How do polynomials generated by the following ideal look like?

Question

I am given the following ideal $I = (3, X^2+X+1)$ in $\mathbb Z[X]$. I want to know how the elements in the ideal look like. For example, can you show some polynomials that belong to $I$, and some that don't? Thank you.

Answer 1

Every polynomial in $I$ is of the form $3f + (X^2+ X + 1)g$ for some $f,g \in \mathbb{Z}[X]$. By choosing $f$ and $g$ you can now create as many examples as you want:

We can choose $f = X$ and $g = 2X + 1$ and get $$2X^3 + 3X^2 + 6X + 1 = 3X + (X^2+X+1)(2X + 1) \in I$$

Pick some $f$ and $g$ for yourself and you will find some element of $I$. The form of the polynomials in $I$ also tells you which polynomials are not in $I$. Here you will have to work a bit more though, as you have to proof that some polynomial cannot be written like that. As Rick pointed out in the comments, it is a good exercise to show that e.g. $1 \not \in I$, i.e. that $1$ is not of the form above. Try to do that and otherwise we can help you.

Answer 2

Any polynomial in $I=\langle 3, x^2+x+1\rangle$ is a linear combination of the polynomials $3$ and $x^2+x+1$, i.e. it looks like $$a(x) \color{red}{(3)}+b(x) \color{red}{(x^2+x+1)}, \qquad \text{ where } \quad a(x),b(x) \in \Bbb{Z}[x].$$ So take $a(x)=x$ and $b(x)=0$, to show that $3x \in I$, take $a(x)=1$ and $b(x)=x-1$, then $x^3+2 \in I$, take $a(x)=-x^2$ and $b(x)=3$, then $3x+3 \in I$.

Try to see if $1 \in I$. If it is, then we can find $a(x),b(x) \in \Bbb{Z}[x]$ such that \begin{align*} 1&=a(x) \color{red}{(3)}+b(x) \color{red}{(x^2+x+1)} \end{align*}

As Gribouillis pointed out, substitute $x=1$, to get $1=3a(0)+3b(0)$. See if you can get some contradiction.