Show that some ideals are prime

Question

I want to show that the ideals $(x,y)$ and $(2,x,y)$ of $\mathbb{Z}[x,y]$ are prime ideals.

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Could you give some hints how we could do that?

Do we have to show that $\mathbb{Z}[x,y]/(x,y)$ and $\mathbb{Z}[x,y]/(2,x,y)$ are integral domains?

How could we do that? Could you give me a hint?

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EDIT:

Do we have that $\phi :\mathbb{Z}[x,y]\rightarrow \mathbb{Z}$ is an homomorphism because of the following? $$$$ Let $f(x,y)\in \mathbb{Z}[x,y]$, then $f(x,y)=\sum_{j=0}^m\sum_{i=0}^{n(j)}\alpha_ix^iy^j$. We have that $$f_1(x,y)+f_2(x,y)=\sum_{j=0}^{m_1}\sum_{i=0}^{n_1(j)}\gamma_ix^iy^j+\sum_{j=0}^{m_2}\sum_{i=0}^{n_2(j)}\beta_ix^iy^j=:\sum_{j=0}^m\sum_{i=0}^{n(j)}\alpha_ix^iy^j=:f(x,y)$$ Then $$\phi (f_1(x,y)+f_2(x,y))=\phi (f(x,y))=f(0,0)=f_1(0,0)+f_2(0,0)=\phi (f_1(x,y))+\phi (f_2(x,y))$$ We also have that $$f_1(x,y)\cdot f_2(x,y)=\left (\sum_{j=0}^{m_1}\sum_{i=0}^{n_1(j)}\gamma_ix^iy^j\right )\cdot \left (\sum_{j=0}^{m_2}\sum_{i=0}^{n_2(j)}\beta_ix^iy^j\right )=:\sum_{j=0}^m\sum_{i=0}^{n(j)}\alpha_ix^iy^j=:f(x,y)$$ Then $$\phi (f_1(x,y)\cdot f_2(x,y))=\phi (f(x,y))=f(0,0)=f_1(0,0)\cdot f_2(0,0)=\phi (f_1(x,y))\cdot \phi (f_2(x,y))$$

Answer 1

Show that $\mathbb{Z}[x,y]/(x,y)$ is isomorphic to $\mathbb{Z}$ using the evaluation map at $0$ and $\mathbb{Z}[x,y]/(2,x,y)$ is isomorphic $\mathbb{Z}/2\mathbb{Z}$ by reducing the coefficents of the polynomial modulo $2$. In both cases consider the kernel of the map and use the first isomorphism theorem.

Answer 2

The first isomorphism is proved by observing that any polynomial in $\mathbb{Z}[x,y]$ can be written as a sum of a constant term, terms involving nontrivial powers of $x$, and terms involving nontrivial powers of $y$. More explicitly, the ideal $(x,y)$ has all polynomials with constant term $0$, i.e., if $f \in (x,y)$ then there exist $p,q \in \mathbb{Z}[x,y]$ such that $f(x,y) = xp(x,y) + yq(x,y)$. Now we note that if $f(x,y)$ is an arbitrary polynomial in $\mathbb{Z}[x,y]$ then we can write it as $$ f(x,y) = a_0 + xp(x,y) + yq(x,y)$$ for some $a_0 \in \mathbb{Z}$ and some $p(x,y), q(x,y) \in \mathbb{Z}[x,y]$. In the quotient mapping $\mathbb{Z}[x,y] \to \mathbb{Z}[x,y]/(x,y)$ the polynomials $xp(x,y) + yq(x,y)$ get sent to zero. Consequently the identification $$ f(x,y) \mod{(x,y)}= a_0 + xp(x,y) + yq(x,y) \mod{(x,y)} \equiv a_0 \mod{(x,y)}$$ induces an isomorphism $\mathbb{Z}[x,y]/(x,y) \cong \mathbb{Z}$ via the mapping $a_0 + (x,y) \mapsto a_0$ for all $a_0 \in \mathbb{Z}$.

For the second isomorphism consider the following isomorphisms for (commutative) rings: if $R$ is a ring and $(a) \subseteq R$ is a principal ideal generated by $a \in R$, then there are isomorphisms $$ (R/(a))[x,y] \cong R[x,y]/(a)$$ so if $p_1, \cdots, p_n$ are any number of polynomials of $R[x,y]$ then $$ \frac{(R/(a))[x,y]}{(p_1, \cdots, p_n)} \cong \frac{R[x,y]}{(a,p_1,\cdots,p_n)}\cong \frac{R[x,y]/(p_1,\cdots,p_n)}{(a)}.$$ Since we already know $\mathbb{Z}[x,y]/(x,y) \cong \mathbb{Z}$, we have $$ \frac{\mathbb{Z}[x,y]}{(2,x,y)} \cong \frac{\mathbb{Z}[x,y]/(x,y)}{(2)} \cong \frac{\mathbb{Z}}{(2)} = \frac{\mathbb{Z}}{2\mathbb{Z}}.$$