# Prime ideals in $R[x]$ that intersects $R$ trivially [duplicate]

We have that $$R$$ is a UFD. Now, I'm trying to show that any prime ideal $$P$$ of $$R[x]$$ that intersects $$R$$ trivially is a principal ideal.

I'm trying to show that P has height $$1.$$ Is this true that P has height $$1$$? But I cannot prove this. If that is the case, then I can easily prove that P is in fact a principal ideal. Any suggestions would be highly appreciated. Thank you.

Here's a more hands-on version Marktmeisters' excellent answer. It assumes the following fact:

Fact: Any two elements of a UFD have a $$\gcd$$.

Let $$f\in P$$ be nonzero and of minimal degree, and let $$c\in R$$ denote the $$\gcd$$ of its coefficients so that $$f=cg$$ for some $$g\in R[x]$$. Because $$P$$ is prime and $$P\cap R=0$$ it follows that $$g\in P$$.

Now let $$h\in P$$ be nonzero, and let $$d=\gcd(g,h)$$. Then $$g=ad$$ and $$h=bd$$ for coprime $$a,b\in R[x]$$. Because $$P$$ is prime it follows that $$d\in P$$ as otherwise $$a,b\in P$$, which implies $$R[x]=(a,b)\subset P$$, a contradiction.

Because $$g$$ is of minimal degree it follows that $$\deg d=\deg g$$ and $$a\in R$$. Then $$a$$ divides all coefficients of $$g$$, hence $$a$$ is a unit. It follows that $$(g)=(d)$$ and hence $$(h)\subset(g)$$. Because $$h\in P$$ was arbitrary it follows that $$P\subset(g)$$, and hence $$P=(g)$$ is principal.

• @James I have corrected a small oversight in my original answer; the original argument failed when starting with a non-primitive $f\in P$. – Servaes Mar 26 at 9:38
• Thanks so much. – James Mar 26 at 10:13

Suppose that $$P$$ is prime in $$R[x]$$ and intersects $$R$$ trivially. We can assume that $$P \neq (0)$$.

Now, let $$S = R \setminus \{0\}$$ and denote by $$K := R_S$$ the localization of $$R$$ at $$S$$; i.e., $$K$$ is the field of fractions of $$R$$. Then, $$R[x]_S = K[x]$$; denote by $$\iota \colon R[x] \hookrightarrow K[x]$$ the canonical inclusion.

Since $$P$$ does not intersect $$S$$ and $$P$$ is prime in $$R[x]$$, we obtain that $$\iota(P)$$ is a non-zero prime ideal of $$K[x]$$. In particular, since $$K[x]$$ is a PID, it is generated by an irreducible polynomial $$f \in K[x]$$.

By assumption $$P \neq (0)$$, so that we will find a polynomial $$0 \neq g \in P$$ of minimal degree.

Claim: We can even choose $$g$$ to be irreducible in $$R[x]$$.

Proof: Assume that all such $$g$$'s were reducible, $$g = g_1 g_2$$, where w.l.o.g. $$g_1$$ is irreducible of positive degree. Since $$g$$ was chosen to be of minimal degree in $$P$$, we obtain that $$\deg(g_1) = \deg(g)$$ and that $$g_2 \in R$$. Since $$P$$ is prime and $$g_2 \notin P$$, we obtain that $$g_1 \in P$$, a contradiction. $$\Box$$

In the following, we identify $$\iota(g)$$ with $$g$$ and $$R[x]$$ with $$\iota(R[x])$$. Since $$g \in \iota(P) = \langle f \rangle_{K[x]}$$, there is a polynomial $$h \in K[x]$$ such that $$g = fh$$.

Claim: $$g$$ generates $$\iota(P) = \langle f \rangle_{K[x]}$$.

Proof: This is essentially GauĂŸ' Lemma. Since $$g$$ was chosen to be irreducible in $$R[x]$$, it is a primitive polynomial. Moreover, since $$f$$ is irreducible in $$K[x]$$, by GauĂŸ' Lemma, we obtain $$h \in K$$. $$\Box$$

Now, the assertion follows from $$\iota(P) \cap R[x] = \langle g \rangle_{K[x]} \cap R[x] = P$$, since this implies $$P = \langle g \rangle_{R[x]}$$.

So let us prove $$\iota(P) \cap R[x] = P$$. The inclusion "$$\supset$$" is clear. For the other inclusion, note that if $$p \in K[x]$$ is such that $$gp \in R[x]$$, by GauĂŸ' Lemma, we have $$p \in R[x]$$.

• @Servaes Oh, I misunderstood something then.. – JVHD2334 Mar 26 at 9:17