I understand that uni-variate polynomial rings with coefficients in a field only have principal ideals. For example, $\mathbb{C}[x]$. But how can I tell if an ideal of integer polynomial ring is principal, please? For example, a textbook claims that "the kernel of the map $\mathbb{Z[x]} \rightarrow \mathbb{Z[i]}$ sending $x \mapsto i$ is the principal ideal of $\mathbb{Z}[x]$ generated by $f=x^2+1$" without any justification. How to show this is true, please?
123400
I understand that uni-variate polynomial rings with coefficients in a field only have principal ideals. For example, $\mathbb{C}[x]$. But how can I tell if an ideal of integer polynomial ring is principal, please? For example, a textbook claims that "the kernel of the map $\mathbb{Z[x]} \rightarrow \mathbb{Z[i]}$ sending $x \mapsto i$ is the principal ideal of $\mathbb{Z}[x]$ generated by $f=x^2+1$" without any justification. How to show this is true, please?
66930
Let $\mathbb M$ be one of the following rings: $\mathbb{R}, \mathbb{Q},  \mathbb{F_9}, \mathbb{C} $. Let $I$ be the ideal generated by $x^4+2x-2$. Is the ring $\mathbb M[x]/I$ a field for some of the above choices for $R$?   I have been trying to work my way around some problems involving ideals, but I get really confused when they get mixed with rings and fields. I'm stuck and don't know how to proceed.
1652
Let $R = F[x,y]/(ax^2+by^2-1)$, where $F$ is a field with $\text{char}(F) \neq 2$ and $a, b \neq 0$. Then $R$ is Dedekind.    My attempt:  $R$ is Dedekind if and only if it is Noetherian, integrally closed, and if every prime ideal is maximal, so it suffices to show these conditions.   Noetherian: I wasn't able to explicitly prove this, but I'm pretty sure that every polynomial in $F[x,y]$ of degree greater than 3 can be expressed as the sum of factors of $ax^2+by^2-1$ and polynomials of degree at most 2, so that $R$ consists of polynomials of degree at most 2. If this is true, then the possible non-units are the possible sums and products of $x, y, x^2, y^2$, and since these are the possible generators of nonzero proper ideals, every ideal is finitely generated, so $R$ is noetherian. Integrally closed: My idea was to show the stronger condition that $R$ is a UFD. Since $ax^2+by^2-1$ is irreducible over $F[x,y]$, $R$ is a domain. Since $ax^2 + by^2 = 1 \in R$, $f \in R^x$ if and only if $f \mid ax^2 + by^2$. But when I tried to use this to show that irreducible elements are prime, I got stuck, and I'm no longer sure $R$ is a UFD, but I can't think of a counterexample either. Every nonzero prime ideal is maximal: Let $P$ be a nonzero prime ideal of $R$. Then, there exists a nonzero ideal $P' \subset F[x,y]$ such that $P'/(ax^2+by^2-1) \cong P$, where this projection comes from the first isomorphism theorem and the canonical projection $\phi: P' \rightarrow P$, where $\text{ker}(\phi) = (ax^2+by^2-1)$. Since the preimage of a prime ideal is also a prime ideal, $P'$ is a prime ideal, and since $$R/P \cong \left(F[x,y]/(ax^2+by^2-1) \right) / \left(P' / (ax^2+by^2-1) \right) \cong F[x,y] / P'$$ by the third isomorphism theorem, it suffices to show that $P'$ is maximal. I feel like $P'$ should be maximal in $F[x,y]$ since $P'$ must contain polynomials of degree at most two, and since these polynomials are "small", the ideal containing them should be "large", but I'm stuck trying to turn this intuition into something formal.
9489
Let $R$ be an integrally closed domain. Let $P$ be a prime ideal of $R[x]$ with the property that $P\cap R=\left\{ 0\right\} $ and $P$ contains a monic polynomial. Prove that $P$ is principal.   Help me a hint to prove.  Thank for any insight.
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Let $K$ be any field. Is it true that:   If $f_1,\dots, f_k \in K[x_1,\dots,x_n]$ are polynomials of degree $1$, then $I=(f_1,\dots, f_k)$ is all $K[x_1,\dots,x_n]$, or, if it is a proper ideal, then is it prime? In particular, is it true that $K[x_1,\dots,x_n]/I$ is isomorphic to a polynomial ring (in the case that $I$ is proper and hence prime by the statement above)?   If they are true, how can i prove these statements in an elementary but rigorous way?  P.S.: Birth of my question: I found on my book that "affine varieties given by linear equations are irreducibile", so I translated (I don't know if I have made it right) this geometric statement into the analogous algebraic version.
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Suppose that $k$ is an algebraically closed field. Then what do the prime ideals in the polynomial ring $k[x,y]$ look like?   As far as I know, the maximal ideals of $k[x,y]$ are of the form $(x-a,y-b)$ where $a,b\in k$. What can we say about the prime ideals? Are there similar results? And what about $k[x,y,z], k[x,y,z,w]$ and so on. Would someone be kind enough to give me some hints or referrence on this topic? Thank you very much!
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Let $R = (\mathbb{Z}/2)[x] $ and $I$ the ideal generated by the polynomial $x^{17} - 1$. Is there a non-zero element of $R/I$ s.t. the square of that element is zero in $R/I$? I know a quotient ring $R/P$ is an integral domain iff $P$ is a prime ideal, and since $R$ is a PID, the prime elements are the irreducible, but $x^{17} - 1 $ is reducible in $R$, hence $R/I$ is not an integral domain. So there are nonzero $a,b \in R/I$ s.t. $ab = 0 $, but $b$ might not be $a$, and that's where I got stuck. I've also tried finding $f \in R$ s.t. $ f^2 = g \cdot (x^{17} - 1)$, but don't know how to attack this methodically.
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Apologies in advance for the naivety of this question.  Let $R$ be a commutative (resp. non-commutative) ring, $S \subset R$, and let $R&#39; = R[x_s (s \in S)]$ be the polynomial ring obtained by adding a formal variable $x_s$ for each $s \in S$.  Let $I$ be the ideal (resp. 2-sided ideal) generated by {$sx_s-1\ |\ s \in S$}.    Does the quotient $R&#39;/I$ yield the localization $S^{-1}R$ in general?  If not, what if we assume $R$ is commutative and/or $S$ is finite?    EDIT: Thanks to Mariano and Arturo for the quick responses!  Both of their responses say that my construction works in the non-commutative case when $R&#39;$ is replaced with a "non-commutative polynomial ring."  But the Wikipedia article on the localization of a ring states that in the non-commutative case, the localization doesn't exist for all prospective sets of units $S$.  What gives?
81375
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 $\mathbb{F}[x]$. I know that the proof of this statement is meant to be very short, but I don't know how to go about it.
91994
I want to prove that if in a UFD every maximal ideal is principal then it is a PID.    My line of attack is: If it is a field i.e. it has no non-zero proper ideal, then we are done. Otherwise consider a non-zero proper ideal. Then, by Zorn's lemma, it is contained in some maximal ideal, hence contained in a principal ideal which is not the whole ring. Now we know that if in a UFD every proper ideal is contained a proper principal ideal, then it is a PID, hence we are done.      Is there any other solution without using Zorn's lemma?
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Just got a question about ideals of $F[x]$ being principal with $F$ being a field.  The proof basically starts assuming $g(x) \in N \subset F[x]$ where $N$ is an ideal and $g(x)$ is a minimal degree element. The rest is just deducing the what the division algorithm says.  But my question is that shouldn't we let $g(x)$ be arbitrary? Why are we choosing the minimal element?
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Two questions listed in Artin under a section called "Gauss's lemma":   Are the kernels of $f(x) \mapsto f(1 + \sqrt{2})$ and $f(x) \mapsto f(\frac12 + \sqrt2)$ (for $f \in \mathbb{Z}[x]$) principal ideals? For $\alpha \in \mathbb{C}$, prove the kernel of $f(x) \mapsto f(\alpha)$ is principal.   What's the general approach for solving such problems? And what (if anything) does it have to do with Gauss's lemma (products of primitive polynomials are primitive)?  For 1, clearly $(x-1)^2 - 2 = x^2 - 2x - 1$ is in the kernel of the first map, and this polynomial is primitive, but I'm unsure how this helps in figuring out if the kernel is principal?
108133
Let R be a commutative ring with identity. If every ideal generated by two elements of R is principal, then can we conclude that R is a PID? Also, if every finitely generated ideal of R is principal, can we conclude that R is a PID?
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I want to prove the  over a field are prime ideals. To be concrete:      For simplicity, let $I=(x_{11}x_{22}-x_{12}x_{21},x_{11}x_{23}-x_{13}x_{21},x_{12}x_{23}-x_{13}x_{22})$ be an ideal of the polynomial ring $k[x_{11},\ldots,x_{23}]$. I have no idea how to prove that $I$ is a radical ideal (i.e. $I=\sqrt{I}$). Could anyone give some hints?   Generally, let $K$ be an algebraically closed field, then $\{A\mid\mathrm{Rank}(A)\leq r\}\subseteq K^{m\times n}$ is an irreducible algebraic set (I first saw this result from ). And I tried to prove this by myself, then I have proved it (when I see the "Segre embedding").      But I have no idea how to show that the "determinantal ideals" are radical ideals (I hope this is true). BTW, is the statement that the determinantal ideals over a field are prime ideals true ?   Thanks.
18120
As the question title suggests, how do I see that every left ideal of a square matrix ring over a field is principal?
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Let $R$ be the ring obtained by taking the quotient of $\mathbb{Z_6}[X]$ by principal ideal $(2X+4)$. Then  1) $R$ has infinite elements  2) $R$ is field  3) $5$ is unit in $R$  4) $4$ is unit in $R$.  My Attempt:  \begin{equation*} \mathbb{Z_6}[X]/(2X+4) = \mathbb{Z_6}[X]/(2(X+2)) = \mathbb{Z_3}[X],  \end{equation*} as $X+2$ has root in $\mathbb{Z_6}$. According to me 1) and 3) are correct but I'm not sure about my answer. Help me. I am not more familiar with polynomial and quotient ring. I searched this problem here but couldn't get this here. If this problem is already asked here then how can I get this?  Thank you in advance.
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The following is a problem from Dummit &amp; Foote.   Let $R$ be an integral domain. Prove that if the following two conditions are true, then $R$ is a principal ideal domain.    Any two non-zero elements $a$ and $b$ in $R$ have a greatest common divisor that can be written as $d=ra+sb$, $r,s \in R$ If $a_1,a_2,\dots$ are non-zero elements of $R$ such that $a_{i+1}\mid a_i$ for all $i$, then there exists a positive integer $N$ such that $a_n$ is a unit times $a_N$ for all $n \ge N$.   I am a bit confused really. Isn't the first condition sufficient? For e.g. an ideal generated by two elements $a,b$ would be a principal ideal, as any two elements have a gcd. This can then be extended to ideals generated by an arbitrary number of elements by induction.   Obviously, there is something wrong in what I am doing. What is the second condition for? I can only think of defining principal ideals generated by the elements in a chain, and then finding a maximal ideal with respect to divisibility. I have no idea.