How to prove that $\mathbb R [x]$ is a UFD I could not believe that my professor asked this on the exam especially since this is intro-level abstract algebra. My idea was to somehow find a norm function and use the inequality $N(r) < N(d)$ in $m = qd + r$. I erased that and said: "since we know that $\mathbb{R}$ is an UFD we can show that $\mathbb{R[x]}$ is an UFD." We only had 30 minutes to do per problem and this was one of the unusually hard problems. 
 A: Here's a more general case which leads to proving the division algorithm for any polynomial ring $F[x]$, where $F$ is a field.
Let $f$ and $g\neq 0$ be polynomials in $R[x]$, for $R$ a commutative ring. Let $\deg(g)=m$, with $b_m$ its leading coefficient. I claim we can write
$$
b_m^j f(x)=q(x)g(x)+r(x)
$$
for $j\in\mathbb{N}$, $q,r\in R[x]$, and $\deg(r)<\deg(g)$. First, if $\deg(f)<\deg(g)$, then simply take $f(x)=0\cdot g+f$, and you're done. Otherwise, suppose $\deg(f)=n\geq m=\deg(g)$, and let $a_n$ be the leading coefficient of $f$. Set
$$
b_mf(x)-a_nx^{n-m}g(x)=f_1(x).
$$
Since the coefficient of $x_n$ in $b_mf(x)$ and $a_nx^{n-m}g(x)$ is $a_nb_m$ in both cases, $\deg(f_1)<\deg(f)$. So by induction on the degree of $f(x)$, one can find $j_1\in\mathbb{N},q_1(x),r(x)\in R[x]$ with $\deg(r)<\deg(g)$ such that
$$
b_m^{j_1}f_1(x)=g(x)q_1(x)+r(x).
$$
Combining these, you find
$$
b_m^{j_1+1}f(x)=b_m^{j_1}a_nx^{n-m}g(x)+g(x)q_1(x)+r(x)=g(x)q(x)+r(x)
$$
where $q(x)=b_m^{k_1}a_n^{n-m}+q_1(x)$.
So in particular, if $b_m$ is a unit, you can multiply both sides of the above equation by $b_m^{-1}$ appropriately many times to find
$$
f(x)=q(x)g(x)+r(x)
$$
for slightly different $q(x)$ and $r(x)$, but still $\deg(r)<\deg(g)$. Thus the division algorithm of course holds for $F[x]$ where $F$ is a field. The following shows how to use the division algorithm to show $F[x]$ is in fact a PID, and BenjaLim's very nice answer shows how to then prove a PID is a UFD. 
Let $I$ be any nonzero ideal. Then there exists a nonzero polynomial $g(x)$ of minimal degree among the nonzero elements of $I$. Then for any polynomial $f(x)\in I$, we can apply the division algorithm to find
$$
f(x)=q(x)g(x)+r(x)
$$
where $r(x)$ has lesser degree than $g(x)$. But then
$$
r(x)=f(x)-q(x)g(x)\in I
$$
so necessarily $r(x)=0$ lest we contradict the minimality of the degree of $g(x)$. So it follows that $I=(g(x))$. Since the zero ideal is obviously principal, and since $\mathbb{R}$ is a domain, so is $\mathbb{R}[x]$, it follows that $\mathbb{R}[x]$ is a PID, thus a UFD.
Note that this holds generally for $F[x]$ for any field $F$. (But only in the case of one indeterminate!)
A: You may prove that $\mathcal{N}(f) = \deg(f)$ is a Euclidean norm on $\mathbb{R}[x]$ (more generally $k[x]$ for any field $k$), thus $\mathbb{R}[x]$ is a Euclidean domain $\Rightarrow$ PID $\Rightarrow$ UFD.
And the logic you wrote down is correct: A field is (trivially) a UFD, since all elements are units, and it is usually proved in classes that $R[x]$ is a UFD if $R$ is a UFD, so your teacher should have given you points for that.
A: I think we can bypass the norm and just prove that  $\Bbb{R}[x]$ being a PID implies that is a UFD. The fact that it is a PID is clear because of the existence of the division algorithm. We now prove that it is a UFD. Now first of all we claim it is a factorisation domain; that is to say that any $f \in \Bbb{R}[x]$ can be written as a unit times a finite product of irreducibles polynomials. For suppose there is an $f$ in such a way. Then in particular because $f = 1\cdot f$ this would mean that $f$ itself cannot be irreducible and hence we can write $f = f_1f_2$ for some polynomials $f_1$ and $f_2$. These cannot be irreducible and so continuing this process we get an ascending chain  of ideals 
$$(f) \subseteq (f_1) \subseteq (f_3) \subseteq \ldots $$
that does not termiminate contradicting $\Bbb{R}[x]$ being a PID. To prove that we have uniqueness of factorisations, suppose
$$f = f_1\ldots f_k = q_1\ldots q_n$$
where the $f_i's$ and $q_m'$ are all irreducible. Now we have $(f_1\ldots f_k) = (f_1)\ldots (f_k) \subseteq (q_1).$
Because $\Bbb{R}[x]$ is a PID  we know TFAE:


*

*$q_1$ irreducible  

*$(q_1)$ is a prime ideal.

*$(q_1)$ is a maximal ideal.


It will now follow that at least one of the ideals in the product $(f_1)\ldots (f_k)$ say $(f_1)$ is contained in $(q_1)$. By maximality 
$$(f_1) = (q_1)$$
and so $f_1 = q_1$. Continuing this way all the $f_i$ are equal to the $q_j$ and so we have proved uniqueness.
