# Is $\{p(x) ∈ \Bbb Q[x]\mid p(0) = 3\}$ an ideal of $\Bbb Q[x]$? [closed]

Is $$\{p(x) ∈ \Bbb Q[x] \mid p(0) = 3\}$$ an ideal of $$\Bbb Q[x]$$?

I don't have any idea of how to start this problem. Any help would be great, thank you in advance!

## closed as off-topic by user21820, Lee David Chung Lin, Cornman, Cesareo, heropupMay 6 at 1:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Lee David Chung Lin, Cornman, Cesareo, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.

• "I don't have any idea how to start..." First thing to do: Write down the definition of "ideal". – GEdgar Apr 21 at 21:46
• – Shaun Apr 21 at 21:47
• Is this set stable by addition? By scalar multiplication? – Bernard Apr 21 at 21:50
• Is $0$ in the ideal? Remember, any ideal of $\Bbb{Q}[x]$ must also be a subgroup under addition of $\Bbb{Q}[x]$. – Chickenmancer Apr 21 at 21:55

Ideals got to be closed under addition. But for $$f,g\in$$ your set, $$(f+g)(0)=f(0)+g(0)=3+3=6\neq3,$$ so $$f+g\notin$$ your set. You set is not an ideal.