Let $R$ be a commutative ring containing a field $k$ as sub-ring. Assume that $R$ is a finite dimensional $k$-vector space. Show that every prime ideal of $R$ is maximal.

My approach:

Let $I$ be an prime ideal of $R$, as $R$ is a finite dimensional vector space, assume $I= \text{span}(\beta_1, \beta_2...\beta_m$). Assume $I$ is not a maximal ideal and $I \subset J$, where $J$ is a proper ideal. Since $J$ is maximal, $R/J$ is a field.

Say, we choose $r_1$, $r_2\in$ $R$, such that $(r_1+J)(r_2+J)=1+J$, since $R/J$ is a field.

Now say $r_1\in I$, so, $r_1=\sum_{i=0}^m a_i\beta_i$, where all $a_i$'s are from field $k$.

As $r_i \in I$, $r_1r_2 \in I$, so $r_1r_2=\sum_{i=0}^m b_i\beta_i$. Where all $b_i$'s are from field $k$.

We have chosen $r_1, r_2$ in such a way $(r_1+J)(r_2+J)= r_1r_2+j_1=1+j_2$.

As $I \subset J$, $J$ has a basis of ($\beta_1, \beta_2...\beta_n$), where ofcourse $n>m$.

Say $j_1 = \sum_{i=0}^n c_i\beta_i$ and $j_2=\sum_{i=0}^n d_i\beta_i$.

Now, $$r_1r_2+j_1=1+j_2\Rightarrow \sum_{i=0}^m b_i\beta_i +\sum_{i=0}^n c_i\beta_i=1+\sum_{i=0}^n d_i\beta_i $$ $$ \Rightarrow \sum_{i=0}^m b_i\beta_i +\sum_{i=0}^n c_i\beta_i -\sum_{i=0}^n d_i\beta_i=1 $$ $$ \Rightarrow 1 \in span(\beta_1, \beta_2... \beta_n)$$ So the multiplicative unity $1 \in J$. Hence, J is not a proper ideal of R, rather $J=R$. Hence $I$ is maximal.

So here we need not to use the property that $I$ is a prime ideal. Therefore I think either I'm missing something, or this method is completely wrong. Please help.

  • 2
    $\begingroup$ I don't see why you think there are $r_i\in I$ with $r_1r_2\equiv1\pmod J$. Surely there aren't such $r_i$? $\endgroup$ Jun 1, 2018 at 5:13
  • $\begingroup$ because every element in the field $R/J$ must have multiplicative inverse. $\endgroup$
    – L--
    Jun 1, 2018 at 5:21
  • 1
    $\begingroup$ Not every element in a field has a multiplicative inverse. $\endgroup$ Jun 1, 2018 at 5:22
  • $\begingroup$ I mean every non zero element. $\endgroup$
    – L--
    Jun 1, 2018 at 5:44
  • 1
    $\begingroup$ Of course here, since $I\subset J$ the $r_i$ become zero in the quotient $R/J$. $\endgroup$ Jun 1, 2018 at 5:47

2 Answers 2


Let $S=R/I$. Then $S$ is also a finite-dimensional $k$-algebra. As $I$ is a prime ideal of $R$ then $S$ is an integral domain.

To prove $I$ is maximal in $R$, it suffices to prove $S$ is a field. Let $a$ be a nonzero element of $S$. Consider the multiplication map $\mu_a:S\to S$ defined by $\mu_a(x)=ax$. As $S$ is an integral domain, $\mu_a$ is an injective map. Also it is a linear map of $k$-vector spaces from $S$ to itself. By the rank-nullity formula, it is also surjective, so there is $b\in S$ with $\mu_a(b)=1$.

  • $\begingroup$ Thanks, I understood this argument but please help me to figure out shortcomings in my method $\endgroup$
    – L--
    Jun 1, 2018 at 5:46
  • $\begingroup$ @ArnabChowdhury In my comment to your original posting I pointed out the first bit of your argument that I didn't understand. $\endgroup$ Jun 1, 2018 at 5:49

My favorite way of attacking problems like this one is somewhat different than, but of course related to, Lord Shark the Unknown's approach.

Before taking that up, however, a few words on our OP Arnab Chowdhury's attempted proof:

The attempt makes it's first wrong turn, I think, where the assumption $r_1 \in I$ is made; certainly there is no error in postulating $r_1, r_2 \in R \setminus J$ such that

$(r_1 + J)(r_2 + J) = 1 + J, \tag 1$

since it is assumed $J$ is a proper maximal ideal; at the very least we might have $r_1 = r_2 = 1$, and if $R$ has any invertible element $s \ne 1$ we can choose $r_1 = s$, $r_2 = s^{-1}$; so up 'til this point things work, in any event; but now the hypothesis $r_1 \in I$ is manifestly problematic, since our OP has also assumed $I \subset J$, so then $r_1 \in J$ which flies in the face of (1) by virtue of the fact that

$r_1 \in J \Longrightarrow r_1 + J \equiv 0 \in R/J. \tag 2$

The contradictory assertions (1) and (2) unfortunately put a crack in the foundation of our OP Arnab Chowdhury's proof, and the logical structure built upon it cannot stand; for example, it is certainly true that

$r_1 \in I \Longrightarrow r_1 r_2 \in I, \tag 3$

but again, this implies $r_1 r_2 \in J$ and thus that

$(r_1 + J)(r_2 + J) = r_1 r_2 + J = J, \tag 4$

another contradiction to (1). At this point, having explained some of the unfortunately fatal flaws in our OP's proof, I will leave of this topic and turn to presenting what I hope is a correct argument of my own:

We first observe that we may regard the field $k$ as a subring of $R$, since the map $k \to R$, $c \to c \dot 1_R$ is manifestly an isomorphism from $k$ onto the subring $k1_R \subset R$; indeed, if $c_1 1_R = c_2 1_R$ then $(c_1 - c_2)1_R = 0$ forcing $c_1 - c_2 = 0$ or $c_1 = c_2$; thus $k \to R, \; c \to c 1_R$ is injective; the other properties which need to be validated to show that $c \to c1_R$ is a homomorphism, namely that

$c_1 + c_2 \to c_1 1_R + c_2 1_R, \tag 5$


$c_1 c_2 \to (c_1 1_R)(c_2 1_R) \tag 6$

follow directly from the vector space axiom for distributivity of scalar multiplication with respect to field addition and ordinary properties of multiplication and of the unit $1_R$ in a commutative ring such as $R$:

$c_1 c_2 \to (c_1 c_2) 1_R = (c_1 c_2) 1_R^2 = c_1 (c_2 1_R) 1_R = c_1 (1_R(c_2 1_R)) = (c_1 1_R)(c_2 1_R); \tag 7$

taking together (5)-(7) we easily see that $c \to c 1_R$ is a homomorphism from $k$ into $R$, and since we have also shown that this map is injective, we conclude that it is in fact an isomorphism from $k$ to the subring $k 1_R$ of $R$; thus $R$ contains a subring $k1_R$ isomorphic to $k$, and since this correspondence is $c \longleftrightarrow c 1_R$ for $c \in k$, we may for all intents and purposes consider vector space scalar multiplication by $c \in k$ to be the same as ring multiplication by $c 1_R \in R$.

The second thing we need note is that all ideals $I$ of $R$ are in fact vector subspaces of $R$ over $k$, a fact which is easy to see since not only are ideals abelian subgroups under ring/vector addition, they are also, being ideals, closed under multiplication by elements of $k$ via the indentification $c \longleftrightarrow c1_R$ developed in the preceding paragraph:

$c \in k, \; i \in I \Longrightarrow ci = c(1_R i) = (c1_R)i \in I; \tag 8$

this shows that ideals $I \subset R$ are in fact vector subspaces, as claimed. Then factor rings of the form $R/I$ inherit both the ring and vector space structure of $I$, and are thus themselves, like $R$, algebras over the field $k$.

These points being made, let $P$ be a prime ideal in $R$; then $R/P$ is an integral domain as well as, as has been seen, a $k$-vector space; furthermore, since $R$ is finite dimensional over $k$, so must be $P$ and also $R/P$; thus $R/P$ is a finite-dimensional algebra over $k$ without divisors of zero; we show every $0 \ne a \in R/P$ is invertible:


$\dim_k R/P = m, \tag 9$

and consider the set

$A = \{ 1_{R/P}, a, a^2, \ldots, a^m \} \subset R/P \tag 9$

of powers of $a$; since $\vert A \vert = m + 1$, it follows from (9) that there must be a linear relation over $k$ amongst the elements $a^i$ of $A$; that is, here exist $c_i$, $0 \le i \le m$, not all zero, with

$\displaystyle \sum_0^m c_i a^i = 0, \tag{10}$

where we take $a^0 = 1_{R/P} = 1_R + P$, the unit of $R/P$. Now I claim that there must be at least two $c_i \ne 0$; otherwise, (10) boils down to

$c_j a^j = 0, c_j \ne 0 \; \text{for some} \; j, 0 \le j \le m; \tag{11}$

if this were true, we could not have $j = 0$, lest (11) read

$c_0 1_{R/P} = 0, \tag{12}$

which forces $1_{R/P} = 0$, contradicting $1_{R/P} = 1_R + P$; thus if (11) were to hold, we must have $j \ge 1$; but then

$c_j a^j = 0 \Longrightarrow a^j = 0 \Longrightarrow a = 0, \tag{13}$

since an integral domain such as $R/P$ has no non-zero nilpotent elements; but this contradicts our assumption that $a \ne 0$; therefore the sum (10) must have at least two non-zero terms; then let $k$ be the least power of $a$ occurring in (10); if $i < k$, $c_i = 0$; then

$a^k \displaystyle \sum_k^m c_i a^{i - k} =\sum_k^m c_i a^i = 0; \tag{14}$

now $a^k \ne 0$ so we have

$\displaystyle \sum_k^m c_i a^{i - k} = 0, \tag{15}$

which may be re-written in the form

$\displaystyle \sum_{k + 1}^m c_i a^{i - k} = -c_k a^0 = -c_k 1_{R/P}, \tag{16}$


$\displaystyle \sum_{k + 1}^m \dfrac{c_i}{-c_k} a^{i - k} = 1_{R/P}, \tag{17}$


$a \displaystyle \sum_{k + 1}^m \dfrac{c_i}{-c_k} a^{i - 1 - k} = 1_{R/P}, \tag{18}$

and thus

$a^{-1} = -\displaystyle \sum_{k + 1}^m \dfrac{c_i}{c_k} a^{i - 1 - k}. \tag{19}$

This shows that every $0 \ne a \in R/P$ is invertible, and thus $R/P$ is a field; we conclude then that $P$ is maximal in $R$.


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