# Non zerodivisors in ideals of polynomial rings [duplicate]

Let $$f$$ be a polynomial of the ring $$R[x_1, \ldots, x_n]$$, with $$R$$ any ring, and let $$\mathrm{cont}(f)$$ be the ideal generated by the coefficients of $$f$$. Why if $$\mathrm{cont}(f)$$ contains a non-zerodivisor of $$R$$, then $$f$$ is a non-zerodivisor of $$R[x_1, \ldots, x_n]$$?
## marked as duplicate by user26857 commutative-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(); } ); }); }); May 22 at 21:36
Hint: When you write that non zero divisor $$a$$ as a linear comnbination, this gives you a recipe for a polynomial $$g$$ such that one coefficient of $$f\cdot g$$ is $$a$$. Now assume $$h\cdot f=0$$. Then also $$hfg=0$$.