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I am a bit confused, as my class is currently on Section 9.2 of Dummit and Foote, and what elements of $R/I$ are of the form of, where $R$ is a polynomial ring.

For example, I am reading that $\mathbb{Z}[x,y]/(x^2,y^2,2)$ are of the form $a + bx + cy + dxy$, where $a,b,c,d \in \mathbb{Z}_2$. How is this deduced?

I suppose there is a definition $R/I = \{ r + I, r \in R \}$, but how does this form $a + bx + cy + dxy$, and can anyone give some other examples of how we can write elements of $R/I$ with a polynomial ring?

How should I be thinking about this more deeply?

Moni145
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1 Answers1

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You can think of quotienting by an ideal as a way of making the elements in the ideal become $0$. A simple example is $\mathbb{Z}/6\mathbb{Z}$ which is just the ring with elements $\{0,1,2,3,4,5\}$. We've made $6 = 0$ which in turn meant that $7 = 1, 8 = 2, ...$. In your example we are making $x^2, y^2$ and $2$ equal $0$. Having $x^2$ be $0$ means that any term of the form $ ax^ny^m = 0$ if $n \ge 2$. We have a similar situation for $y^2$. This means that your polynomial can only have the form $ a + bx + cy + dxy $ since the higher order terms will be $0$. The reason why the coefficients are in $Z_2$ is because we have quotiented by $2$ which means $2 = 0$. Hence all even numbers are actually $0$ since they can be written as $2n = 2\cdot 0 = 0$. All odd numbers are just equal to $1$ since they can be written as $2n + 1 = 2 \cdot 0 + 1 = 1$.

Pedro Amaral
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    This is actually really helpful. Thank you for this, I'm not sure why I haven't seen this kind of explanation anywhere else. – Moni145 Jan 18 '21 at 05:41
  • One other note: elements of the quotient ring are technically equivalence classes, i.e. $[4]\in\mathbb{Z}/6\mathbb{Z}\cong\mathbb{Z}_6$ is not the same element as $4\in\mathbb{Z}_6$, because ${\dots,-2,4,10,\dots}\in[4]$. (haha I forgot I don't have my usual macros here) – Nico Terry Jan 18 '21 at 06:55