0
$\begingroup$

Let $k$ be a field. Consider the polynomial ring $R=k[x_1,\dots,x_n]$. We know by the Hilbert Basis theorem that $R$ is noetherian. So any ideal $I$ of $R$ is of the form $I=\langle f_1,\dots,f_k\rangle$. Is it true that if $f_1, \dots, f_k$ are irreducible, then $I$ is prime ?

Conversely, if the ideal $I$ is prime, does it imply that $f_1, f_2\dots, f_k$ are irreducible ? (Assume that $k$ is the smallest integer such that $I$ can be generated by $k$ elements. Otherwise, we can find trivial counterexamples such as $\langle x\rangle=\langle x, x^2\rangle$ )

  • 0
    I found the answer for second question. Assume that $f_i$ is not irreducible for some $i$. Then $f_i=gh$ for some nonconstant polynomials $g$ and $h$. Then $g,h\notin I$ (If $g$ or $h$ is in $I$, then it is in the ideal $I'=\langle f_1, \dots f_{i-1}, f_{i+1}, \dots, f_k\rangle$ $\text{ }\implies\text{ }$ $f_i\in I'$ $\text{ }\implies\text{ }$ $I$ can be generated by fewer than $k$ elements, which is a contradiction to the minimality of $k$) – 2017-02-09
  • 1
    I think the answer to your first question is no. Consider $R = k[x, y]$, and $f_{1} = x, f_{2} = y^{2}-x$. Then both $f_{1}$ and $f_{2}$ are irreducible in $R$, but $k[x, y]/\langle x, y^{2}-x \rangle \cong k[y]/\langle y^{2} \rangle$ is not a domain. – 2017-02-09
  • 1
    @learning_math The argument in your comment is wrong, but the conclusion is somehow valid: you can replace the reducible generators by ireeducible ones. – 2017-02-09
  • 0
    For the first question, take a look here: https://math.stackexchange.com/questions/427263/sequence-of-irreducible-polynomials-in-kx-1-x-n-generates-a-prime-i – 2017-10-30

0 Answers 0