Let $K$ be any field. Is it true that:

  • If $f_1,\dots, f_k \in K[x_1,\dots,x_n]$ are polynomials of degree $1$, then $I=(f_1,\dots, f_k)$ is all $K[x_1,\dots,x_n]$, or, if it is a proper ideal, then is it prime?

  • In particular, is it true that $K[x_1,\dots,x_n]/I$ is isomorphic to a polynomial ring (in the case that $I$ is proper and hence prime by the statement above)?

If they are true, how can i prove these statements in an elementary but rigorous way?

P.S.: Birth of my question: I found on my book that "affine varieties given by linear equations are irreducibile", so I translated (I don't know if I have made it right) this geometric statement into the analogous algebraic version.

  • $\begingroup$ Maybe you know that an ideal is prime if and only if the quotient by this ideal is a domain $\endgroup$
    – Dario
    Oct 30, 2017 at 16:25
  • $\begingroup$ Yes, in fact my strategy was to prove that $K[x_1,...,x_n]/I$ is isomorphic to some polynomial ring over $K$, so thaht this last one is a domain, hence $I$ is prime. The question was: How can i take a polynomial ring to be isomorphic to $K[x_1,...,x_n]/I$? $\endgroup$
    – Minato
    Oct 30, 2017 at 16:35
  • 1
    $\begingroup$ If you know Groebner basis theory - then a Groebner basis of $I$ ends up having all linear terms, with at most one generator having leading monomial $x_i$ for each $i$. From there, it's easy to see the quotient is isomorphic to $K[x_j]$ where $j$ ranges over values such that $x_j$ is not a leading monomial of any generator. (The Groebner basis calculation ends up just being equivalent to a Gaussian elimination on the coefficient matrix.) $\endgroup$ Oct 30, 2017 at 21:45

1 Answer 1


Yes, this is true. Here's probably the easiest way to prove it. Let $V$ be the vector space of linear polynomials in $K[x_1,\dots,x_n]$. Note that $\{1,x_1,\dots,x_n\}$ is a basis for $V$. Let $T:V\to V$ be any invertible linear map such that $T(1)=1$. Note that $T$ then induces an automorphism $T^*$ of the ring $K[x_1,\dots,x_n]$ by $T^*f(x_1,\dots,x_n)=f(Tx_1,\dots,Tx_n)$. This is an automorphism because it has inverse $f(x_1,\dots,x_n)\mapsto f(T^{-1}x_1,\dots,T^{-1}x_n)$. (To prove these operations really are inverse, note that when restricted to $V$, they are just $T$ and $T^{-1}$, since $T(1)=1$ and $T^{-1}(1)=1$. So their compositions when restricted to $V$ are the identity, which means their compositions on all of $K[x_1,\dots,x_n]$ are the identity since it is generated by $V$ as a ring. Geometrically, you can think of $T$ as an affine linear automorphism $K^n\to K^n$ and then $T^*$ is the induced automorphism of $K[x_1,\dots,x_n]$ when you think of polynomials as functions on $K^n$.)

Now let $I\subseteq K[x_1\dots,x_n]$ be an ideal generated by linear polynomials. Let $W=I\cap V$, which by assumption generates $I$. If $1\in W$, then $I$ is the entire ring, so we may assume $1\not\in W$. Let $y_1,\dots,y_r$ be a basis for $W$. Then $1,y_1,\dots,y_r$ are linearly independent, so there exists a linear automorphism $T$ of $V$ such that $T(1)=1$ and $T(y_i)=x_i$ for $i=1,\dots,r$.

Now consider the induced automorphism $T^*$ of $K[x_1,\dots,x_n]$. Since $T^*(y_i)=x_i$, $T^*$ maps $I$ to the ideal $(x_1,\dots,x_r)$. So, since $T^*$ is an isomorphism of rings, it suffices to answer your question in the special case $I=(x_1,\dots,x_r)$.

But in that case, your question is very easy, since the quotient $K[x_1,\dots,x_n]/(x_1,\dots,x_r)$ is just isomorphic to the polynomial ring $K[x_{r+1},\dots,x_n]$. So the quotient is indeed isomorphic to a polynomial ring, and in particular the ideal is prime since polynomial rings are domains.

  • $\begingroup$ A stupid question: to obtain a linear automorphism $T$ of $V$ such that $T(1)=1$ and $T(y_i)=x_i$ for $i=1,\dots,r$ i have to complete the basis (for $W$ ) $y_1,\dots,y_r$ to a basis for $V$ picking elements from the basis $1,x_1,\dots,x_n$ and than define $T$ such that is verified the above condition and such that T is bijective, right? Or there is something else? $\endgroup$
    – Minato
    Nov 1, 2017 at 18:31
  • $\begingroup$ More or less. You extend $1,y_1,\dots,y_r$ to a basis for $V$ however you want, and then extend $1\mapsto 1$, $y_i\mapsto x_i$ to a bijection between that basis and $1,x_1,\dots,x_n$. $\endgroup$ Nov 1, 2017 at 18:43

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