A bottom-up approach to Hilbert’s Basis Theorem

The proof that $I_{z}$ is an ideal follows from my previous remark. So then suppose there exists an ideal $I$ that cannot be written as an $I_{z}$. $I\neq\{0\}$ because then $I=I_{0}$, so there exists at least one nonzero integer in $I$. Consider the set of nonzero positive integers in $I$. This set must be nonempty; an empty set would imply that the nonzero integer $z$ we found earlier is negative, but $z*(-1)\in I$—a contradiction. By the Well-Ordering Principle, find the smallest positive nonzero integer $z\in I$.

Because $I\neq I_{z}$, there must exist an element $z^{\prime}\in I$ that is not in the form $z*r$ for any $r\in\mathbb{Z}$. Suppose it is positive (if it is negative multiply it by $-1$ and it will still be in the ideal). We know that $z^{\prime}>z$ because $z$ is the smallest positive integer. Now, observe that $z^{\prime}+(-z)$ is still in the ideal and also will not be in the form $z*r$. Repeat until the result is less than $z$. The result will be in $I$, positive and nonzero (because it is not in the form $z*r$). Thus, we have found a smaller positive nonzero integer in $I$ that is smaller than $z$—a contradiction. ∎

First, we prove that that the only ideals of $k$ are the field itself and the zero ideal. Let $I$ be an arbitrary ideal.

1. Case 1:

   $I$ contains no non-zero elements. Then, $I=\{0\}$.

2. Case 2:

   $I$ contains a non-zero element. Let $i$ be this element. Then, because $k$ is a field, there exists a multiplicative inverse of $i$, denoted by $i^{-1}$. By the property of ideals, $i*i^{-1}=1\in I$. Any element $j\in k$ is in the ideal now because $1*j=j\in I$, so then $I=k$.

Second, we prove that if the only ideals of a ring $R$ are the ring and the zero ideal, the ring is a field. Let $r$ be a non-zero element in our commutative ring $R$. Construct a set $I=\{rq\mid q\in R\}$. It is easy to check that this is an ideal. Because this ideal contains a nonzero element $r$, it cannot be the zero ideal, so by our given, the ideal must be the entire ring. Then, there exists an element $r^{\prime}$ such that $r(r^{\prime})=1$ since $1$ is in our ideal. Thus, $r$ has an inverse. ∎

First, we show Definition 2.9 from Definition 2.3. We need to find a ring $S$ and a function $\phi$ as described. We need $\phi$ such that for all $i\in I$, $\phi(i)=0$. Furthermore, we need $\phi$ such that adding $i\in I$ to any $r\in R$ will not change the computed result (i.e., $\phi(r)=\phi(r+i)$).

Let $r$ be an element in $R$. Define $\phi\colon r\mapsto X_{r}:=\{r+i\ :\ i\in I\}$. Define $R/I=\{X_{r}\mid r\in R\}$.

We now define an equivalence relation between elements in $R$. We say that $r\sim q$ for $r,q\in R$ if and only if $\phi(r)=\phi(q)$. It follows from properties of an ideal that $\sim$ is an equivalence relation. Notice that, under this definition, $\phi(i)=\phi(0)\;\forall\;i\in I$.

Define addition and multiplication as follows: for $\phi(r),\phi(q)\in R/I$, let $\phi(r)+\phi(q)=\phi(r+q)$ and $\phi(r)*\phi(q)=\phi(r*q)$. Both of these are well-defined. Consider, for example, addition. Let $r_{1},r_{2},q_{1},q_{2}$ be elements in $R$ with $r_{1}\sim r_{2}$ and $q_{1}\sim q_{2}$. Let $i,j$ be elements in $I$ such that $r_{1}=r_{2}+i$ and $q_{1}=q_{2}+j$. Then,

We conclude that equivalent inputs implies equivalent outputs under our addition operation. Multiplication follows similarly as well.

Under these operations, $R/I$ forms a ring. To finish the proof, define $S=R/I$ and $\phi$ as defined above. By our definition of $\phi$, we already know that $\phi(r_{1}+r_{2})=\phi(r_{1})+\phi(r_{2})$ and $\phi(r_{1}*r_{2})=\phi(r_{1})*\phi(r_{2})$. It suffices to show that $\phi(1_{R})=1_{S}$ where $1_{S}$ is the multiplicative identity of $R/I$. The proof follows from the definition.

Second, we show Definition 2.3 from Definition 2.9. As discussed, we must prove only three statements:

1. (a)

   Closure with respect to addition: let $i_{1},i_{2}\in I$. We know $\phi(i_{1}+i_{2})=\phi(i_{1})+\phi(i_{2})=0+0=0$. Then, $i_{1}+i_{2}\in I$ because $\phi(i_{1}+i_{2})=0$.

2. (b)

   Existence of additive inverse: $\phi(0)=0$ so $0\in I$.

3. (c)

   If $i\in I$ and $r\in R$ then $i*r\in I$: let $i\in I,r\in R$. $\phi(i*r)=\phi(i)*\phi(r)=0*\phi(r)=0$. Then, $i*r\in I$ because $\phi(i*r)=0$.

∎

Let $I$ be an ideal generated by a prime natural number $p$. Suppose $I$ is not prime. Then, there exists an element $k\in I$ and $i,j\in I$ such that $i*j=k$ but $i\not\in I$ and $j\not\in I$. This means that $k$ is divisible by $p$ but neither $i$ nor $j$ are divisible by $p$, which is impossible.

The zero ideal is prime because no integer divides $0$.

We already showed that all ideals of $\mathbb{Z}$ are generated by one element, and so it is only necessary to look at ideals generated by nonprime elements. Let $I$ be an ideal generated by a nonprime natural number $n$. Because $n$ is not prime, $n=\ell*m$ for some $\ell,m$ not divisible by $n$, so $\ell$ and $m$ are not in $I$; then, $I$ is not prime. The ideal $I$ generated by $-n$ is the same as the one generated by $n$ so the same logic applies. ∎

First, we prove that Definition 2.11 is equivalent to Definition 2.13.

Let $I$ be a prime ideal by Definition 2.11. Suppose that Definition 2.13 does not hold. By Definition 2.9, there must exist a set $S$ and $\phi$ such that $I$ is the kernel of $\phi$. Then, take $S$ and $\phi$ that Definition 2.13 refers to. Consider two cases:

1. Case 1:

   $S=\{X_{0}\}$. $I$ is not equal to $R$. Then, there exists an element $r$ in $R$ that is not in $I$, and $\phi(r)\neq X_{0}$.

2. Case 2:

   $S$ has a zero-divisor. Let $s$ be a zero-divisor that is not equal to $0$ with a corresponding $t$ that is not equal to $0$ such that $s*t$ is equal to $0$. By construction, $\phi$ is surjective, so there exist $r,q\in R$ such that $\phi(r)=s$ and $\phi(q)=t$. Then,

   $$0=s*t=\phi(r)*\phi(q)=\phi(r*q),$$

   so $r*q\in I$. However, $r\not\in I$—otherwise $\phi(r)=0$ (and similar argument for $s$). Then, $I$ is not prime according to Definition 2.11, so we have arrived at a contradiction.

Second, we prove that Definition 2.13 is equivalent to Definition 2.11.

For sake of contradiction, let $I$ be an ideal that is not prime by Definition 2.11. By Definition 2.9, there exists a set $S$ and $\phi$ such that $I$ is the kernel of $\phi$. Consider two cases:

1. Case 1:

   $I=R$. Then, every element $i\in I$ maps to $0$, so $S=\{X_{0}\}$.

2. Case 2:

   There exist elements $r,q\in R$ such that $r*q$ is in $I$ but neither $r$ nor $q$ is in $I$. Then,

   $$0=\phi(r*q)=\phi(r)*\phi(q),$$

   but neither $\phi(r)$ nor $\phi(q)$ are equal to $0$ because neither $r$ nor $q$ are in $I$. Then, both $\phi(r)$ and $\phi(q)$ are zero-divisors of $S$ so we have arrived at a contradiction.

∎

(i) $\Rightarrow$ (ii): Take a chain of ideals, and let $I$ be the union of them all. For $I$, there exist $x_{1},\dots,x_{n}$ that generate $I$. All of these $x_{i}$ should appear at some point in the chain of ideals. When they appear, they stay in the chain because each ideal contains all previous ideals. Because there are finitely many $x_{i}$, we can find an $I_{\ell}$ for some $\ell$ that finally contains all of them. It is clear that $I_{\ell}=I$ because they are generated by the same elements. To prove the chain terminates, let $k$ be a natural number bigger than $l$. $I_{l}$ is a subset of $I$ and therefore is a subset of $I_{k}$. $I_{k}$ is a subset of $I_{l}$ because we have a chain of ideals. Then, $I_{l}=I_{k}$, so the chain terminates.
  (ii) $\Rightarrow$ (iii): At the ideal at which the chain terminates, that ideal is not contained in any other.
  (iii) $\Rightarrow$ (i): Let $I$ be an ideal and $\Sigma=\{J\subset I\mid J\text{ is a finitely generated ideal}\}$. By our given, $J$ has an element $J_{0}$ that is not contained in any other. Suppose that $J_{0}\neq I$. Then, there exists $x\in I,x\not\in J_{0}$. Let $J_{0}^{\prime}$ be the ideal generated by all the finitely many generators in $J_{0}$ and also $x$. $J_{0}^{\prime}$ is still finitely generated and contains $J_{0}$, contradicting our assumption that $J_{0}$ is not contained in any other. Then, $I=J_{0}$ is finitely generated. ∎

Follows from field operations. ∎

$(1)$ and $(2)$ follow from the definition.

For $(3)$, note that $I(X)$ is defined as the set of all polynomials that vanish on $X$, so these polynomials will vanish on $x\in X$. For the second part of $(3)$, we know that $X=V(I(X))$; then, $X=V(S)$ for $S=I(X)$, meaning that $X$ is an algebraic set. Next, let us prove that $X$ is an algebraic set implies that $X=V(I(X))$. $X=V(S)$ for some $S$. Suppose, for sake of contradiction, that $I(X)$ does not contain $S$. Then, there exists a polynomial $f\in S,f\not\in I(X)$. We know that $f(x)=0$ for all $x\in X$ by definition, so then $f\in I(X)$—a contradiction. Because $S\subset I(X)$, $V(I(X))\subset V(S)=X$ so $V(I(X))=X$ (we already know $X\subset V(I(X))$ because of the first part of $(3)$).

For $(4)$, let $S$ be such a set and let $f\in S$. By definition, we know that $f$ vanishes at the points at which $S$ vanishes. ∎

Both follow from the definition. ∎

1. (1)

   Let $X=V(S)$ and $Y=V(T)$ be algebraic sets for some sets of polynomials $S$ and $T$. Indeed, $X\cup Y=V(ST)$, where $ST$ denotes the set of all products of an element of $S$ by an element of $T$. Let $x$ be an element in $X\cup Y$. Then, $x\in X$ or $x\in Y$. Without loss of generality, suppose $x\in X$. Then, $f(x)=0$ for all $f\in S$ so every function in $ST$ vanishes on $x$. Now, let $x\in V(ST)$. Without loss of generality, suppose that $x\not\in X$. Then, there is an $f\in S$ such that $f(s)\neq 0$. All polynomials $fg\in ST$, where $g\in T$, vanish on $x$, so then $g(t)=0$ for all $g\in T$ and $x$ vanishes on $Y\subset X\cup Y$.

   By induction, the union of finitely many algebraic sets is an algebraic set as well.

2. (2)

   Let $\{X_{\alpha}=V(S_{\alpha})\}$ be a collection of algebraic sets indexed by $\alpha$. Then, $\bigcap_{\alpha}X_{\alpha}=V(\bigcup_{\alpha}X_{\alpha})$. Let $x\in\bigcap_{\alpha}X_{\alpha}$. Then, $x\in X_{\alpha}$ for all $\alpha$ so $x\in V(S_{\alpha})$ for all $\alpha$. Now, let $x\in V(\bigcup_{\alpha}X_{\alpha})$. By definition, $x\in V(X_{\alpha})$ for every $\alpha$ so $x\in X_{\alpha}$ for every $\alpha$. Then, $x\in\bigcap_{\alpha}X_{\alpha}$.

∎

$X=V(I(X))$ by Proposition 3.8. $I(X)$ is finitely generated by Corollary 3.12. So, we have $X=V((f_{1},\dots,f_{n}))$. $V((f_{1},\dots,f_{n}))=V(f_{1})\cap\dots\cap V(f_{n})$ follows from the definition of the variety of a set of polynomials. ∎

Suppose that the chain does not terminate. Then, $I(X_{1})\subset I(X_{2})\subset\dots$ does not terminate—a contradiction because $k[X_{1},\dots,X_{n}]$ is Noetherian. ∎

Let $\Sigma$ be the set of all algebraic sets in $k^{n}$ that do not have property $*$. If $\Sigma$ is empty, our proof is done. Suppose that $\Sigma$ is not empty. Then, there exists an algebraic set $X$ that does not contain any other algebraic set in $\Sigma$ (otherwise we have a descending chain of algebraic sets that does not terminate). $X$ is not irreducible by our given. Then, let $X_{1},X_{2}$ be algebraic subsets such that $X_{1},X_{2}\subsetneq X$ and $X=X_{1}\cup X_{2}$. One of $X_{1},X_{2}$ cannot have property $*$; otherwise $X$ has property $*$ (by unioning all finite irreducible polynomials making up $X_{1}$ and $X_{2}$) so then $X\not\in\Sigma$. Thus, at least one of $X_{1},X_{2}$ is in $\Sigma$, so there does exist an algebraic set in $\Sigma$ that is contained in $X$—a contradiction. ∎

The chain of ideals $(X_{1})\subset(X_{1},X_{2})\subset\dots$ never terminates. ∎

Let us prove that if $X$ is irreducible, $I(X)$ is prime. Suppose that $I(X)$ is not prime. Then, there exist polynomials $f_{1},f_{2}\not\in I(X)$ such that $f_{1}f_{2}\in I(X)$. Let $I_{1}$ be the ideal generated by $f$ and all generators of $I$. $I(X)\subsetneq I_{1}$ so $V(I_{1})\subsetneq V(I(X))=X$ because $X$ is an algebraic set. The same logic applies for $f_{2}$. So we have that $V(I_{1}),V(I_{2})\subsetneq X$. Let $x\in X$. We know that $f_{1}f_{2}(x)=0$, so either $f_{1}(x)=0$ or $f_{2}(x)=0$, so $x\in V(I_{1})$ or $x\in V(I_{2})$. Then, $X\subset V(I_{1})\cup V(I_{2})$ so $X$ is reducible.

Let us prove now that if $I(X)$ is prime, $X$ is irreducible. Suppose that $X$ is not irreducible. Let $X_{1},X_{2}\subsetneq X$ such that $X=X_{1}\cup X_{2}$. Because $X_{1}\subsetneq X$, $I(X)\subsetneq I(X_{1})$, there exists $f_{1}$ such that $f_{1}\in I(X_{1})$ but $f_{1}\not\in I(X)$ (and a similar $f_{2}$ for $I(X_{2})$). $f_{1}f_{2}$ vanishes at all points in $X$ so $f_{1}f_{2}\in I(X)$, so $I(X)$ is not prime. ∎