# Ideals of the form $(X_1-a_1,\ldots,X_n-a_n)$ in a not algebraically closed field

In my commutative algebra class I was given the following exercise:

Give an example of a field $$k$$ and ideals $$\mathfrak{a}= (X_1-a_1,\ldots,X_n-a_n), \mathfrak{b} = (X_1-b_1,\ldots,X_n-b_n) \subseteq k[X_1,\ldots,X_n]$$, such that $$\mathfrak{a} = \mathfrak{b}$$ but $$(a_1,\ldots,a_n) \neq (b_1,\ldots,b_n)$$.

Hint: Take a field, which is not algebriacally closed.

I have a feeling this exercise might be unsolvable, because I think I have proven that such ideals can't exist. My proof attempt goes as follows:

Let $$\mathfrak{a}$$ and $$\mathfrak{b}$$ be defined as above. Suppose $$\mathfrak{a} = \mathfrak{b}$$ but $$(a_1,\ldots,a_n) \neq (b_1,\ldots,b_n)$$. Without loss of generality we may assume $$a_1 \neq b_1$$.

From $$X_1 - a_1 \in \mathfrak{a}$$ and $$X_1 - b_1 \in \mathfrak{b} = \mathfrak{a}$$ it follows that $$b_1 - a_1 = (X_1-a_1) - (X_1 - b_1) \in \mathfrak{a}$$. Hence, $$\mathfrak{a}$$ contains a constant non-zero polynomial. This is, however, a contradicition to the fact that every polynomial in $$\mathfrak{a}$$ vanishes at the point $$(a_1,\ldots,a_n) \in k^n$$.

Thus, we necessarily have $$\mathfrak{a} \neq \mathfrak{b}$$ or $$(a_1,\ldots,a_n) = (b_1,\ldots,b_n)$$.

Is my proof wrong or is this exercise, in fact, unsolvable?

• You are right, the statement looks false. Another explanation: if $a_1\ne b_1$ then $x_1-a_1$ doesn't vanish at the point $(b_1,...,b_n)$ and so can't belong to the ideal $(x_1-b_1,...,x_n-b_n)$. These ideals are always maximal ideals of $k[x_1,...,x_n]$, and they are always distinct. The only difference $k$ being algebraically closed makes is that then these are actually all the maximal ideals.
– Mark
Jul 23, 2022 at 10:20

Your argument is correct and the exercise is wrong as stated. Here is what I think was intended: if $$k$$ is a field which is not algebraically closed, then any point $$a = (a_1, \dots a_n) \in \bar{k}^n$$ over the algebraic closure defines a maximal ideal of $$k[x_1, \dots, x_n]$$
$$m_a = \{ f : f(a_1, \dots a_n) = 0 \}$$
with residue field the algebraic extension $$k[a_1, \dots a_n]$$ of $$k$$ generated by the coordinates $$a_i$$. In this setting it can happen that $$m_a = m_b$$ even if $$a \neq b$$, and it's a nice exercise to determine exactly when this occurs: it turns out to be exactly when $$a$$ and $$b$$ are in the same orbit of the action of the absolute Galois group $$G = \text{Gal}(\bar{k}/k)$$.
The simplest case of this to play around with occurs when $$n = 1$$, where it's possible to cleanly characterize $$m_a = \{ f : f(a) = 0 \}$$ as being the ideal generated by the minimal polynomial of $$a$$. Then we have $$m_a = m_b$$ iff $$a$$ and $$b$$ have the same minimal polynomial (which is exactly the condition that they're in the same Galois orbit). For example we can take $$k = \mathbb{R}, a = i, b = -i$$, where in both cases the corresponding maximal ideal is generated by $$x^2 + 1$$.