These days, I have some basic problem in abstract algebra. I know that in any integral domain, any prime element must be an irreducible element. Moreover, if $A$ is a UFD, then an element $a \in A$ is a prime if and only if it is an irreducible element.
Question 1
If $A$ is an arbitrary integral domain, does there exist a criterion to check that whether an irreducible polynomial over $A$ is prime or not?Question 2
The following arguments are based on an irreducible polynomial being monic. Is the following arguments correct? If not, who can give me an example that a monic irreducible polynomial over domain $A$ is not a prime? Thanks a lot!
(Update by myself) Ah!!!!!!!! Dear all: for the second question, I find a error! Because there are 2 assumptions of $f_1(y)$, I require that f1 must be monic and have minimal degree… But I don't know whether r(y) and r′(y) are monic or not!!!!!!
Let $A$ be an integral domain and $K$ be it fractional field. Suppose $f \in A[y]$ is a monic irreducible polynomial. We fix one algebraic closure of $K$, denoted by $\overline{K}$. Because $\overline{K}$ is algebraic closed, we can find $\alpha \in \overline{K}$ such that $f(\alpha) = 0$. We also define $f_1(y) \in A[y]$ to be a monic polynomial of minimal degree so that $f_1(\alpha) = 0$. We note that $f_1(y)$ exists because $f(y)$ is a monic polynomial which satisfies $f(\alpha) = 0$. Because $f_1(y)$ is monic, we can applying division algorithm on $f$ divided by $f_1$. In other words, we can find $q(y), r(y) \in A[y]$ such that
\begin{equation*}
f(y) = f_1(y)q(y) + r(y),
\end{equation*}
where either $r(y) = 0$ or $\deg(r(y)) < \deg(f_1(y))$. Hence, we have
\begin{equation*}
0 = f(\alpha) = f_1(\alpha)q(\alpha) + r(\alpha) = r(\alpha).
\end{equation*}
By the assumption of minimal degree of $f_1(y)$ as defined above, it forces that $r(y) = 0$. We deduce that $f(y) = f_1(y)q(y)$. Because $f(y)$ is irreducible, either $f_1(y)$ or $q(y)$ is a unit in $A[y]$. But it is impossible that $f_1(y)$ is a unit since $f_1(\alpha) = 0$. It implies that $q(y)$ is a unit in $A[y]$. Because both $f(y)$ and $f_1(y)$ are monic, we deduce that $f(y) = f_1(y)$.
If $g(y) \in A[y]$ is a polynomial such that $g(\alpha) = 0$, by the same reason, by applying division algorithm (since $f(y)$ is monic), we can find $q'(y), r'(y) \in A[y]$ such that
\begin{equation*}
g(y) = f(y)q'(y) + r'(y),
\end{equation*}
where either $r'(y) = 0$ or $\deg(r'(y)) < \deg(f(y))$. Similarly as above, we also have $r'(\alpha) = g(\alpha) = 0$. Because $f(y) = f_1(y)$, we deduce that $r'(y) = 0$ whence $g(y) = f(y)q'(y) \in (f(y))$. Finally, we conclude that $A[y] / (f(y)) \simeq A[\alpha] \subseteq \overline{K}$. In particular, $(f(y))$ is a prime ideal in $A[y]$.
Best Answer
Here is a general criterion that is useful. Let $\rm\,D\,$ be a domain with fraction field $\,\rm K.$
$$\rm f\,\ is\ prime\ in\ D[x]\iff f\,\ is\ prime (= irreducible)\ in\ K[x]\ and\,\ f\,\ is\ superprimitive $$
$$\rm where\,\ f\,\ is\ {\bf superprimitive}\ in\ D[x]\,\ :=\,\ d\,|\,cf\, \Rightarrow\, d\,|\,c\,\ \ for\ all\,\ c,d\in D^*$$
For proofs of this and related results see the freely downloadable paper below (which succinctly summarizes results of the author's thesis under Kaplansky).
Hwa Tsang Tang, Gauss’ lemma, Proc. Amer. Math. Soc. 35 (1972), 372-376