Proving a subset is an ideal in Q[x,y]

Question

I am tasked to prove that the subset:

$K:= \{P\in \mathbb{Q}[x,y]|P(1,0)=P(0,3)=0\}$

is an ideal in $\mathbb{Q}[x,y]$.

Now I understand the steps to show a subset is an ideal but I'm not sure how to apply them or how to interpret K.

Firstly, is K representing the ideal $\langle x,3y \rangle$?

Am I correct in thinking I need to test if $P +Q \in K$ for $P,Q \in k$ and if $PQ \in K$ for $P \in K$ and $Q \in \mathbb{Q}[x,y]$? I'm very confused as to find out how to complete these calculations.

Any advise is greatly appreciated, I want to complete these equations myself but the lecture notes I am following are really not very helpful.

Thanks

Answer 1

Yes, those are the steps to do in order to show $K$ is an ideal.

If $P,Q\in K$ and $R=P+Q$, then $$ R(1,0)=P(1,0)+Q(1,0)=0, \qquad R(0,3)=P(0,3)+Q(0,3)=0 $$ Similarly if $P\in K$ and $Q\in\mathbb{Q}[x,y]$.

Note that $x\notin K$ and $3y\notin K$. What you should consider are $x-1$ and $y-3$, but they aren't in $K$, so some adjustments are to be made. Hint: use the fact that, for any polynomial $P(x,y)\in\mathbb{Q}[x,y]$ you can write \begin{align} P(x,y)&=(x-1)Q_1(x,y)+R_1(y)\\ &=(y-3)Q_2(x,y)+R_2(x) \end{align}

Answer 2

Considering the ring $R\triangleq\mathbb Q [x,y]$ is commutative, in order to show $K\triangleleft R$ one must show the following:
1. $K\neq\emptyset$
2. $\forall a,b\in K \implies a+b\in K$
3. $\forall a\in K \implies (-a)\in K$
4. $\forall a\in K,r\in R \implies ar\in K$
Showing that $K\neq \emptyset$ is easy.
Now, consider $a,b\in K$. This means $a=P(x,y) \in \mathbb Q[x,y]$ such that $P(1,0)=P(0,3)=0$. Same can be written about $b=S(x,y) \in \mathbb Q[x,y]$ such that $S(1,0)=S(0,3)=0$. By definition $a+b =P(x,y)+S(x,y)$ implying that $a+b\in K$ because $P(1,0)+S(1,0)=P(0,3)+S(0,3)=0+0=0$.
In the same way you can prove the third condition.
The last one is usually the condition that defines the ideal. Consider $r=P(x,y)\in \mathbb Q[x,y]$ being any polynomial and let $a=S(x,y) \in K $ therefore $S(1,0)=S(0,3)=0$. Also by definition $ar=S(x,y)\cdot P(x,y)$ and one can argue that $S(1,0)\cdot P(1,0) = 0 \cdot P(1,0) = 0 = 0 \cdot P(0,3) = S(0,3) \cdot P(0,3)$. From this you can conclude that $ar\in K$ and therefore $K\triangleleft R$.