# $(x_1-a_1, x_2-a_2)$ is a maximal ideal of $K[x_1,x_2]$ [duplicate]

$K$ is field. $a_1$,$a_2$ elements of $K$. Show that $(x_1-a_1,x_2-a_2)$ is a maximal ideal of $K[x_1,x_2]$.

$K[x_1,x_2]$ is UFD so if $K[x_1,x_2]/(x_1-a_1,x_2-a_2)$ is field then $(x_1-a_1,x_2-a_2)$ is maximal ideal.

If I can show that $K[x_1,x_2]/(x_1-a_1,x_2-a_2)$ isomorphic to $K$, we can verify that $(x_1-a_1,x_2-a_2)$ maximal ideal of $K[x_1,x_2]$.

## marked as duplicate by user26857, Ken Duna, Arnaud D., C. Falcon, Namaste abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 2 '17 at 0:22
• Hint: $$\phi:K[x_1,x_2]/\langle x_1-a,x_2-b\rangle \to K\;,\;\phi(f(x,y):=f(a,b)\;\ldots$$ – DonAntonio Nov 3 '13 at 22:06
• $K[x,y]/(x-a_1, y-a_2)$ is just $K[x,y]$ modulo the relations $x=a_1$ and $y=a_2$. Of course it is isomorphic to $K$. You can fill in the details. – TBrendle Nov 3 '13 at 22:08
Consider the following map (slightly modified the suggestion of @DonAntonio in comments): $$\phi:K[x,y]\to K,\quad \phi(f(x,y)):=f(a,b)$$ Then show that its kernel is just the ideal $(x-a,\,y-b)$ and its image contains $1$, so it is surjective, then use the first isomorphism theorem.