Linked Questions

0
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3answers
5k views

If $F$ is a field show that $F[x]$ is not a field. [duplicate]

I know that $ax=1$ has a solution in $F$ so that every element must be a unit but then I'm not sure how to proceed.
1
vote
2answers
398 views

Let $F$ be a field. Could the ring $F[x]$ be a field? [duplicate]

$F$ is a field, so by definition, $F$ is a commutative ring with unity in which every non-zero element is a unit. Then, $F[x]$ is a set of polynomials in which the coefficients come from $F$, so all ...
0
votes
3answers
105 views

Is $R[x]$ never a field? [duplicate]

I know that if $F[x]$ is a PID then $F$ is a field. Now $F[x]$ being a field implies that $F[x]$ is a PID, so $F$ is a field. Anyway, I tried to prove that $F$ is a field right away and the following ...
1
vote
2answers
394 views

examples such that (i)$F[x]$ is not a field, (ii) $F[x]$ is also a field. [duplicate]

Let $F$ be a field.Then we know that $F[x]$ is a Euclidean Domain.But can someone give me few examples such that (i)$F[x]$ is not a field, (ii) $F[x]$ is also a field. Thanks for your kind help.
0
votes
1answer
120 views

Disproving that $\mathbb R[x]$ is a field [duplicate]

The task is to determine if $\mathbb R[x]$, which represents the set of all polynomials with real coefficients, is a field. My response is that it is not, since I feel that not all of these ...
0
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0answers
57 views

Yes/No: Is $\mathbb{Q}[y]$ is field? [duplicate]

We know that $\mathbb{Q} [x,y]/(x)$ is isomorphic to $\mathbb{Q}[y]$. My question is that Is $\mathbb{Q}[y]$ is field ? Yes/No My attempt: I think yes, because we know $ \mathbb{Q}[x] $ ...
0
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0answers
37 views

Prove that if $R$ is a commutative ring, then $R[x]$ is never a field [duplicate]

Problem: Prove that if $R$ is a commutative ring, then $R[x]$ is never a field. My attempt: We prove that by contradiction. Suppose $R[x]$ is a field, then we have $a_0 + a_1x + \dots +a_nx^n \in R[x]...
9
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2answers
4k views

Why is $Q[\pi]$ not a field?

I am having trouble seeing how to apply the definition of transcendental to see this. Thanks!
10
votes
6answers
6k views

The ideal $I= \langle x,y \rangle\subset k[x,y]$ is not principal [closed]

The ideal $I= \langle x,y \rangle\subset k[x,y]$ is not a principal ideal. I don't know how to consider it. Any suggestions?
14
votes
5answers
26k views

Finding inverse of polynomial in a field

I'm having trouble with the procedure to find an inverse of a polynomial in a field. For example, take: In $\frac{\mathbb{Z}_3[x]}{m(x)}$, where $m(x) = x^3 + 2x +1$, find the inverse of $x^2 + 1$. ...
2
votes
2answers
3k views

Understanding the kernel of the evaluation map

Let $R$ be a non-zero commutative ring. Prove that the ideal $(x)$ of $R[x]$ is prime if and only if $R$ is an integral domain. I'm working on the left-to-right direction right now. I know that $R[x]/...
3
votes
2answers
474 views

gcd of $x$ and $2$ in $\mathbb Z[x]$

In $\mathbb Z[x]$, $x$ and $2$ have gcd $1$. But they cannot be expressed as the linear combination of two polynomials. Then assuming that $1=2f(x)+xg(x)$ we are supposed to arrive at a ...
2
votes
3answers
663 views

Subfields of Rings

I am currently working through an undergraduate class in Galois Theory. I have come across a question that I am unsure about. Can a ring that is not a field, have a subring that satisfies the ...
4
votes
3answers
221 views

Basic Question Concerning Elementary Functions

I have a homework problem which says If $f(x) = \dfrac{x^2 - x}{x - 1}$ and $g(x) = x$. Is it true that $f = g$? What do they mean by saying is it true that $f = g$? Aren't these two functions ...
2
votes
5answers
127 views

Why isn't $\frac{1}{x}$ a polynomial?

Why isn't $\frac{1}{x}$ a polynomial? Does it directly follow from definition? As far as I know, polynomials in $F$ are expressions of the form $\sum_{i=0}^{n} a_ix^i$, where $a_i\in F$ and $x$ is a ...

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