A finite dimensional algebra over a field has only finitely many prime ideals and all of them are maximal

Question

Let $K$ be a field and let $R$ be a $K$-algebra with unity which is finite dimensional as a $K$-vector space. Prove that $R$ has only finitely many prime ideals all of which are maximal. (Hint: Chinese remainder theorem)

My attempt: We have $R \simeq K^n$, for some $n$, as vector spaces. I claim that there are finitely many ideals of $K^n$. Let $I$ be an ideal of $K^n$ and $(a_1, a_2,...,a_n) \in I$. For every entry $a_i$, we have two choices, either $a_i=0$ or not. So we get $2^n$ many cases, and any case give a particular ideal of $K^n$ which are distinct from others.

Is my proof correct or am I missing some point? If it is true, how can we give a proof by using the Chinese remainder theorem? Thanks!

Answer 1

A finite dimensional (commutative or not) algebra over a field is an artinian ring, because it certainly satisfies the descending chain condition on left or right ideals.

Let $A$ be a commutative artinian ring and let $P$ be a prime ideal. Then $A/P$ is an artinian domain, which is a field: if $D$ is an artinian domain, consider $d\in D$ and the chain of ideal $(d)\supseteq(d^2)\supseteq\dots\supseteq(d^n)\supseteq\dotsb$. This chain stabilizes, so $(d^n)=(d^{n+1})$ for some $n$, which means that $d^n=td^{n+1}$. Since $D$ is a domain, $1=td$ or $d=0$. Therefore all prime ideals of $A$ are maximal.

Consider now a minimal element in the family of products of finitely many maximal ideals, let it be $M_1\cdots M_k$. If $M$ is a maximal ideal, then $$ MM_1\cdots M_k\subseteq M_1\cdots M_k $$ so, by minimality, $MM_1\cdots M_k=M_1\cdots M_k$ and $$ M_1\cdots M_k\subseteq M $$ Since $M$ is prime, it follows that $M_i\subseteq M$, for some $i=1,\dots,k$; therefore $M_i=M$.

Here we don't use any general fact about artinian rings, just that any nonempty set of ideals has a minimal element (which is general theory of lattices). The Chinese remainder theorem can be used to get an upper bound on the number of maximal ideals; let $M_1,\dots,M_k$ be the set of (pairwise distinct) maximal ideals; then the Jacobson radical $J(A)=M_1\cap\cdots\cap M_k=M_1\cdots M_k$ by coprimality; so $$ A/J(A)\cong A/M_1\times\cdots\times A/M_k $$ by the CRT. If $A$ is a finite dimensional $K$-algebra, then $k\le\dim_K A/J(A)\le\dim_K A$. The upper bound can be reached: the product algebra $K^n$ has exactly $n$ maximal ideals.

Answer 2

I have the impression that the underlying additive structure is that of a $K$-vector space. But can one conclude that $R=K^n$ (isn't the ring $\mathbb{F}_p\times \mathbb{F}_{p^2}$ also a $\mathbb{F}_p$-algebra?). In any case, $R$ is a direct sum of fields $\oplus^k_{i=1} F_i $, so every ideal is of the form $\oplus_{i\in S} F_i$ with $S\subseteq\{1,\ldots,k\}$, and the maximal ideals are those where $S=\{1,\ldots,i-1,i+1,\ldots,k\}$.

Answer 3

Now that the mild ambiguity is resolved, let's address the question.

We begin by showing maximality. Consider a non-trivial prime ideal, $\mathfrak{p}$ of $R$. If $x\not\in\mathfrak{p}$ is fixed, then considering the $\mod \mathfrak{p}$ reductions of $1, x,\ldots, x^{\dim_K(R)}=x^m$ we have a linear dependence by finite dimensionality of $R$. And since $R/\mathfrak{p}$ is an integral domain, we know that this algebraic relation

$$c_0+c_1x+\ldots +c_mx^m$$

has $c_0\ne 0$, otherwise $\overline{x}\in R/\mathfrak{p}$ would be a zero divisor, so that the reduction, $\overline{1}\in R/\mathfrak{p}$ satisfies

$$\overline{1}=c_0^{-1}(-c_m\overline{x}^m-\ldots-c_1\overline{x})$$

verifying the existence of an inverse for $x\mod \mathfrak{p}$, since

$$\overline{1}=\overline{x}\cdot\bigg(-c_0^{-1}(c_m\overline{x}^{m-1}+c_{m-1}\overline{x}^{m-2}+\ldots + c_2\overline{x}+c_1)\bigg)$$

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If there are no more than $\dim_K(R)$ prime ideals we are done, so assume otherwise. Then we take a collection, $\{\mathfrak{p}_1,\ldots,\mathfrak{p}_n\}$, of $n=\dim_K(R)+1$ distinct prime ideals of $R$. Since all prime ideals are maximal, $\mathfrak{p}_i+\mathfrak{p}_j=R$ when $i\ne j$. Then by the CRT we can find $x_1,\ldots, x_n$ such that

$$x_k\equiv\delta_{ik} \mod \mathfrak{p}_i,\quad 1\le k\le n$$

Necessarily the $x_i$ span the quotient $K$-algebra

$$R/(\mathfrak{p}_1\cdot\ldots\cdot\mathfrak{p}_n)\cong R/\mathfrak{p}_1\oplus\ldots\oplus R/\mathfrak{p}_n$$

treated as a vector space, which is of dimension at least $\dim_K(R)+1$, implying there is a surjective $K$-algebra homomorphism from $K^m\to K^{M}$ for some $m<M$. However algebra homomorphisms are also linear maps, which means we have a vector space of lower dimension surjecting onto one of higher dimension. Hence there are at most $\dim_K(R)$ prime ideals, i.e. finitely many.

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Remarks ($1$)In fact any such system of congruences over the finitely many prime ideals has $\gcd(\{x_i\})=1$, i.e. the $x_i$ are coprime.

($2$) Really we could have stopped after solving the system of congruences, as it already implies that the $x_i$ are linearly independent over $K$. To see this we can assume, WLOG, that

$$x_1=\sum_{i=1}^{\dim_K(R)+1}r_ix_i$$

Then $x_1\in \left(x_2,\ldots, x_{\dim_K(R)+1}\right)$, since the latter is an algebra ideal, a contradiction to the congruences.

Answer 4

Another proof: first, note that a prime ideal always arises as the kernel of a map $R\to L$ to an algebraically closed field ($L=\overline{Q(R/P)}$). But $R$ is finite over $K$, so any map $R\to L$ factors: $R\to \overline{K} \to L$. We have only finitely many choices for where to send a generating set, hence finitely many prime ideals.

But we also have the fact that a finite-dimensional integral domain over a field must be a field itself (because multiplication by a nonzero element is injective, hence surjective). So the image of $R$ in $\overline{K}$ is always a field, i.e. every prime ideal of $R$ is maximal.

Answer 5

Since $R$ is finitely generated as a $K$-module, $R$ is integral over $K$ and $R$ is finitely generated $K$-algebra. Hence $\dim R=0$ and by Hilbert basis theorem $R$ is Noetherian ring. Hence $R$ is an Artinian ring and the result follows.