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I am reading the book the book Introduction to commutative algebra by Atiyah and Macdonald. I have two questions On Page 51.

On Line 5 of Page 51, it is said that the zero-divisors in $A/\mathfrak{q} \cong k[y]/(y^2)$ are all the multiples of $y$. I think that $A/\mathfrak{q} \cong k[y]/(y^2)=\{a+by : a, b \in k\}$. Let $a \neq 0$ and $a+by \in A/\mathfrak{q}$. We have $y (a+by)=ay+by^2 = ay \neq 0$. So $y$ is not a zero-divisor of $A/ \mathfrak{q}$. But it is said that $y$ is a zero-divisor of $A/\mathfrak{q}$. I am confused.

On Line 12 of Page 51, it is said that $A/\mathfrak{p} \cong k[y]$. Here $A=k[x, y, z]/(xy-z^2)$ and $p=(\bar{x}, \bar{z})$, $\bar{x}=x+(xy-z^2), \bar{z}=z+(xy-z^2)$. I think that $A/\mathfrak{p} = k[y]/(xy-z^2)$. Why $A/\mathfrak{p} \cong k[y]$? Thank you very much.

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LJR
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1 Answers1

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Your reasoning is wrong: Put $a = 0$ and $b = 1$ to get $y \cdot y = y^2 = 0$ and so $y$ is a zero divisor of $k[y]/(y^2)$. For your second question, the quotient by $\mathfrak{p}$ is isomorphic to $$k[x,y,z]/(xy - z^2, x,z) \cong k[x,y,z]/(xy,x,z) \cong k[y].$$