apparent mistake in description of $(x^2,xy,y^2, \alpha x + \beta y) \subseteq K[x,y]$ in Geometry of Schemes

Question

I am reading the section II.3 about non-reduced Geometry of Schemes by Eisenbud and Harris. I am confused about something about something basic and very ring-theoretical. You can find it in item (i) on page 59.

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He claims that the ideal $\mathfrak{a} \ := \ (x^2,xy,y^2, \alpha x + \beta y) \subseteq K[x,y]$ can be described as $$ \mathfrak{b} \quad \quad := \quad \quad \{ f(x,y) \in K[x,y] \quad | \quad f(0,0) = 0 \quad \text{ and} \quad \beta {\partial f \over \partial x}(0,0) \stackrel{*}{=} \alpha {\partial f \over \partial y}(0,0) \} $$

I don't see why this description holds. First of all any $f \in \mathfrak{a}$ vanishes two times at $(0,0)$ so quite trivially $\mathfrak{a} \subseteq \mathfrak{b}$, but obviously $(*)$ doesn't make vanish $f$ twice so "$\supseteq$" does not seem to hold.

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So here is my question:

• Is the book wrong or am I just being stupid?
• If the book is wrong, what is the nature of this mistake? Could it be fixed by replacing $\mathfrak{a}$ by another ideal?

I hope you can help me with this because I really enjoyed reading untill I got stuck on this. Thanks in advance for your advice.

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Edit

Thanks to your comments I realized that I have been stupid. I totally understand why what I said above is wrong. Could you sketch an argument why the description should be valid? I think working explicitly with $f(x,y) = \sum_{i,j} a_{ij}x^iy^j$ could become cumbersome. Would it be the best way to go?

Answer 1

The ideal $I=(x^2,xy,y^2)$ consists of all polynomials in $k[x,y]$ such that

1) constant term $= 0,$

2) linear term $= 0$.

Therefore $\mathfrak{a}=I+(\alpha x + \beta y)$ consists of all polynomials in $k[x,y]$ such that

i) constant term $=0,$

ii) linear term $=\gamma(\alpha x+\beta y)$ for some $\gamma \in k.$

For any $f \in k[x,y],$ we observe that:

A) $f(0,0) = $ the constant term of $f(x,y),$

B) $\displaystyle \frac{\partial f}{\partial x}{(0,0)} = $ the coefficient of the $x$-term in $f(x,y),$

C) $\displaystyle \frac{\partial f}{\partial y}{(0,0)} = $ the coefficient of the $y$-term in $f(x,y).$

It therefore follows that $f \in k[x,y]$ belongs to $\mathfrak{a}$ if and only if

(i') $f(0,0) = 0,$

(ii') $\alpha\displaystyle \frac{\partial f}{\partial y}{(0,0)}=\beta\frac{\partial f}{\partial x}{(0,0)}.$

Hence, in your notation, we have $\mathfrak{a}=\mathfrak{b}.$

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Hope this helps! Please let me know if you need any more details :)