Understanding the concept of polynomial ideals

Question

I am not able to understand the fundamental concept behind a polynomial ideal. What I have so far in terms of $I$ being an ideal of a ring is:

• for each $f, g \in I$, we have $-f$ and $f+g \in I$, and
• for each $f \in I$ and $g \in \mathbb{K}[X_1,...,X_n]$, we have $f\cdot g \in I$.

The second bullet point being the most interesting part, as it maps a polynomial of the ring into the ideal subset. Now I am trying to construct an example making use of following:

$\langle f_1,...,f_s \rangle=\bigg \lbrace \sum\limits_{i=1}^s h_if_i | > h_1,...,h_s \in \mathbb{K}[X_1,...,X_n] \bigg \rbrace$

So my example is:

$I=\langle f_1,f_2 \rangle=\langle x+y,x^2 \rangle \subseteq \mathbb{K}[x,y] = \bigg \lbrace h_1\cdot (x+y) + h_2\cdot x^2 | h_1,h_2 \in \mathbb{K}[x,y] \bigg \rbrace$

Now if I take an arbitrary polynomial $g$ out of my ring, e.g. $g = x^3$, it should be mapped into the ideal subset by multiplying it with any members of my ideal set, i.e. $g\cdot f_1$ or $g\cdot f_2$. Now if I do it here I get:

$f_1 \cdot g = (x+y)x^3 = x^4+yx^3$

But how is $x^4+yx^3$ being part of $I$? Do I have to choose $h_1=x^3$ in order to do so? I feel being confused by the whole concept of an ideal and missing the fundamental point here. Is an ideal some sort of a basis for algebraic structures?

Answer 1

Let's work over a commutative ring $R$. It might also help to think of ideals $I \subset R$ as precisely those subsets that arise as kernels of homomorphisms, i.e., as those subsets that induce a ring structure on the quotient $R/I$. Indeed the second bullet point that you mention is the one that makes the product operation in $R/I$ well-defined. Check that the kernel of a homomorphism between two rings is an ideal; conversely, any ideal $I \subset R$ arises as the kernel of the quotient map $R \to R/I$.

Once you pass to the quotient $R/I$, the generators of $I$ become relations that elements in the ring must satisfy. For example, if $R = \mathbb{K}[x,y]$ and $I = \langle x+y, x^2 \rangle$, the relation $y = -x$ in $R/I$ means that every term where some instances of $y$ appear can be reduced to a term where only $x$ appears. More formally, the map $R/I \to \mathbb{K}[x]/\langle x^2 \rangle$ that sends $ax+by+c$ to $(a-b)x + c$ is an isomorphism between $R/I$ and the algebra of "dual numbers" $\mathbb{K}[x]/\langle x^2 \rangle$.

Also, properties of ideals correspond to properties of the quotient ring. For example, $I$ is prime iff $R/I$ is an integral domain, and $I$ is maximal iff $R/I$ is a field. Taking your example $I = \langle x,y \rangle$, which as we said is a maximal ideal, we have the relation $x = y = 0$ in the quotient ring, so passing to the quotient kills all the variables and leaves only the constants. Thus $R/I \cong \mathbb{K}$ really is a field.

Answer 2

You are asking about the purpose of ideals; a very good question!

To understand the interest of this concept, do the following exercise.

Suppose $(R, +, *)$ is a commutative ring, and $J$ is a subset of $R$.

Call two elements in $R$ equivalent if their difference lies in $J$.

Denote by $R/J$ the quotient set of $R$ by this equivalence relation.

1. What condition should $J$ satisfy so that the $+$ operation descends to $R/J$ and $(R/J, +)$ becomes a group? Answer: $J$ should be a subgroup for this to work. (Really $J$ should be a normal subgroup, but since $+$ is commutative, all subgroups are normal.)

2. What extra condition should $J$ satisfy so that the $*$ operation also descends to $R/J$ and $(R/J, +, *)$ becomes a ring? Answer: $J$ should be an ideal.

So although the definition of ideals looks somewhat arbitrary if stared at bluntly, once you do the exercise above it becomes clear why these properties are interesting, and clearly we need a name for subsets of a ring satisfying these properties.