# Proof that for primitive $p(x) \in D[x]$, $p(x)$ irreducible in $F[x]$ $\Leftrightarrow$ $p(x)$ irreducible in $D[x]$

Following on from this post (apologies for bad etiquette, seemed cleaner to repost with new information), I want to prove the assertion that for a UFD $$D$$, $$F$$ the field of fractions of $$D$$ and $$p(x) \in D[x]$$,

$$p(x)$$ irreducible in $$F[x]$$ $$\Leftrightarrow$$ $$p(x)$$ irreducible in $$D[x]$$

I had struggled to understand the relevance of the primitive property of $$p(x)$$, but found from other questions that e.g. $$2x+4$$ is reducible over $$\mathbb{Z}[x]$$, but not over $$\mathbb{Q}[z]$$, which was something that escaped me, since I thought irreducibility for polynomials was slightly different to elements of a UFD.

So does the following proof seem correct (this is not a hw Q, just something I'm playing with):

First, this lemma without proof:

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 $$deg~g(x)=deg~g_1(x)$$.

Then to the main result:

$$\Rightarrow$$:

Let $$p(x)$$ be irreducible over $$F[x]$$. Then $$p(x)=a g(x)$$, where $$a\in F$$, $$g(x) \in F[x]$$, and $$deg~g(x) = deg~p(x)$$.

Suppose $$p(x)=f_1(x)g_1(x)$$ with $$f_1(x)$$,$$g_1(x)$$ $$\in D[x]$$, and let the content of $$f_1(x)$$, $$g_1(x)$$ be $$a_1, b_1$$ respectively. Since $$p(x)$$ is primitive, we have $$1=a_1b_1$$ so $$a_1$$, $$b_1$$ are units in $$D$$.

Since $$D[x]\subset F[x]$$, we must then have either $$f_1(x)$$ a constant and $$deg~g_1(x) = deg~p(x)$$ , or vice versa. If $$f_1(x)$$ is a constant, it is $$a_1$$ and thus a unit in $$D$$, so $$p(x) = a_1 g_1(x)$$ is irreducible in $$D[x]$$.

$$\Leftarrow$$:

Let $$p(x)$$ be irreducible in $$D[x]$$. Suppose $$p(x)=f(x)g(x)$$, with $$f(x),g(x)\in F[x]$$. The lemma tells us that we must have $$p(x)=f_1(x)g_1(x)$$ with $$f_1(x),g_1(x)\in D[x]$$ with the appropriate degrees.

Since $$p(x)$$ is irreducible, then either $$f_1(x) =a_1 \in D$$ and $$deg~g_1(x) = deg~p(x)$$, or vice versa. Again, using the fact that $$p(x)$$ is primitive tells us that $$a_1$$ must be a unit in $$D$$.

Since the degrees of the factors are the same in $$F[x]$$ and $$D[x]$$, we must have that $$f(x)$$ is a constant (and since $$F$$ is a field, a unit) and $$deg~g(x) = deg~p(x)$$. Hence $$p(x)$$ is irreducible in $$F[x]$$.

I realize it may be simpler to directly prove the contrapositive statement but this is how the logic flows in my head.

• Is $F$ the field of fractions of $D$? Aug 8, 2020 at 12:30
• Yes, should state that more clearly! Aug 8, 2020 at 12:55