# Abstract Nonsense

## Polynomial Rings in Relation to Euclidean Domains, PIDs, and UFDs (Pt. I)

Point of Post:

In this post we discuss how polynomial rings act when their underlying rings are PIDs, Euclidean domains, and UFDs. The high points being the proof that $k[x]$ is a Euclidean domain for fields $k$ and $R[x]$ is a UFD when $R$ is a UFD.

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Motivation

We have mentioned several times that polynomial rings occupy a central role in modern mathematics. Consequently, studying them should occupy no small amount of our time. In this post we shall prove some previously made assertions about when polynomial rings are Euclidean domains, PIDs, and UFDs and some of the consequences of thes results.

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Polynomial Rings Over Fields are Euclidean Domains

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We have mentioned several times now that polynomial rings $k[x]$ where $k$ is a field are Euclidean domains. We now prove this result:

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Theorem: Let $k$ be a field, then for every $a(x),b(x)\in k[x]$ there exists unique $q(x),r(x)\in k[x]$ such that $b(x)=q(x)a(x)+r(x)$ and $r(x)=0$ or $\deg(r(x))<\deg(a(x))$.

Proof: If $b(x)=0$ we may take $q(x)=r(x)=0$ and if $\deg(b(x))<\deg(a(x))$ then we may take $q(x)=0$ and $r(x)=b(x)$. So, from now on we assume that $b(x)\ne 0$ and $\deg(b(x))\geqslant \deg(a(x))$. We proceed then by induction on $\deg(b(x))=n$. If $n=0$ then $\deg(a(x))=0$ and the rest follows since $k$ is a field. So, assume the result is true for $\deg(b(x)) and let $\deg(b(x))=n$, say $b(x)=b_0+\cdots+b_nx^n$. Let us be given $a(x)=a_0+\cdots+a_kx^k$ with, by assumption, $n\geqslant k$. Consider then the polynomial $c(x)=b(x)-b_na_k^{-1}x^{n-k}a(x)$. We know that $c(x)$ is of degree less than $n$ and so by the induction hypothesis there exists $q'(x),r(x)$ with $c(x)=q'(x)a(x)+r(x)$ with $r(x)=0$ or $\deg(r(x))<\deg(a(x))$. Letting then $q(x)=q'(x)+b_na_k^{-1}x^{n-k}$ gives us that $b(x)=q(x)a(x)+r(x)$ with $r(x)=0$ or $\deg(r(x))<\deg(a(x))$. This completes the induction. $\text{ }$ To see that this is unique suppose that $q(x)a(x)+r(x)=b(x)=q'(x)a(x)+r'(x)$ we see then that $a(x)(q(x)-q'(x))=r'(x)-r(x)$ and since the degree of the right hand side is strictly less than the degree of $a(x)$ we must have that $q(x)-q'(x)=0$ and so $q(x)=q'(x)$ from where it easily follows that $r(x)=r'(x)$. $\blacksquare$

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With this we can now state:

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Theorem: Let $k$ be a field, then $(k[x],\deg)$ is a Euclidean domain and $\deg$ satisfies the $d$-inequality.

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This gives us the unproven part of:

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Theorem: Let $R$ be a ring. Then, $R[x]$ is a PID or a Euclidean domain if and only if $R$ is a field.

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Now, the mere existence of this division algorithm allows us to conclude some important things about polynomials over a field:

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Theorem: Let $k$ be a field and $p(x)\in k[x]$. Then, $x-a\mid p(x)$ if and only if $p(a)=0$.

Proof: Evidently if $x-a\mid p(x)$ then $p(a)=0$. Conversely, suppose $p(a)=0$. We know that $p(x)=(x-a)q(x)+r$ where $r$ is a constant. Since $r=(a-a)q(a)+r=p(a)=0$ we may conclude that $r=0$ and so $p(x)=(x-a)q(x)$. $\blacksquare$

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From this and the fact that the degree function is additive allows one to conclude the following:

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Corollary: Let $k$ be a field and $p(x)\in k[x]$ be non-zero. Then the number of distinct roots in $k$ of $p(x)$ does not exceed $\deg(p(x))$.

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We are now in position to prove a long-discussed theorem. Namely:

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Theorem: Let $R$ be an infinite integral domain and $\text{Fun}(R)$ the ring of all functions $R\to R$ with the usual operations. Then the obvious map $R[x]\hookrightarrow \text{Fun}(R)$ is an embedding.

Proof: We have already remarked that this map is a unital homomorphism of rings, and so it suffices to prove that it’s injective. This is equivalent though to showing that if $p(x)=a_0+\cdots+a_nx^n$ satisfies $p(a)=0$ for all $a\in R$ then $a_0=\cdots=a_n=0$. To see this we proceed in two steps. $\text{ }$ First, we prove this result for the field of fractions $\text{Frac}(R)$ of $R$. But, this is obvious, if $p(x)\in\text{Frac}(R)[x]$ is identically zero as a function on $\text{Frac}(R)$ then $p(x)$ has infinitely many roots in $\text{Frac}(R)$ and so by the previous corollary must be zero.

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Suppose now that $p(x)\in R[x]$ is identically zero as a function on $R$ we claim that it is identically zero as a function on $\text{Frac}(R)$. But, this is obvious since then $p(x)$ would have infinitely many roots in $R\subseteq\text{Frac}(R)$. Thus, $p(x)$ is the zero polynomial. $\blacksquare$

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So, in an integral domain $R$ viewing a polynomial as a set of coefficients and as a function are the same thing.

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References:

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

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

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

October 24, 2011 -

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