7
$\begingroup$

Background

I have been confused about a particular definition in the textbook for my abstract algebra class, Ideals, Varieties, and Algorithms by Cox, Little, and O'Shea. It is frustrating because I feel that I have a partial grasp on what the definition is trying to say, but when it comes down to it I simply find my confused. So, instead of suffering by myself for any longer with this definition, I have decided to turn to you lovely folks to help me understand this seemingly simple concept.


What I Understand

First off, I understand (at least in the context of this book) what an ideal is. The definition my book gives for an ideal is

Definition. A subset $I\subseteq k[x_1,..., x_n]$ is an ideal if it satisfies:

(i) $0\in I$.

(ii) If $f,g\in I$, then $f+g\in I$.

(iii) If $f\in I$ and $h\in k[x_1,...,x_n]$, then $hf\in I$.

I find this to be a simple, easy to understand definition. My problem, however, arises a few lines later when they define an ideal generated by a finite number of polynomials.


What I Don't Understand

And now, I give you the definition that has been causing me an incredible amount of confusion and frustration.

Definition. Let $f_1,...,f_s$ be polynomials in $k[x_1,...,x_n]$. Then we set $$\langle f_1,...,f_s \rangle=\Big\lbrace \sum_{i=1}^s h_if_i \ | \ h_1,...,h_s\in k[x_1,...,x_n] \Big\rbrace.$$

I know what you are thinking. How does he not understand this? I wish I knew the answer to that question, but in the meantime, can someone please help me visualize what this set looks like? I understand that $\langle f_1,...,f_s \rangle$ is an ideal, but I don't understand its structure, if that makes sense. In other words, I can't visualize this definition in a way that makes sense to me. The authors do make a slightly helpful note, saying that "we can think of $\langle f_1,...,f_s \rangle$ as consisting of all 'polynomial consequences' of the equations $f_1=f_2=...=f_s=0$."

To elaborate a little more on my confusion, what I'm asking for is a less "compact" definition. When I read this definition, for whatever reason the only thing I can come up with is $$f_1h_1+f_2h_2+...+f_sh_s.$$ But that doesn't make sense, because $\langle f_1,...,f_s \rangle$ is supposed to generate a set, not just a single polynomial.


As always, thank you all for your time. If you find this to be a stupid or silly question, then I'm sorry to have disappointed you -- I'm a slow learner, and I get hung up on stupid things sometimes.

Oh, and Happy Halloween!

$\endgroup$
4
  • 1
    $\begingroup$ Suggestion: Try looking at one or two specific examples, such as: (1) In $k[x,y]$, look at $\langle x^2, y^2 \rangle$. This is equal to the set $\{ x^2 h_1 + y^2 h_2 \mid h_1, h_2 \in k[x,y]\}$. Some elements include $x^2 \cdot 1 + y^2 \cdot 0$, $x^2 \cdot 0 + y^2 \cdot 1$,... $x^2 \cdot (x y^2 - 3 xy) + y^2 \cdot (2x^3 - 5)$, etc. (2) Look at examples in $\mathbb{Z}$, e.g., $\langle 6,4 \rangle = \{ 6 h_1 + 4 h_2 \mid h_1, h_2 \in \mathbb{Z} \}$. Can you prove this is equal to the set of even numbers? $\endgroup$ – Zach Teitler Nov 1 '17 at 4:55
  • $\begingroup$ Makes alot more sense when you put it like that! Thank you @ZachTeitler! $\endgroup$ – Thy Art is Math Nov 1 '17 at 5:00
  • 1
    $\begingroup$ One more suggestion. Make sure you do Exercise 1.4.2, or at least try to do it. The result in Exercise 1.4.2 gets used constantly, over and over, throughout the rest of the book, including in a lot of exercises. It's the main way to show that two ideals are equal, or that one ideal is contained in another. So it is extremely useful. Also it's good practice with the definition of ideal, and proving a subset relationship. (There are a lot of other good exercises, I just want to highlight that particular one. I made sure to assign it in the class I'm teaching from IVA this semester.) $\endgroup$ – Zach Teitler Nov 1 '17 at 5:49
  • 1
    $\begingroup$ Read the Rings chapter in Algebra by Hungerford. He describes ideals in full generality very well (non-commutative and commutative). $\endgroup$ – user494247 Nov 11 '17 at 9:57
4
$\begingroup$

Would it help to see the see the set with explicit polynomials in place of the $f_1, f_2, \dots, f_n$?

For example, lets look at an explicit example when $n = 2$, so an ideal generated by 2 polynomials. Also, we'll work in a polynomial ring in two variables over $k$, i.e. $k[x,y]$.

Here is the ideal generated by the polynomials $x^2 - 1$, $yx+x$.$$\langle \, x^2 - 1, \, yx+x \, \rangle = \{ \, f\cdot(x^2 - 1) + g\cdot(yx + x) \, | \, f,g \in k[x,y] \, \}.$$

So the elements of the set are any polynomial that can be written in the form $f \cdot (x^2 - 1) + g \cdot (yx+x)$. But we can choose $f,g$ to be $\textit{any}$ polynomial we want. For example, we know $x^2 - 1$ itself is in that set because we can choose $f = 1, g = 0$. We also know that the polynomial $x^3-x+yx ^2 + x^2$ is in the set, because we can choose $f = x, g = x$.

If you are familiar with linear algebra you can maybe, in a way, draw a connection between an ideal generated by polynomials and the span of a set of vectors. You can think of it as, the set of all things that can be made from the objects defining it.

$\endgroup$
3
  • 1
    $\begingroup$ Also, is your class an Algebraic Geometry class or an Abstract Algebra class? I think IVA is a strange book to teach general algebra out of. $\endgroup$ – Prince M Nov 1 '17 at 7:38
  • $\begingroup$ Sadly I was absolutely terrible at linear algebra but seeing the explicit polynomials made this make so much more sense, so thank you for the help. The class is labeled rather generally as "abstract algebra," so you are probably right that the course should be called "algebraic geometry". $\endgroup$ – Thy Art is Math Nov 1 '17 at 15:30
  • $\begingroup$ Just a note that the authors of IVA do write in the preface: "For instance, the book could serve as a basis of a second course in undergraduate abstract algebra, but we think that it just as easily could provide a credible alternative to the first course." $\endgroup$ – J W Jan 3 at 15:16
4
$\begingroup$

Perhaps a good way to understand this (or to understand anything, for that matter) is to look at examples.

The definition of ideal makes sense in any ring, so let’s look at $\Bbb Z$ first. Every ideal, you soon persuade yourself, is a set $d\Bbb Z\subset\Bbb Z$, that is, just the set of all multiples of a given number $d$. What if you try to take two numbers and look at $\langle d_1,d_2\rangle\subset\Bbb Z$? Please do this with specific numbers. Like what about $d_1=8$ and $d_2=6$? You want to describe the totality of all numbers writable in the form $8m+6n$. You rapidly see that this is an ideal in the sense of the definition, and you see that the set is equal to $2\Bbb Z$. This is basic number theory.

Now do the same thing with a ring of polynomials, but in just one variable, $R=\Bbb Q[x]$. One proves (you prove it, with Euclidean division of polynomials) that every ideal of $R$ is of form $fR$, where $f$ is a well-chosen polynomial. In fact, for a nonzero ideal $I$, you take $f$ to be a nonzero element of $I$ of least degree. So, what is $\langle f_1,f_2\rangle$, when $f_1$ and $f_2$ are two polynomials in $x$ that are given? Just as with numbers, it’s the ideal $gR$ where $g$ is the greatest common divisor of $f$ and $g$. Convince yourself that $\langle x^3-1,x^2-2x+1\rangle$ is the set of all multiples of $x-1$.

Things are no longer so simple when you have polynomials in more than one indeterminate. But at least you get some insight by looking at the simplest cases, and I hope you see that the set of all possible $h_1f_1+\cdots+h_nf_n$ is an ideal.

$\endgroup$
1
  • 2
    $\begingroup$ @ThyArtisMath Note that the ring of polynomials in one variable is treated very explicitly and thoroughly in section 1.5 of the textbook Ideals, Varieties, and Algorithms. So, by all means feel free to think about that example and take a shot at proving some things about it, but don't feel like you have to work it out on your own. The textbook goes over all that. $\endgroup$ – Zach Teitler Nov 1 '17 at 5:40
4
$\begingroup$

You were close.

The ideal $(f_1,...,f_s)$ of $k[x_1,..., x_n]$ is the set of all polynomials which can be expressed as $h_1f_1+\cdots +h_sf_s$, for some $h_1,...,h_s \in k[x_1,..., x_n]$.

In other words, "linear combinations" of $f_1,...,f_s$ where the "linear coefficients" are arbitrary elements $h_1,...,h_s$ of $k[x_1,..., x_n]$.

Regarding the author's point about common zeros . . .

If $\bar{k}$ is an algebraically closed extension of $k$, and if there is some $a = (a_1,...,a_n) \in \left(\bar{k}\right)^n$ such that $f_1(a) = \cdots = f_s(a) = 0$, then $a$ is automatically a zero of any polynomial of the form $h_1f_1+\cdots +h_sf_s$, hence $a$ is a common zero for all members of the ideal $(f_1,...,f_s)$.

In particular, if you've identified a common zero $a \in \left(\bar{k}\right)^n$ for $f_1,...,f_s$, then if $g \in k[x_1,..., x_n]$ is such that $g(a) \ne 0$, it follows that $g$ is not in the ideal $(f_1,...,f_s)$.

An important, nontrivial result is a partial converse: If $g \in k[x_1,...,x_n]$ is such that $g,g^2,g^3,...$ are not in the ideal $(f_1,...,f_s)$, there is some $a \in \left(\bar{k}\right)^n$ such that $a$ is a common zero of $f_1,...,f_n$, but $a$ is not a zero of $g$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.