Linked Questions

0 votes
3 answers
10k 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.
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0 votes
3 answers
955 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 ...
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  • 1,493
2 votes
2 answers
2k 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 ...
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1 vote
2 answers
622 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.
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  • 49
0 votes
1 answer
153 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 ...
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0 votes
0 answers
196 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]...
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  • 973
0 votes
0 answers
82 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] $ ...
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1 vote
1 answer
30 views

Let $\Bbb R$ be a field of all real numbers. Prove that $x+1$ is not a unit in $\Bbb R[x]$. [duplicate]

Let $\Bbb R$ be a field of all real numbers. Prove that $x+1$ is not a unit in $\Bbb R[x]$. Attempts: Suppose that $x+1$ is a unit in $\Bbb R[x]$. Then, there is exists $g(x) \in \Bbb R[x]$ such that $...
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  • 2,178
0 votes
1 answer
49 views

What techniques can be used to show that a multiplicative inverse doesn't exist? [duplicate]

Example: To show that the polynomials of a finite field is itself not a field, I need to show that a multiplicative inverse does not exist. What are the general techniques that can be used to show ...
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152 votes
2 answers
35k views

Example of infinite field of characteristic $p\neq 0$

Can you give me an example of infinite field of characteristic $p\neq0$? Thanks.
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  • 5,223
18 votes
6 answers
40k 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 $x^2 + 1$. ...
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10 votes
6 answers
9k 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?
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  • 187
10 votes
2 answers
6k views

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

I am having trouble seeing how to apply the definition of transcendental to see this. Thanks!
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  • 4,081
14 votes
2 answers
12k views

Proof of the polynomial division algorithm

The theorem which I am referring to states: for any $f, g$ there exist $q, r$ such that $f(x)=g(x)q(x)+r(x)$ with the degree of $r$ less than the degree of $g$ if $g$ is monic. The book I am using ...
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  • 5,847
3 votes
2 answers
4k 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]/...
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  • 3,613

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