A prime ideal

**electricjaguar** · February 18th 2009, 11:14 AM

Hi everyone, this is my first post. Hope you don't mind that I'm using it to ask a homework question.

Here's the problem:
Let $I = \{f\in\mathbb{Z}\left[x\right] : f(0) = 0\}$ .

Show $I$ is a prime ideal.
Show $I$ is not maximal.

For 1), I let $f(x),g(x)\in\mathbb{Z}\left[x\right]$ and $f(x)g(x)\in I$ ; that is, $f(0)g(0) = 0$ . Because $\mathbb{Z}\left[x\right]$ is an integral domain, it has no zero divisors; thus, either $f(0)$ or $g(0)$ must be $0$ and hence in $I$ . Is this right?

I wasn't sure about 2); can I take $J = \{f\in\mathbb{Z}\left[x\right] : f(0) = 0$ or $f(\frac{1}{2}) = 0\}$ ?

Thanks, Sam

**clic-clac** · February 18th 2009, 12:37 PM

Hi

1) Yep

2) $2x\in J\ \text{and}\ -2x+1\in J\ ,$ but...

Try to work with an ideal generated by two elements.

**electricjaguar** · February 18th 2009, 06:07 PM

Thanks for your help clic-clac. I'm still a little confused, though.

The polynomials such that f(0) = 0 are all polynomials with constant term 0. So I need to find an ideal that contains these and a polynomial with a non-zero constant term. So could I take $\{f(x)\in\mathbb{Z}\left[x\right] : f(0) = 0\} \cup (2)$ ?

**electricjaguar** · February 18th 2009, 09:11 PM

Actually I think (2) would be the whole ring. Hmm... how can I find a generating element that includes all functions that vanish at 0 without getting the whole ring?

**clic-clac** · February 19th 2009, 12:07 AM

Your new answer isn't so far from a solution: the idea of "adding" $2$ , which wasn't in your first ideal, is good.

But a reunion of ideals is, in general, not an ideal. The construction that transforms a reunion of ideals into an ideal is the ideal generated by a set.

To simplify notations, do you agree that $\{f(x)\in\mathbb{Z}[x];\ f(0)=0\}$ is the ideal $(x)$ in $\mathbb{Z}[x]$ ?

Since $(x)\cup(2)$ is not a ideal, we can consider $<(x)\cup(2)>,$ which is also written $(x,2),\ (x)+(2)$ or $x\mathbb{Z}[x]+2\mathbb{Z}[x]$

$(x,2)$ contains $(x)$ and $2$ so it strictly contains $(x)$ . The question now is $(x,2)\neq\mathbb{Z}[x]\ ?$

So take a $p(x)\in (x,2),$ i.e. $\exists a,b\in\mathbb{Z}[x],\ p(x)=xa(x)+2b(x).$ Is $p(x)=1$ possible? (That will answer the problem, since for any ideal $\mathfrak{A},\ 1\in\mathfrak{A}\Leftrightarrow \mathfrak{A}=\mathbb{Z}[x]$ )

**electricjaguar** · February 19th 2009, 03:22 PM

I was about to say what about b(x) = 1/2, then I realized we're working in Z[x]. Thank you so much clic-clac!

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