# Corollary to lemma on primitive polynomials in a UFD

I've been following this excellent resource on abstract algebra and became stuck on proving a corollary to a lemma (18.26 in the text):

Let $$D$$ be a UFD and $$F$$ its field of fractions. Suppose that $$p(x)\in D[x]$$ and $$p(x)=f(x)g(x)$$, where $$f(x)$$ and $$g(x)$$ are in $$F[x]$$. Then $$p(x)=f_1(x)g_1(x)$$, where $$f_1(x)$$ and $$g_1(x)$$ are in $$D[x]$$. Furthermore, $$deg~f(x)=deg~f_1(x)$$ and $$deg~g(x)=deg~g_1(x)$$.

The corollary is that:

Let $$D$$ be a UFD and $$F$$ its field of fractions. A primitive polynomial $$p(x)$$ in $$D[x]$$ is irreducible in $$F[x]$$ if and only if it is irreducible in $$D[x]$$.

My problem is that I can prove both directions, but nowhere do I use the fact that $$p(x)$$ is primitive:

$$\Rightarrow$$ : Let $$p(x)$$ be irreducible in $$F[x]$$, then it must be that $$p(x)$$ must take the form $$a \cdot g(x)$$, for $$a\in F$$, $$g(x)\in F(x)$$. Now suppose $$p(x) = f_1(x) g_1(x)$$, with $$f_1(x)$$, $$g_1(x)$$ $$\in D[x]$$. But since every polynomial in $$D[x]$$ is in $$F[x]$$, we must have that $$f_1(x) \in F[x]$$ a constant (which is some multiple of $$a$$), and $$deg~g_1(x)=deg~g(x)$$ (or vice versa). Hence $$p(x)$$ is irreducible in $$D[x]$$.

$$\Leftarrow$$: Let $$p(x)$$ be irreducible in $$D[x]$$. Suppose $$p(x)=f(x)g(x)$$, where $$f(x)$$ and $$g(x)$$ are in $$F[x]$$. Then by the lemma above, we have $$p(x)=f_1(x)g_1(x)$$, where $$f_1(x)$$,$$g_1(x)$$ $$\in D[x]$$ and have the appropriate degrees. Since $$p(x)$$ is irreducible in $$D[x]$$, then we must have $$f_1(x)$$ a constant, and $$deg~g_1(x)=deg~p(x)$$ (or vice versa). Then $$deg~f(x)=0$$, $$deg~g(x)=deg~p(x)$$, and so $$p(x)$$ is irreducible in $$F[x]$$.

I'm probably missing something obvious so if some kind soul could point this out to me, I'd be grateful!

EDIT: Incorporating @TokenToucan's hint in the comments, for the $$\Leftarrow$$ part, I append:

Let $$f_1(x)=a_1 (\neq 0) \in D$$, and let the content of $$g_1(x)$$ be $$b_1$$. Since $$p(x)$$ is primitive, we have $$1=a_1 b_1$$, hence $$a_1$$ is a unit.

But my issue with this is that while I use the fact that $$p(x)$$ is primitive, I no longer need to use the lemma that the corollary depends upon! I could just do:

$$\Leftarrow$$: Let $$p(x)$$ be irreducible in $$D[x]$$ .Then we have $$p(x)=f_1(x)g_1(x)$$. Since $$p(x)$$ is irreducible in $$D[x]$$, then we must have $$f_1(x)$$ a constant, and $$deg~g_1(x)=deg~p(x)$$ (or vice versa). Let $$f_1(x)=a_1 (\neq 0) \in D$$, and let the content of $$g_1(x)$$ be $$b_1$$. Since $$p(x)$$ is primitive, we have $$1=a_1 b_1$$, hence $$a_1$$ is a unit. Since $$D\subset F$$, we have $$p(x) = a_1 g_1(x)$$ with $$a_1 \in F$$ and $$g_1(x) \in F[x]$$ so $$p(x)$$ is irreducible over $$F[x]$$.

• In your second half, just because $deg(f)=0$ makes zero does not mean you are done. You need to show that it is a unit. Eg $4(x+1)$ over the integers will have $f(x)=4$. You need to use primitivity to rule that out.
– user208649
Aug 5, 2020 at 19:55
• Thanks @TokenToucan! Aug 5, 2020 at 20:44
• Sorry, @TokenToucan, I just confused myself again. By postulate $f(x)$ is in $F[x]$, so if $f(x)$ is some constant $c$, then surely $c$ is a unit? All I have claimed is that $f_1(x) \in D[x]$ is a constant, which means that its correspondent in $f(x) \in F[x]$ is a constant, but not the same constant as $f_1(x)$, and since $f(x)$ takes its coefficients from a field this constant has to be a unit. Aug 5, 2020 at 22:18
• It is a unit in the field but not necessarily a unit in D, and the latter is what you need. Otherwise you have a nontrivial factorization.
– user208649
Aug 5, 2020 at 22:49