# Ideals with two inputs

What does it mean if an Ideal $I$ is generated by two inputs. Like, let $I = (p, x^2 +1)$. And then, for a Ring $R$, what does $R/I$ mean?

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Is $R = F[x]$ some polynomial ring, over a field or domain? Are you interested in understanding ideals generally, or, instead, interested only in how to compute in this particular type of quotient ring $R/I\, ?\ \ \$ – Bill Dubuque Nov 27 '12 at 21:42

An ideal $I$ of a ring (commutative, with unit) $R$ is said to be generated by a set $M \subseteq I$, written $I = (M)$, if $I$ is the smallest ideal containing $M$, $I$ can be written in terms of $M$ as $$I =\left\{\sum_{i=1}^n r_i m_i \biggm| n \in \mathbb N, r_i \in R, m_i \in M\right\}$$ If $M = \{m_1, \ldots, m_n\}$ is finite, one writes $I = (m_1, \ldots, m_n)$, in your case you have $$(p, x^2 + 1) = \left\{ a \cdot p + b \cdot (x^2 + 1) \bigm| a,b \in R \right\}.$$ For a ring $R$ and an ideal $I$, one denotes by $R/I$ the set of cosets of $I$, that is $R/I = \{r + I \mid r \in R\}$ where $r+ I = \{r + i \mid i \in I\}$. $R/I$ is a ring with the representativewise defined operations $$(r+I)(s+I) := rs + I, \qquad (r+I)+(s+I) := (r+s) + I.$$

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If $I = (g_1, g_2)$ it means that every element in $I$ is of the form $r_1 g_1 + r_2 g_2$ for $r_1, r_2$ in the ring $R$.

The factor ring $R/I$ is the set of cosets of $I$. This means that two things in $R/I$ are considered equivalent if they differ by an element in $I$.

It's easiest to understand by looking at an example. Take $R = \mathbb Z$ and the ideal $(2)$. Then what are the cosets of $I$? Consider $0$ in $\mathbb Z$. Then all elements differing by an even number from it are considered equivalent. Let's denote this set $\bar{0}$. This is an element of $\mathbb Z / (2)$.

Next, let's consider $1$ in $\mathbb Z$. Then again, everything differing by an even number is equivalent to $1$. That is to say, all odd numbers are equivalent. Let's denote them $\bar{1}$.

Now we are done because there are no elements left in $\mathbb Z$ hence we have found $R/I = \mathbb Z / (2) = \{ \bar{0}, \bar{1} \}$.

Hope this helps.

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