# Tag Info

4

I recommend that you don't use the phrase "ideal in"; use the phrase "ideal of". Why? Because it is meaningless to say that some set $X$ "is an ideal" without defining a context, i.e., what ring is it an ideal of. The set of even integers $$I=\{\ldots,-4,-2,0,2,4,\ldots\}$$ is an ideal of the ring $\mathbb{Z}$, but it is not an ideal of the ring ...

4

Just keep going with using $X-a$ and $Y-b$ to translate stuff. $(X-a)(Y-b)+b(X-a)+a(Y-b)=XY-aY-bX+1+bX-1+aY-1=XY-1$

3

If $a\in N$ and $r\in R$, then $(ar)^n=a^nr^n=0$ by commutativity of $R$, so $ar\in N$. Next suppose that $a,b\in N$. Then there are $n,m$ for which $a^n=b^m=0$. Since the ring is commutative, we may apply the Binomial theorem to obtain$$(a+b)^{n+m}=\sum_{i=0}^{n+m}\binom{n+m}{i}a^ib^{n+m-i}.$$ Now observe that either $i\geq n$ or $n+m-i\geq m$ (if neither ...

3

Yes! $F[X]/(f(x))$ ($F$ field ) is a field if and only if $f(x)$ is irreducible in $\mathbf{F}[X]$. Then what you've done is correct!

3

Hint: If $I_k\not\subset P$ for all $k\in\{1,\ldots,n\}$, then there is at least one $r_k\in I_k\setminus P$ for all $k\in\{1,\ldots,n\}$. What can you say about the product $r_1\cdots r_n$?

3

As you noticed, a polynomial $p$ is in $I$ if and only if its coefficients $a_0$ and $a_1$ of $1$ and of $X$ vanish. This happens exactly when $p$ is divisible by $X^2$.

3

To show that $B/A$ is an ideal in $R/A$, or $B/A \triangleleft R/A$ as our OP Bozidar Vulicevic writes it, is really not at all difficult; we merely need to follow through the details inherent in the definitions. A modular approach is indeed possible, as our colleague Bernard has shown, but here I take a somewhat more first-principles, ...

3

Let $x\in R$ a zero-divisor. We want to show that $x$ belongs to a minimal prime ideal. Let $y\in R$, $y\ne 0$ such that $xy=0$. If $x$ does not belong to any minimal prime, then from $xy=0\in\mathfrak p$ for $\mathfrak p$ a minimal prime, we get $y\in\mathfrak p$, so $y$ belongs to any minimal prime hence it is nilpotent and thus $y=0$, a contradiction. ...

3

The answer is yes. Suppose $I\nsubseteq J$, and let $x\in I\setminus J$. Consider the ideal $(J:x)=\{a\in R:ax\in J\}$. Notice that $(J:x)\subsetneq R$, and let $P$ be a prime ideal such that $(J:x)\subseteq P$. As you observed, from $I_P=J_P$ it follows that there is $u\in R\setminus P$ such that $ux\in J$, that is, $s\in(J:x)$, a contradiction. The other ...

3

Start by supposing that $a\mathbb Z[x] + x\mathbb Z[x] = (f)$ for some $f\in \mathbb Z[x]$. On the one hand, we must have $a = gf$ for some $g\in \mathbb Z[x]$, and since $$\deg(fg) = \deg(f)+\deg(g) \ge deg (f),$$ we must have $\deg(f) = 0$ or $f=0$. When can this happen?

3

Try a proof by contradiction. Suppose there exists $f\in\mathbb{Z}[x]$ with $a\mathbb{Z}[x]+x\mathbb{Z}[x]=(f)$ Then we must have $a\in(f)$. What does this tell us about $f$?

2

Let $Z(R)$ denote the set of zero divisors of $R$, and let $\{P_i\}$ denote the set of minimal primes. Suppose that $x\in R$ is a zero divisor. Choose some nonzero $y\in R$ such that $xy = 0$. Since $R$ is reduced, the intersection of all primes is just $\{ 0\}$. However, every prime contains a minimal prime so $\cap_i P_i = 0$ as well. Hence $y\not\in P_i$ ...

2

The definition of an ideal P being prime is, given any ideals A and B such that AB $\subset P$, either A $\subset P$ or B $\subset P$. Thus, if $\bigcap_1^{n+1} I_i \subset P$, then $(\bigcap_1^n I_i)I_{n+1} \subset \bigcap_1^{n+1} I_i \subset P$ so that either $\bigcap_1^n I_i \subset P$ or $I_{n+1} \subset P$. If the former, use induction hypothesis. If ...

2

So the Minkowski bound is indeed generated by the prime ideals dividing $(2),(3),(5),(7)$. Then the minimal polynomial of $\alpha = \sqrt{-41}$ is $x^2 + 41$ which reduces modulo $p$ as \begin{align} x^2 + 41 &\cong (x+1)^2 \pmod{2} \\ x^2 + 41 &\cong (x+1)(x-1) \pmod{3} \\ x^2 + 41 &\cong (x-2)(x-3) \pmod{5} \\ x^2 + 41 &\cong (x+1)(x-1) ...

2

The following argument is correct but is based on the false premises that $[P_2]=[P_3]$ and $5$ is inert and is hence invalid: $P_2$ is ramified (as $-41 \equiv 3 \pmod 4$), so it has order 1 or 2 in the class group. $P_2=(2,1+\sqrt{-41})$ which is non-principal so it has order 2. Let $K=\mathbb Q(\sqrt{-41})$ and let $[P_i]$ denote the class containing ...

2

More concisely, for part a), observe that $\mathbb{Z}/2\mathbb{Z}[x]/(x^2+1)$ is not an integral domain, because $(\overline{x+1})(\overline{x+1})=\overline{x^2+2x+1}=\overline{x^2+1}=0$ in the quotient ring. And $\overline{x+1}$ is non-zero element. If it is not an integral domain, it surely cannot be a field.

2

Consider the set $S = \{X+\alpha Y \mid \alpha\in k\} \subset (X,Y)$. If an ideal $I$ contains two distinct elements of $S$, $X+\alpha Y$ and $X+\beta Y$, then it contains $\frac{1}{\beta-\alpha}((X-\alpha Y)-(X-\beta Y)) = Y$, and thus contains $X$ as well, so $(X,Y) \subset I$. Since an ideal properly contained in $(X,Y)$ contains at most one element of ...

2

Let's note $I$ is the intersection of the primes $\langle x,y \rangle$ and $\langle z\rangle$ of $\mathbb{R}[x,y,z]$. Now is easy to conclude using the following facts: A prime ideal $P$ is radical (The quotient is a domain and so is reduced) The radical of finite intersection of ideals is the intersection of the radicals (A containment is easy, for the ...

2

There are generally two cases to consider. If you define everything in a very clever way, they are actually just one case, but I think it is helpful to distinguish here. Case 1: $R$ contains zerodivisors. In this case, let $x\in R$ be a zerodivisor, i.e. $x\ne 0$ and there is $0\ne y\in R$ with $xy=0$. Since you require that the ideal generated by both ...

2

Let P be the prime ideal $P \subseteq \Bbb{Z}[x]$ and let it be not a principal ideal, so $P$ contains two non zero elements $f,g$ with no common factor in $\Bbb{Z}[x]$. We can prove this easy lemma- Lemma- $f,g$ has no common factor in $\Bbb{Q}[x]$. Sketch- If they have a common factor, say $h\in \Bbb{Q}[x]$ with degree $\ge1$, then writing $h=ah_0$ ...

2

If it were principal, generated by a polynomial $f$, this polynomial would have to divide $2$, hence be a constant (because over an integral domain, $\,\deg fg=\deg f+\deg g$) which is either a unit, or associated with $2$. However, it can't be a unit, since $(2,x)$ is a maximal ideal in $\mathbf Z[x]$. Indeed, $$\mathbf Z[x]/(2,x)\simeq\mathbf Z/2\mathbf ... 2 Think about residue fields. By the Nullstellensatz, the residue field L=F[x_1,\dots,x_n]/M is a finite extension of F, and the residue field of K[x_1,\dots,x_n] at any maximal ideal is K. From this, you can show that there is a bijection between maximal ideals extending M and embeddings of L into K (extending the given embedding F\to K). 2 Recall I prime ideal in ring R \iff R/I is a integral domain. \\$$\frac{\mathbb{Z}[\sqrt{10}]}{\langle 2,\sqrt{10} \rangle} \cong \mathbb{Z}/\langle 2 \rangle$$2 Let f\in R[X_1,\dots,X_n] such that c(f), the ideal generated by the coefficients of f equals R. Let I\subset R an ideal. We want to prove that f is a non-zero divisor on R[X_1,\dots,X_n]/IR[X_1,\dots,X_n]. First notice that R[X_1,\dots,X_n]/IR[X_1,\dots,X_n]\simeq (R/I)[X_1,\dots,X_n]. Then suppose the contrary: there is a non-zero ... 1 What you're using when you define the conjugate \bar{z} = a - b\sqrt{10} and then multiply by it is that the norm |a+b\sqrt{10}| = a^2 - b^2 10 is well-defined and multiplicative. Your proof looks good. The other solutions are also good but if you're practicing this stuff or doing it as homework then I like your way of doing it. 1 If you know something about the norm of an ideal, the following would be a straightforward argument. Check that$$\langle 2, \sqrt{10} \rangle^2 = \langle 2\rangle$$so the norm of the ideal \langle 2, \sqrt{10} \rangle is 2. In particular, \Bbb{Z}[\sqrt{10}]/\langle 2, \sqrt{10} \rangle = \Bbb{F}_2, and this implies that it is a prime ideal. 1 Let R be a commutative ring and consider three R-modules A \subseteq B \subseteq C. Then B/A \subseteq C/A and we have the following Proposition. (C/A) \big/ (B/A) \simeq C/B. Proof. Let \varphi \colon C/A \to C/B be the map defined (on the cosets) by \varphi(x + A) = x + B. Then \varphi is a well defined module homomorphism with ... 1 You know that r(x)\in\mathbb Q[x] and Lr(x)\in P for some L\in\mathbb Z, L\ne0. Now multiply Lr(x) by a non-zero integer a such that ar(x)\in\mathbb Z[x], and find Lar(X)\in P. Then ar(x)\in P and \deg ar(x)=\deg r(x)<\deg m(x). You have to be careful when start with m\in P, \deg m\ge1 of minimal degree to suppose that m is ... 1 Usually this is proved for prime ideals, but it holds for any ideal. If I is generated by n elements, and P is a minimal prime over I, then \operatorname{ht}P\le n. But \operatorname{ht}I=\min_{I\subseteq P}\operatorname{ht}P, so the conclusion follows. 1 @CPM Thank you for your link. I just have obtained an elementary proof. In a principal ideal domain, the relation (a)\cap ((b)+(c)) = (a)\cap (b) + (a)\cap (c)  holds. I'll show the \subset side. Let (b)+(c) = (d) and (a)\cap (d) = (e) and we can write b=b'd, c=c'd with sb'+tc'=1 and  e=fa = gd Then$$ e = e \cdot 1 = e(sb'+tc') = ...

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