## Gauss’s Lemma

In the proof that $R[x]$ is a UFD if $R$ is one, we need the following fact, which is often called “Gauss’s Lemma.”

**Theorem:** Let $R$ be a UFD. Then if a polynomial $p(x)\in R[x]$ is reducible in $K(R)[x]$, it is reducible in $R[x]$.

If $R=\Z$, then what this is saying is that if a polynomial can be factored in a nontrivial way in $\Q[x]$ – meaning using polynomial factors whose coefficients have denominators – then it can be factored in $\Z[x]$, meaning without denominators. More precisely, if $p(x)\in R[x]$ and \(p(x)=A(x)B(x)\) where $A(x)$ and $B(x)$ are in $K(R)[x]$, then there are elements $a$ and $b$ in $R$ such that $aA(x)$ and $bB(X)$ are in $R[x]$ and $abA(X)B(X)=p(x)$.

To see that there is some content to this, let $R=\Z[\sqrt{-3}]$. Consider the polynomial $x^2-x+1\in R[x]$. This polynomial factors in $Q(\sqrt{-3})[x]$ as \(x^2-x+1=(x-\rho)(x-\bar{\rho})\) where \(\rho = \frac{1+\sqrt{-3}}{2}.\) So this polynomial factors in $K(R)[x]$. But it cannot factor in $R[x]$ because it’s monic and the roots don’t lie in $R[x]$.

This means $R$ is not a UFD and in fact the idea generated by $2$ and $1+\sqrt{-3}$ is not principal.

However, the ring $\Z[\rho]$ *is* a PID.

**Proof of Gauss’s Lemma:** Given $p(x)$ in $R[x]$, where $R$ is a UFD, assume $p(x)=a(x)b(x)$ where both factors are in $K(R)[x]$. Let $d$ be a common denominator so $dp(x)=a_1(x)b_1(x)$ where $a_1$ and $b_1$ are in $R[x]$. Now, since $R$ is a UFD, we can factor $d$ into a product of irreducibles $\pi_{i}$. Formally speaking the proof is on the number of irreducible factors of $d$. If $d$ is a unit, then our factorization is already over $R[x]$, so suppose our result is true for $d$ with at most $n$ irreducible factors. In other words, if we have $dp(x)=a(x)b(x)$ with $d$ having at most $n$ irreducible factors, then there is an expression $p(x)=a’(x)b’(x)$ with $a’$ and $b’$ in $R[x]$. Now suppose we have an expression $dp(x)=a_1(x)b_1(x)$ where $d$ has $n+1$ factors. Let $\pi$ be one them.

Since $R$ is a UFD, the ideal $\pi R$ is prime and therefore $(R/\pi R)[x]$ is an integral domain. Since $a_1(x)b_1(x)\equiv 0\pmod{\pi R[x]}$, one of them must be zero; say $a_1(x)$. That means all the coefficients of $a_1(x)$ are divisible by $\pi$ so we can divide $a_1(x)$ by $\pi$ and get $a_2$ which still has cofficients in $R[x]$. Now we have $(d/\pi)p(x)=((a(x))/\pi)b(x)$. By induction we get the factorization of $p(x)$ over $R$.

**Remark:** A polynomial $p(x)$ over $\Z$ is called primitive if its coefficients are relatively prime. This theorem is sometimes expressed over $\Z$ by saying that the product $p(x)q(x)$ of two *primitive* polynomials is primitive.

## Eisenstein’s Criterion

**Theorem:** Suppose $R$ is an integral domain. If $f(x)=a_0+a_1 x+\ldots + x^n\in R[x]$ is a monic polynomial and $P$ is a prime ideal of $R$ such that $a_i\in P$ for $i=0,\ldots, n-1$, and if $a_0\not\in P^2$, then $f(x)$ is irreducible.

**Proof:** Suppose $f(x)=a(x)b(x)$. Then $f(x)\equiv x^n\equiv a(x)b(x)\pmod{P}$, This means that the constant terms of $a(x)$ and $b(x)$ have product zero mod $P$ so must be in $P$. But then the constant term of $f$ would be in $P^2$.