Affine Algebraic Sets - D&F Chapter 15, Section 15.1 - Example 2

[Math Amateur]

I am reading Dummit and Foote Ch 15, Commutative Rings and Algebraic Geometry. In Section 15.1 Noetherian Rings and Affine Algebraic Sets, Example 2 on page 660 reads as follows: (see attachment)

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(2) Over any field k, the ideal of functions vanishing at $(a_1, a_2, ... ... ... a_n) \in \mathbb{A}^n$ is a maximal ideal since it is the kernel of the surjective ring homomorphism from $k[x_1, x_2, ... ... x_n]$ to the field k given by evaluation at $(a_1, a_2, ... ... ... a_n)$.

It follows that $\mathcal{I}((a_1, a_2, ... ... ... a_n)) = (x - a_1, x - a_2, ... ... ... , x - a_n)$

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I can see that $(x - a_1, x - a_2, ... ... ... , x - a_n)$ gives zeros for each polynomial in $k[ \mathbb{A}^n ]$ - indeed, to take a specific example involving $\mathbb{R} [x,y]$ we have for, let us say, a particular polynomial $g \in \mathbb{R} [x,y]$ where g is as follows:

$g(x,y) = 6(x - a_1)^3 + 11(x - a_1)^2(y - a_2) + 12(y - a_2)^2$

so in this case, clearly $g(a_1, a_2) = 0$ ... ... ... and, of course, other polynomials in $\mathbb{R} [x,y]$ similarly.

BUT ... ... I cannot understand D&Fs reference to maximal ideals. Why is it necessary to reason about maximal ideals.

Since I am obviously missing something, can someone please help by explaining what is going on in this example.

Another issue I have is why do D&F write $\mathcal{I}((a_1, a_2, ... ... ... a_n))$ with 'double' parentheses and not just $\mathcal{I}(a_1, a_2, ... ... ... a_n)$?

Would appreciate some help.

Peter

Note - see attachment for definition of $\mathcal{I}(A)$

 

[R136a1]

The ideal $I=\mathcal{I}((a_1,...,a_n))$ is the set of all polynomials vanish on the point $\mathbf{a} = (a_1,...,a_n)$.

The reasoning is as follows. Note that the polynomials $X_i-a_i$ certainly vanish on this point, as is easily checked. So $X_i-a_i\in I$ certainly. Thus the ideal generated by these polynomials is in $I$ too. So $(X_1-a_1,...,X_n-a_n)\subseteq I$. But perhaps $I$ contains more! Here is our information about maximality comes in, because if $I$ contains more, it must be the entire ring $k[X_1,...,X_n]$. But this can not be since the constant polynomial $1$ does not vanish on the set.

 

[Math Amateur]

Excellent! Now see the link to maximal ideals!

But will reflect on this further so I am sure I have understood all angles of the theory

Thank you so much for this critical help ... I can now progress with more confidence!

Peter