Hot answers tagged ideals
10
The motivation is to characterize what subsets of a ring can be the kernel of homomorphisms.
An equivalent way is to define congruences as for the integers: $x \equiv y \bmod I$ iff $x-y \in I$. Then congruences are compatible with the ring operations iff $I$ is an ideal. Being compatible allows one to define a natural ring structure for the set of ...
5
The link gap between principal maximal ideals and principal rings is bridged by two theorems (and their combination):
(Kaplansky): For a commutative Noetherian ring $R$, $R$ is a principal ideal ring iff every maximal ideal is principal.
(Cohen): For a commutative ring $R$, $R$ is Noetherian iff every prime ideal is finitely generated.
...
4
The notion of ideal comes from a generalization of modular arithmetic. It is a refinement (by Dedekind) of Kummer's notion of ideal number, which arose from attempts to prove Fermat's Last Theorem (or some special cases).
When you have a number $n\in {\bf Z}$, then the multiples of $n$ form an ideal in ${\bf Z}$ (a principal ideal denoted by $(n)$ or $n{\bf ...
3
In groups, we have the notion of a normal subgroup. If $G$ is a group, then we have the result that every normal subgroup $N$ is the kernel of some homomorphism $\phi: G \rightarrow H$.
If $R$ is a ring and $\psi: R \rightarrow S$ is a ring homomorphism, its kernel $I$ will certainly be a subring of $R$, but it will also have the property that if $r \in R$ ...
3
The other solutions have the best elementary solutions, so I'll try to give a distinct way.
The Jacobson radical $J(R)$ of $R$ contains all nilpotent ideals and is contained in all maximal ideals.
By these conditions, $M\subseteq J(R)\subseteq M$. Thus, the Jacobson radical is a maximal ideal. Thus, there cannot be any different maximal ideals, since there ...
3
For the first one, it is called the radical of an ideal, denoted $\sqrt{I}$, see Radical of an ideal. More precisely, suppose $x, y \in \sqrt{I}$, i.e. $x^n \in I, y^m \in I$, how about $(xy)^n, (x + y)^{m + n}, (rx)^n$?
For the second one, consider the ring of matrices. For example $Mat(2, \mathbb{R})$. $0$ is an ideal, and
$$
\begin{pmatrix}
0 & 1 ...
3
A starter would be the observation that prime ideals of an extension field are always above prime ideals of the ground field. In this case, the extension field is $F$, and the ground field is $\mathbb Q$, where all prime ideals are principal ideals generated by prime numbers. Therefore we are reduced to looking for the prime dieal factorisations of primes in ...
2
Prove that
$$aR_1\cap\cdots\cap aR_t\cap A=aA.$$
"$\supseteq$" is clear.
"$\subseteq$" it's easy: take $x\in aR_1\cap\cdots\cap aR_t\cap A$. Then $a^{-1}x\in R_1\cap\cdots\cap R_t$. Since $aR_l=R_l$ for all $l\neq 1,\dots,t$ it follows that $a^{-1}\in R_l$, so $a^{-1}x\in R_l$, and that's all.
2
To prove this result, you must prove along the way that $\mathbb{Z}[i]$ is a PID, to guarantee that you have not missed any non-principal ideals. One cheap way to get around your restriction is to implicitly use the implication
$$\text{Euclidean domain} \Longrightarrow \text{PID}.$$
In the case at hand, we obtain a Euclidean algorithm on $\mathbb{Z}[i]$ as ...
2
There's no need to use any dimension theory. Rather, the proof is by induction. The case $n = 1$ is well known, so I won't repeat it here.
Now, suppose $\mathfrak{m}$ is a maximal ideal of $A = K [x_1, \ldots, x_n]$. By a generalised Nullstellensatz, we know that $A / \mathfrak{m}$ is a finite field extension of $K$; in particular, the image of $x_n$ in $A ...
2
One direction is easy: if $R\cong R_1\times\dots\times R_n$, finding ideals satisfying (a), (b) and (c) is trivial.
Let's do the converse. Conditions (b) and (c) say that $R=I_1\oplus I_2\oplus\dots\oplus I_n$ as abelian groups. This is a standard argument, so I'll not talk about it here.
In particular, there is an additive isomorphism
$f\colon I_1\times ...
1
The addition table tells us that $a$ is the additive identity, i.e. $a=0$. The multiplication table tells us that $b$ is the multiplicative identity, i.e. $b=1$ so $R$ is a commutative ring with identity and we immediately get that $R$ and $(a)$ are ideals.
What are the non-trivial additive subgroups? We notice that $c+c=a=0$ and $d+d=a=0$ so $\{a,c\}$ and ...
1
In general, given $I=(s_1,\dots,s_n)\lhd\mathcal{O}_K$ in the ring of integers of some number field $K$, we can factor it into primes to try and work out its norm.
You can see the answer here which explains the rational behind it, but the important points are that:
If a prime $\mathfrak{p}\mid I$, then $s_1,\dots,s_n\in\mathfrak{p}$. Let ...
1
Just to give this an answer: it seems that the problem was incorrectly stated in the source. As written the claim is not true. E.g. if $B= \mathbb C$ and $A=\mathbb C[x]$, then l(B,A)=0, which is prime, but there are many rings intermediate between A and B, e.g. $\mathbb C[x^2]$. Note that there is nothing special about the choice of $\mathbb C$; it could ...
1
$I[[X]]=I A[[X]]$ holds if and only if for every at most countable subset $S \subseteq I$ there is some finite subset $B \subseteq I$ such that $(S) \subseteq (B)$. Clearly this holds when $I$ is finitely generated. But the converse is not true. There are examples with $|I|=\aleph_1$, the smallest uncountable cardinality. I think ...
1
The situation for the maximal ideals is clear cut since the greatest lower bound (their intersection, the Brown-McCoy radical) happens to be the biggest superfluous ideal of the ring. Any ideal below it is also superfluous, and it contains all superfluous ideals.
Following this, we consider the intersection of all prime ideals (The "lower nilradical", ...
1
Your definition of ideal is correct, so now we need to show that $J$ satisfies the definition. So we ask the following questions:
Is $J$ closed under addition? That is, if $r_1(x)a(x)+s_1(x)b(x)$ and $r_2(x)a(x)+s_2(x)b(x)$ are in $J$, is their sum also in $J$?
Is $J$ closed under multiplication by elements in $F[x]$? That is, if $p(x)\in F[x]$, is the ...
1
You must note that the statement is not true for countable field $\mathbb Q$. Write $\mathbb Q$ as $\{a_1, a_2, \ldots, a_n, \ldots \}$. Then in $\mathbb Q[X]$ define $I_k = (X-a_1)(X-a_2) \ldots (X-a_n)$. Here no $V(I_k)$ is $\mathbb Q$.
Now consider the field is uncountable and $n=1$. Then each $V(I)$ is either empty or finite set or the full space $F$. ...
1
Your intuition is right: The ideal $\left(X_1,X_2,...,X_n\right)$ of $K\left[X_1,X_2,...,X_n\right]$ cannot be generated by less than $n$ elements.
Proof: Let $M$ be the ideal $\left(X_1,X_2,...,X_n\right)$ of $K\left[X_1,X_2,...,X_n\right]$. We need to show that this ideal $M$ cannot be generated by less than $n$ elements.
Assume the contrary. That is, ...
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