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 $A[y]$.
