# Abstract Nonsense

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

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

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Polynomial Rings as UFDs

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So, we have stated earlier that a polynomial ring is a UFD if and only if its underlying ring is a UFD. But, why is this true? Certainly if the polynomial ring is a UFD then the underlying ring must be a UFD (if this isn’t obvious we shall prove it shortly) it is the other direction that is difficult.  There seems to be an obvious way one might try to approach this. Namely, if $R$ is a UFD we know that $\text{Frac}(R)[x]$ is a UFD and since $R[x]\subseteq\text{Frac}(R)[x]$ we can factor any element of $R[x]$ into irreducibles in $\text{Frac}(R)[x]$ . Hopefully, we can pass this factorization down. This is the content of Gauss’s Lemma:

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Lemma(Gauss): Let $R$ be a UFD and let $p(x)\in R[x]$. Let $k=\text{Frac}(R)$. If $p(x)$ is reducible in $k[x]$ then $p(x)$ is reducible in $R[x]$.

Proof: Since $p(x)$ is reducible in $k[x]$ we can write $p(x)=a(x)b(x)$ with $a(x),b(x)\in k[x]$. Multiplying through by an appropriate constant $m$ we have that $mp(x)=A(x)B(x)$ where now $A(x),B(x)\in R[x]$. If $m\in R^\times$ then $p(x)=m^{-1}A(x)B(x)$ and we’re done, so assume that $m\notin R^\times$. Since $R$ is a UFD and $d$ is a non-zero non-unit we can write it as a product of irreducibles, say $m=p_1\cdots p_n$. Since $p_1$ is irreducible (and so prime) we have that $(p_1)\in\text{Spec}(R)$. We know then by previous discussion that $p_1R[x]$ is a prime ideal in $R[x]$. and $R[x]/p_1R[x]\cong (R/(p_1))[x]$ is an integral domain. But, note that $A(x)B(x)+R[x]/p_1R[x]=0+R[x]/p_1R[x]$ and since $R[x]/p_1R[x]$ is an integral domain we may conclude that $A(x)$ or $B(x)$ is in $R[x]/p_1R[x]$ which implies, if we say that $A(x)\in R[x]/p_1R[x]$, that $p_1^{-1}A(x)\in R[x]$. So, $p_2\cdots p_n p(x)=(p_1^{-1}A(x))B(x)$. Continuing in this way allows us to write $p(x)=C(x)D(x)$ with $C(x),D(x)\in R[x]$ (of the form $\displaystyle C(x)=\prod_i p_i^{-1}A(x)$ and $\displaystyle D(x)=\prod_j p_j^{-1}B(x)$ where $i,j$ run over some two subsets of $[n]$). $\blacksquare$

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Note that in all reality the only difference between factorization in $R[x]$ and $\text{Frac}(R)(x)$ come from units in the coefficients. With this in mind we define the content of a polynomial $a_0+\cdots+a_nx^n=p(x)\in R[x]$, denoted $c(p)$, to be the ideal $(a_0,\cdots,a_n)$.  With this idea and Gauss’s lemma it becomes clear that:

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Theorem: Let $R$ be a UFD and let $p(x)\in R[x]$. If $c(p)=R$ then $p(x)$ is irreducible in $R[x]$ if and only if $p(x)$ is irreducible in $\text{Frac}(R)[x]$.

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We now wish to prove what we set out to do:

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Theorem: A ring $R$ is a UFD if and only if $R[x]$ is a UFD.

Proof: Suppose first that $R$ is a UFD and let $k=\text{Frac}(R)$. Write $p(x)$ as $dp'(x)$ where $d$ is a greatest common divisor of the coefficients of $p(x)$ (i.e. a generator of $c(p)$). We know that $d$ can be factored uniquely into irreducibles in $R$ and so it suffices to show that $p'(x)$ can be factored uniquely. From this it’s clear that in general, we can assume that $p(x)$ has content $R$ (i.e. the coefficients are coprime) and that they are non-constant.

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So, let $p(x)$ be non-unit non-zero polynomial in $R[x]$ with content $R$. Since $p(x)$ is non-zero and non-unit in $k[x]$ we know that $p(x)$ can be factored into irreducibles in $k[x]$.  But, we know that each of the factors in this factorizations are just irreducibles in $R[x]$ times a unit in $k$. But, since our original polynomial has content one we have by the previous theorem that each of the factors are irreducible in $R[x]$. Thus, every element of $R[x]$ can be factored into irreducibles.

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It now remains to show that every irreducible element of $R[x]$ is prime. This is mostly just busy work, and can be found in [2] on page 308-309.

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So, now let’s formally prove that $R[x]$ being a UFD implies that $R$ is a UFD. We already know that $R[x]$ being an integral domain implies that $R$ is an integral domain and so it suffices to prove that every irreducible in $R$ is prime and that every element of $R$ can be written as a product of irreducibles. Now, suppose that $a\in R$ is irreducible, then clearly $a$ is irreducible in $R[x]$ (since the only way to factor it is as constant terms etc.) and so by assumption $(a)$ is prime in $R[x]$. Thus, $R[x]/(a)$ is an integral domain, but $R[x]/(a)\cong (R/(a))[x]$ and so $(R/(a))[x]$ is an integral domain. Thus, $R/(a)$ is an integral domain and so $(a)$ is prime in $R$. Now,  let $a\in R$ be arbitrary. Since $R[x]$ is a UFD we can factor $a$ as a product of irreducibles $p_1\cdots p_n$ in $R[x]$, but note that being irreducible in $R[x]$ implies being irreducible in $R$, and since $R[x]$ is an integral domain (so that degree is additive) we know that $p_1,\cdots,p_n\in R$. Thus, we can factor any $a\in R$ as the product of irreducibles.$\blacksquare$

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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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