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I am working on this problem and I was wondering if anyone would be able to help me with it.

The problem states:"Let F be a field and let J be an ideal in F[x]. Prove that J is prime if its generator is irreducible over F."

I am not sure what "...irreducible over F" means. I've done research and I keep finding problems related to polynomials being irreducible.

I know that a field is a commutative ring with unity in which every non zero element is invertible. An ideal is a nonempty subset that is closed under addition, negatives and it absorbs products. And I know that prime means that "If ab is in J, then a is in J or b is in J.

Any help is greatly appreciated! Thanks!

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It means irreducible as an element of the ring $F[x].$ See en.wikipedia.org/wiki/Irreducible_element –  Andrew Dec 13 '12 at 1:04
    
Thank you Andrew! –  RedPotatoe Dec 13 '12 at 5:17
    
Glad to help!{} –  Andrew Dec 13 '12 at 5:18
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1 Answer

A polynomial is irreducible over $F$ if it is an irreducible element of $F[x]$. Here are are some definitions that you may or may not need:

  • If R is a commutative ring, a nonzero element $x\in R$ is a unit if it is invertible. For example, in a field every nonzero element is a unit. The units in $\mathbb{Z}$ are $\pm1$.
  • An element $x\in R$ that is nonzero and not a unit is irreducible if whenever $x=ab$ with $a,b\in R$ then either $a$ or $b$ is a unit. For instance, the irreducible elements of $\mathbb{Z}$ are the prime numbers and their negatives.

Here are some examples for polynomials over a field:

  • $x^2-1$ is reducible over $\mathbb{R}$ because $x^2-1=(x+1)(x-1)$ and neither $x+1$ nor $x-1$ is a unit in $\mathbb{R}[x]$.
  • $x^2+1$ is irreducible over $\mathbb{R}$. (Maybe you should try to prove this!)
  • $x^2+1$ is reducible over $\mathbb{C}$ because $x^2+1=(x+i)(x-i)$ and $x+i$ and $x-i$ are not units in $\mathbb{C}[x]$.
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