,x_n] are precisely the polynomials.

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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.


July 19, 2011 - Posted by | Algebra, Ring Theory | , , , , , , ,


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