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6

Let $R$ be the $\mathbb Q$ vector space generated by all ordinals of cardinality less than $c$ together with the ordinal $c$. Let multiplication be given by intersection. Let $I$ be the ideal generated by all countable ordinals. Since the continuum hypothesis is undecidable, it is undecidable whether $I$ is a maximal ideal.


5

Not sure if this is satisfactory, but division rings only have trivial left and right ideals. Therefore finite products of noncommutative division rings are counterexamples.


4

Note that you get $P_1=P_1P_2\subsetneq P_2$ , so $P_1$ is not maximal.


4

Note that $P_1P_2 \subseteq P_2$, so if $P_1 = P_1P_2$ then by maximality $P_1 = P_2$.


4

An ideal of $2\mathbb{Z}$ must be an additive subgroup, to begin with; so it is of the form $2k\mathbb{Z}$. Any such subset (with $k\ge0$ for a unique representation) is indeed an ideal, as it is readily checked. Saying that $2k\mathbb{Z}\subsetneq 2l\mathbb{Z}\subsetneq 2\mathbb{Z}$ means $l>1$ and $l\mid k$ with $k\ne l$. Thus the maximal ideals are ...


4

$I$ is a prime ideal since clearly $k[x,y,z]/(z-1,x^2-y) = k[x]$.


4

It's a standard argument in commutative algebra. Consider the set of all ideals that contain $I$ and avoid $S$, ordered by inclusion. By Zorn's lemma there are maximal elements (since unions are upper bounds) and you can prove that an ideal maximal via inclusion among those avoiding $S$ and containing $I$ turns out to be prime.


3

Define $$\phi: R\to Rx\le M\;,\;\;\phi(r):=rx$$ prove the above is a (left) $\;R$- module homomorphism, and now use the first isomorphism theorem.


3

Ideals are not generalized numbers It is true that ideals were introduced by Dedekind as "ideal numbers" in order to restore unique factorization in certain rings where it fails. These rings are now called Dedekind rings and indeed it is a fruitful point of view to treat ideals in these rings as generalized numbers. However these rings have Krull dimension ...


3

$\langle 14, \sqrt{-2} \rangle = \langle \sqrt{-2} \rangle$. That's because $\gcd(14, \sqrt{-2}) = \sqrt{-2}$. You already had all the information to get the answer yourself, so I'm not sure what threw you off. Maybe the fact that, as you yourself noted, $\langle 7, \sqrt{-2} \rangle = \langle 1 \rangle$, and since $14 = (-1) \times (\sqrt{-2})^2 \times 7$, ...


3

In a PID, every ideal is principal. That includes every ideal generated by a pair of elements. So such an ideal $(a, b)$ must always be generated by a single element $c$, which therefore must be the greatest common divisor $\gcd(a, b)$ (up to a unit). This is one definition of $\gcd(a, b)$, in fact. In $\mathbb{Z}$ you get the $\gcd$ in the usual sense. The ...


3

You are correct about the definition of $\mathbb{Z}[x]$. Maximal in this context means "maximal among proper ideals". The ideal $J$ that you suggested is in fact all of $\mathbb{Z}[x]$. To see that $I$ is maximal among proper ideals, suppose that another ideal $K$ contains $I$. Since $I$ is the set of polynomials with even constant term, either $K=I$ or ...


3

We assume that by $R[t]$ you mean the ring of polynomials with real coefficients. But assuming that $R$ is a field is enough. Edit: OP has clarified that it is the reals. That $U$ is an ideal is a straightforward verification of properties. We show now that any ideal $J$ which is a proper extension of $U$ is the whole ring. Let $f(t)$ be an element of $J$ ...


3

A ring is called a left duo ring if every left ideal is also a right ideal. (It's called a duo ring if every one-sided ideal is two-sided.) There are many rings which are duo on one side but not commutative. As already discussed, division rings and finite products of them are duo rings. But putting them aside, you could also make a duo ring by taking your ...


3

Suppose that $F$ is a field, and $I$ is a non-zero ideal in $F$; show that $1_F\in I$ and then that $I=F$. The second equivalence is from the fourth isomorphism theorem for rings.


3

It makes life easier if we add: (c) The function $R \to R/I \times R/J$ has $(0,1)$ in its image. Notice that (c) means that there is some $i \in R$ such that $i \equiv 0 \bmod I$ and $i \equiv 1 \bmod J$, i.e. $i \in I$ and $i-1 \in J$. In other words, $1 \in I+J$. This shows that (b) and (c) are equivalent. It is trivial that (a) implies (c). Conversely, ...


2

Yes. In fact for any ring homomorphism $\phi:R\to S$, and any prime $P\subset S$, the preimage $\phi^{-1}(P)$ is also prime. In your situation, $\phi$ is the inclusion of $\mathbb{R}[x_1,\ldots,x_n]$ into $\mathbb{C}[x_1,\ldots,x_n]$.


2

$x \in rad(I) \Leftrightarrow x^n \in I,$ for some $n \geq 1.$ So, $x + I \in rad(I)/I \Leftrightarrow x^n + I = 0$ in $R/I$ for some $n \geq 1 \Leftrightarrow (x + I )^n = 0$ in $R/I$ for some $n \geq 1 \Leftrightarrow x + I \in \mathfrak{N}(R/I).$


2

This is straightforward but confusing. $\subseteq$: Suppose $x \in V(\sum I_a)$. By definition $f(x)=0$ for all $f \in \sum I_a$. But the generators of $\sum I_a$ is the union of the generators of each $I_a$, so in particular $f(x)=0$ for each generator of $I_a$. This means that $x \in V(I_a)$ for all $a$. So $ x \in \cap_a V(I_a)$. $\supseteq$: This is ...


2

Suppose $I$ is an ideal in $R$. Clearly $I[x]$ is a subring of $R[x]$ since $I$ must be a subring of $R$. By definition, $\forall r\in R\;\forall x\in I\;rx,xr\in I$. Then since the coefficients of the product of two polynomials result from products of the coefficients, it is clear that $\forall p\in R[x]\;\forall q\in I[x]\;pq,qp\in I[x]$.


2

would i be right in saying that it needs to have an inverse in R as well if it is to have one in the quotient? Interestingly, yes. If $x+I$ is a unit, there exists $y$ such that $xy-1\in I$ (this is just the translation of what it means to be a unit in $R/I$.) Then $xy-1=n$ for some nilpotent element, and rearranging we get $xy=n+1$. Since the right ...


2

Here is a more structural approach to supplement André Nicolas's excellent answer. Let $R$ be a field. Consider the ring homomorphism $\phi: R[t] \rightarrow R$ given by $\phi(f(t)) = f(0)$. Like all evaluation maps, it is straightforward to see that $\phi$ is indeed a homomorphism. It is also straightforward to see that $\phi$ is surjective; note, for ...


2

Once an ideal in $\Bbb{Z}[x]$ contains both $2$ and $3$, it contains $\operatorname{gcd}(2, 3) = 1$. Then the ideal is all of $\Bbb{Z}[x]$. It's not that you can't have such an ideal; it's just redundant. You can describe it with fewer generators (and you should).


2

Let $x \in V(I \cap J) \Rightarrow f(x) = 0, \forall f \in I \cap J.$ Suppose there exists $h \in I$ such that $h(x) \neq 0.$ The for any $g \in J, g(x) = 0,$ because $hg \in I \cap J.$


2

Assume $x$ is neither in $V(I)$ nor $V(J)$ (i.e. $x \notin V(I)\cup V(J)$). That means that there is an $f\in I$ and a $g\in J$ such that $f(x) \neq 0$ and $g(x) \neq 0$. We have $fg \in I\cap J$, but $fg(x) \neq 0$, and therefore $x \notin V(I\cap J)$. This shows that $x \notin V(I)\cup V(J)$ implies $x \notin V(I\cap J)$, which is the contrapositive of ...


2

Regarding the first equality: Note that if $p\subset m$ then by definition $A_m\subset A_p$ and $A_p\cap A_m=A_m$. Regarding the second equality: Let $s/t\in K$. Since $A$ is an integral domain, there is a prime ideal $p$ such that $t\not\in p$ and so $s/t\in A_p$.


2

Let $f(x) \in F[x]$ be the irreducible polynomial of degree $n$, $g(x)$ be a nonzero polynomial of degree less than $n$. Since $F$ is a field, $F[x]$ is a PID, in fact a Euclidean domain. Then $J = \langle f(x) \rangle$ is a maximal ideal, as it is generated by an irreducible element. We know that $J \subset I$ since $I$ contains $f(x)$, so it follows that ...


2

The ideals in $\mathbb{Q}[x]/I$ correspond to the ideals in $\mathbb{Q}[x]$ that contain $I$. These are principal ideals generated by the divisors of $x^2(x^2+x+1)$. There are $(2+1)\cdot(1+1)$ such divisors. Find them. This is similar to finding the ideals of $\mathbb{Z}/I$ when $I=(p^2 q)$, where $p$ and $q$ are prime numbers.


2

Let $R=\mathbb C[x_1,\dots,x_9]$ and $I=(x_1x_9-x_4x_8,x_4x_6-x_7x_9,x_2x_5-x_3x_9,x_2x_3-x_5x_6).$ Macaulay2 gives $$\sqrt I=(x_4x_8-x_1x_9, x_4x_6-x_7x_9,x_2x_5-x_3x_9,x_2x_3-x_5x_6, x_1x_5x_6^2-x_5x_6x_7x_8).$$


2

If $1-x$ is not invertible, then $(1-x)$, the ideal generated by $1-x$, is strictly contained in $R$ (why?) and therefore it is contained in a maximal ideal (why?).



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