2
votes
2answers
40 views

For which values of $a\in\mathbb ℤ/3\mathbb ℤ$ is the quotient $\mathbb ℤ/3\mathbb ℤ[x]/(x^3+x^2+ax+1)$ a field?

I'm trying to solve the following problem: Determine for which values of $a\in\mathbb{Z}/3\mathbb{Z}$ the quotient $Q_a=(\mathbb{Z}/3\mathbb{Z})[x]/(x^3+x^2+ax+1)$ is a field. I see two options: ...
8
votes
3answers
174 views

Polynomial irreducible - maximal ideal

I have a couple of ideals which I wonder if I correctly classify as maximal/prime ideal. $I_1 = \langle 2x^2 + 9x -3\rangle$, $I_2 = \langle x - 1\rangle$ $\mathbf 1)$ Is $I_1$ a maximal ideal in ...
0
votes
2answers
34 views

Let $R$ be a commutative ring with identity. Let $M$ be an ideal such that every element of $R-M$ is a unit. Then $R/M$ is a field.

Let $R$ be a commutative ring with identity. Let $M$ be an ideal such that every element of $R$ not in $M$ is a unit. Then $R/M$ is a field. I am solving this question of NBHM 2011. To solve this is ...
0
votes
3answers
55 views

Checking if something is a unit

Check if $\mathbb{Z}_5/x^2 + 3x + 1$ is a field. Is $(x+2)$ a unit, if so calculate its inverse. I would say that this quotient ring is not a field, because $<x^2 + 3x + 1>$ is not a maximal ...
1
vote
3answers
85 views

The kernel of homomorphism of a local ring into a field is its maximal ideal?

I have a question about the proof of Theorem 3.2. of Algebra by Serge Lang. In the theorem $A$ is a subring of a field $K$ and $\phi:A \rightarrow L$ is a homomorphism of $A$ into an algebraically ...
2
votes
2answers
597 views

Prove that every nonzero prime ideal is maximal in $\mathbb{Z}[\sqrt{d}]$

$d \in \mathbb{Z}$ is a square-free integer ($d \ne 1$, and $d$ has no factors of the form $c^2$ except $c = \pm 1$), and let $R=\mathbb{Z}[\sqrt{d}]= \{ a+b\sqrt{d} \mid a,b \in \mathbb{Z} \}$. ...
2
votes
3answers
99 views

Ideals in a real/complex number field?

Considering a real or complex number field (with traditional addition and multiplication) I see no ideals besides $\mathbb{R}$ and $\{ 0\}$ or $\mathbb{C}$ and $\{ 0 + 0i\}$. Quick web search gave no ...
0
votes
1answer
111 views

subring of rational numbers and its ideal

Let $p$ be a prime number. For any $p$ the subring $\mathbb{Q}_p$ of of the field of rational numbers is defined: $\mathbb{Q}_p=\{\frac{a}{b}|a,b\mbox{ are integers, $p$ does not divide $b$}\}$ Let ...
0
votes
1answer
66 views

Why is (gcd(f,g)) = (f,g)?

f and g are polynomials of F[X]. I can't see why (f,g) = (gcd(f,g)) ? (the ideal that f and g are the generators, and the ideal that the gcd is the generator). gcd(f,g) = a*f+b*g , for specific a ...
2
votes
2answers
185 views

Find a non-principal ideal (if one exists) in $\mathbb Z[x]$ and $\mathbb Q[x,y]$

I know that $\mathbb Z$ is not a field so this doesn't rule out non-principal ideals. I don't know how to find them though besides with guessing, which could take forever. As for $\mathbb Q[x,y]$ I ...
3
votes
2answers
178 views

Show that this set of polynomials is ideal in F[x]

In $\mathbb{F}[x]$, where $\mathbb{F}$ is a field, let $J$ be the set of elements of polynomials that have coefficients that add to zero (so $a_0 + a_1 + ... + a_n = 0$). Show that $J$ is an ideal of ...
1
vote
1answer
99 views

If a field contains the complex field, then it is $\mathbb{C}$

This question is originated from a book by Gaal, (Linear Analysis and Representation Theory). Theorem 7 from section 6, chapter 1 reads as follows, and I quote: "Theorem 7: Let $A$ be a complex, ...
4
votes
2answers
103 views

What is the field of definition of an invariant ideal?

Let $K/k$ be a finitely generated field extension, such that $k=K^G$ for some (possibly infinite) set $G$ of automorphisms of $K$. Now, consider the extension of polynomial rings $$ ...
5
votes
3answers
509 views

Help with proof that $\mathbb Z[i]/\langle 1 - i \rangle$ is a field.

I have been having a lot of trouble teaching myself rings, so much so that even "simple" proofs are really difficult for me. I think I am finally starting to get it, but just to be sure could some one ...
4
votes
2answers
97 views

Where do I use the fact that $F$ is algebraically closed in this proof?

I have to do the following. Let $F$ be an algebraically closed field. $I\in F[X_1,...,X_n]$ an ideal. Denote by $S(I)$ the subset in $F^n$ consisting of all $n$-tuples $(a_1,...,a_n)\in F^n$ such that ...
4
votes
4answers
121 views

Why do $f$ and $f'$ generate all of $K[X]$?

I have been studying Marcus' Number Fields. I am stuck on a remark in Appendix 2, page 258. He says: A monic irreducible polynomial $f$ of degree n over $K$ (a subfield of $\mathbb{C}$) splits into n ...
3
votes
1answer
94 views

Equivalence of Valuations - Trouble Understanding Proof

I want to complete the proof of the following theorem. Here is what I have got so far: Theorem Every non-euclidean valuation $v$ on a number field $K$ is equivalent to $v_{\mathfrak p}$ for some ...
4
votes
3answers
413 views

fields are characterized by the property of having exactly 2 ideals [duplicate]

Possible Duplicate: A ring is a field iff the only ideals are $(0)$ and $(1)$ Michael Artin's Algebra in the introduction of maximal fields, there was a sentence stated that fields are ...
4
votes
3answers
82 views

For an ideal $I$ of $K[x]$, $K[x]/I$ is finitely generated iff $I$ is nonnull

In a book on rational series, a blunt statement is made to the effect that: For $K$ a field, $I$ an ideal of $K[x]$, $K[x]/I$ is finitely generated iff $I$ is nonnull. The statement elaborates ...
4
votes
1answer
522 views

Are distinct prime ideals in a ring always coprime? If not, then when are they?

Essentially as the title suggests - in some commutative ring $K$ (with 0,1), if we have 2 distinct proper prime ideals $\mathfrak{p}_1 \neq \mathfrak{p}_2$, is it necessarily the case (or if not, when ...
1
vote
0answers
142 views

Prime ideal splitting in field extension and its normal closure

The question is: Let L / K be a finite (not necessarily Galois) extension of algebraic number fields and N / K the normal closure of L / K. Show that a prime ideal p of K is totally split in L if and ...