# Does $\forall n,d\!\in\!\mathbb{N}$ $\forall$ field $\mathbb{F}$ exist an irreducible $f\!\in\!\mathbb{F}[x_1,\ldots,x_n]$ of degree $d$?

how can one show (hopefully in an elementary manner) that there exist irreducible polynomials of arbitrary degree and number of variables over arbitrary field?

thank you

P.S. induction?

EDIT: ehm, sorry, I meant for $n\geq2$, i.e. in at least two variables.

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If the field is algebraically closed then only polynomials of degree $1$ are irreducible. – Asaf Karagila Mar 2 '11 at 18:04
Almost a duplicate: math.stackexchange.com/questions/24425/… – Qiaochu Yuan Mar 2 '11 at 20:02

This is false. Any irreducible polynomial in one variable over an algebraically closed field (such as $\mathbb{C}$) has degree 1.
For $n \ge 2$, the answer is yes: just take the irreducible polynomial $y-x^d$.
Proof: If $x_1\!-\!x_2^d\!=\!f(\bf x)g(\bf x)\!=\!(\sum_{\bf a}\alpha_{\bf a}\bf x^{\bf a})(\sum_{\bf b}\beta_{\bf b}\bf x^{\bf b})\!=\!\sum_{\bf a,\bf b}\alpha_{\bf a}\beta_{\bf b}\bf x^{\bf a+\bf b}$, then $fg$ would include a monomial with $x_1x_2$, a contradiction. $\blacksquare$ – Leon Nov 6 '11 at 9:30