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?

  • 3
    $\begingroup$ 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. $\endgroup$
    – Mark
    Jul 23, 2022 at 10:20

1 Answer 1


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 $x^2 + 1$.


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