# Abstract Nonsense

## 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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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.

July 19, 2011 -