So my class on ring theory recently began, and I'm having a bit of trouble understanding ideals that are generated by polynomials. For an arbitrary ring, I know the definition of such an ideal, that is, for $R$ a ring and $a \in R$, $(a) = \{ \sum_{i=1}^{n} r_ias_i \mid r_i,s_i \in R \}$. So, then, this set is essentially just all the possible sums of different combinations of $r_ias_i$? Say, for $n=1$, $r_1as_1 \in (a)$, where $r_1,s_1$ just "run through" all the elements in $R$?
Specifically, I had a previous homework problem concerning the following ring of polynomials: $R = [\mathbb{Z}/2\mathbb{Z}](t)$. I know what this means as a set, but I had trouble understanding this ideal: $f = f(t) = t^2 + t + 1 \in R$ and the ideal being $(f)$. The question was regarding $R/(f)$, which I know is the set $\{r + (f) \mid r \in R \}$, but I didn't even know where to start due to my lack of understanding of $(f)$. Any help would in understanding would be welcomed
$\endgroup$3 Answers
$\begingroup$Put in simple words,
In any ring $R$ (with unit!), the principle ideal $(a)$ generated by $a\in R$ consists of nothing but the multiples of $a$.
I give here just a heuristic:
I like to think of the quotient ring $R/(a)$ as having the same elements of $R$ but the equality is modified so that all elements of $(a)$ be equal to zero in the quotient. Observe that it is enough to require $a=0$, because $ras=0$ and $\sum r_ias_i=0$ already follows by the properties of equality.
In the case of polynomial ring, a factor like $R[t]/(f)$ with, say $f=t^n-a_{n-1}t^{n-1}-\dots-a_1t-a_0$ will always be represented by the set of polynomials of degree $<n$, because in the quotient ring we have $$t^n=a_{n-1}t^{n-1}+\dots+a_1t+a_0$$ So that, any time a $t^n$ appears, it can be replaced by a degree $<n$ polynomial in $R[t]/(f)$.
$\endgroup$ 2 $\begingroup$When the ring is commutative, as in your case, things are way simpler and easier:
$$\forall\,a\in R\;,\;\;(a):=\{ra\;;\;r\in R\}$$
And thus in your very particular case, the ideal generated by a polynomial $\;f(x)\in R[x]\;$ is just the set of all products of $\;f\;$ by elements of $\;R[x]\;$ .
$\endgroup$ 4 $\begingroup$A two-sided ideal $I$ of a ring $R$ is naturally the kernel of the surjective homomorphism $$ R \twoheadrightarrow R/I. $$
Take your example, where $R = \Bbb{Z}/2\Bbb{Z}[t]$ and $I = (f) = (t^2 + t + 1)$. In the quotient (the image of the map), the polynomial $f$ is identified with $0$. But if $f = 0$, then $af = a0 = 0$ and $fb = 0b = 0$ for any $a, b \in R$. The fact that the ideal must be closed under multiplication by any element in the ring on either side is forced by the desire for the ideal to be a kernel.
Since $t^2 + t + 1 = 0$ in $R/I$, any time that you see $t^2$, you can replace it by $-(t + 1)$ or just $t + 1$, as the ground field is $\Bbb{Z}/2\Bbb{Z}$.
Try working out the sum $(at + b) + (ct + d)$ and product $(at + b)(ct + d)$ of two generic elements in $R/I$ to get a sense of the structure of this ring. Hint: it's not very big.
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