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The chiral superfield condition is $\boxed{{\bar{D}}_{\dot{\alpha }}\Phi =0}$ for both dotted components. Use left Grassmann derivatives with the displayed supersymmetric covariant derivatives. The odd product rule gives $\partial (\theta {\sigma }^{\mu }\bar{\theta })/\partial {\bar{\theta }}^{\dot{\alpha }}=-(\theta {\sigma }^{\mu }{)}_{\dot{\alpha }}$, soThe chirality constraint therefore says that $\Phi$ is independent of $\bar{\theta }$ at fixed $y$. The Grassmann algebra has only two independent $\theta$ components, so every cubic product of them vanishes. Its general even scalar superfield consequently has a scalar coefficient, a term linear in $\theta$ with a Weyl spinor coefficient, and one quadratic term. Choosing the conventional component normalization gives the chiral-superfield component expansionHere $\varphi$ is a complex scalar field, $\psi$ an odd Weyl spinor and $F_{\mathrm{aux}}$ a complex nonpropagating auxiliary field. Conversely this expansion is annihilated by $\bar{D}$ because $y$ and $\theta$ are. At fixed ordinary $x$, the same result is the terminating translation $\Phi =e^{i\theta {\sigma }^{\mu }\bar{\theta }{\partial }_{\mu }}[\varphi (x)+\sqrt{2}\theta \psi (x)+(\theta \theta )F_{\mathrm{aux}}(x)]$; this retains the printed contraction convention without changing the order in the barred square.

For a local function of scalar superfields, initially allow $f(\Phi ,{\Phi }^{†})$. The chain rule givesFor arbitrary chiral component fields the last factor is not constrained to vanish, so chirality requires $f_{{\Phi }^{†}}=0$. Thus $\boxed{f\text{ must be holomorphic in the chiral fields}}$ on its domain, with no generic conjugate-field dependence. Its ordinary holomorphic derivatives must exist where its nilpotent expansion is evaluated; a pole or branch point there does not define a regular local superfield. Using $(\theta \psi {)}^2=-\frac{1}{2}(\theta \theta )(\psi \psi )$ gives the explicit holomorphic closure of chiral superfields:All three component coefficients are evaluated at $y$. For several chiral superfields, replace $f^{\prime }\psi$ by $f_i{\psi }_i$ and $f^{\prime \prime }\psi \psi$ by $f_{ij}{\psi }_i{\psi }_j$. This requirement concerns a function that is to be chiral for arbitrary field configurations; a special constant field configuration does not impose the same functional constraint.

The ungauged two-derivative supersymmetric action is obtained from a real Kähler potential and a holomorphic superpotential:The first integral is a D-term; the second is an F-term. The Kähler metric $K_{i\bar{j}}$ must be positive for positive scalar kinetic energy. The auxiliary part is $K_{i\bar{j}}{F}_i{\bar{F}}_{\bar{j}}+W_i{F}_i+{\overline{W}}_{\bar{j}}{\bar{F}}_{\bar{j}}$, so eliminating it gives the scalar potential $V=K^{i\bar{j}}{W}_i{\overline{W}}_{\bar{j}}$ in the bosonic sector.

For the single-field Wess–Zumino model, take the canonical Kähler potential $K={\Phi }^{†}\Phi$ and spacetime signature $(+---)$. Component superspace integration givesSince $W^{\prime }=m\varphi +g{\varphi }^2$ and $W^{\prime \prime }=m+2g\varphi$, the auxiliary equation is $F_{\mathrm{aux}}=-(m\varphi +g{\varphi }^2{)}^{\ast }$. Completing its square leaves $-|W^{\prime }{|}^2$ in the Lagrangian. Hence the tree-level effective potential isFor nonzero $m,g$, its two zero-energy supersymmetric vacua are $\varphi =0$ and $\varphi =-m/g$. The potential is nonnegative because it is an absolute square.

The Trilinear scalar vertices in the Wess–Zumino model come fromWith all field legs incoming, expanding $e^{iS}$ and differentiating each cubic monomial givesThe factor $2!$ counts the two identical legs; there is no $1/2!$ in ${\mathcal{L}}_3$ to cancel it. A factor enforcing four-momentum conservation is implicit. These labels specify fields in an all-incoming Feynman rule; physical outgoing particles are assigned by crossing. There are no ${\varphi }^3$ or $({\varphi }^{\ast }{)}^3$ scalar vertices from this on-shell potential.

Equivalently, if $m,g$ are real and $\varphi =(A+iB)/\sqrt{2}$ with canonically normalized real scalars $A,B$, then ${\mathcal{L}}_3=-mgA(A^2+B^2)/\sqrt{2}$. Differentiation gives $\boxed{V_{AAA}=-3\sqrt{2}img,V_{ABB}=-\sqrt{2}img}$, with vanishing $AAB$ and $BBB$ vertices. This is the same pair of complex-field interactions in a real basis.

For the non-renormalization argument, promote $m,g$ to nondynamical chiral spurions. Define one ordinary $U(1)$ and one R-symmetry by the Wess–Zumino spurion charge assignment:

Each superpotential term is neutral under the ordinary symmetry. Under the R-symmetry, both terms have charge two, which is canceled by the charge $-2$ of $d^2\theta$. In particular, $m$ is R-neutral as required. The scalar, spinor and auxiliary components of $\Phi$ have R-charges $1,0,-1$. These are formal symmetries when the couplings transform as spurions; replacing them by fixed nonzero numbers need not leave an actual continuous field symmetry.

A local loop-corrected Wilsonian effective action superpotential $F_{\mathrm{eff}}(\Phi ,m,g)$ must be holomorphic in all three chiral variables and obey the same charges. The ratio $z=g\Phi /m$ is neutral under both symmetries, while $m{\Phi }^2$ has the required charges and mass dimension three. Therefore, on a patch with $m\ne 0$, the most general symmetry-allowed form iswith $h$ holomorphic on its domain. This form follows directly for a monomial ${\Phi }^a{m}^b{g}^c$: ordinary charge gives $a-2b-3c=0$ and R-charge gives $a-c=2$, hence $a=c+2$, $b=1-c$. Each allowed monomial is $g^c{m}^{1-c}{\Phi }^{c+2}$ and already has mass dimension three. Charge symmetry alone leaves $h$ undetermined.

To constrain perturbative loops, retain the original elementary chiral superfield and a nonzero infrared cutoff, rather than integrating out the entire field. A regular perturbative superpotential can be expanded about $g=0$, so $h(z)={\sum }_{n\ge 0}{c}_n{z}^n$ andTerms with $n\ge 2$ have poles at $m=0$. They are excluded by regularity of the local Wilsonian coefficients in the massless limit with the infrared cutoff fixed. At $g=0$ the theory is Gaussian, so the quadratic coefficient remains $c_0=1/2$. The term linear in $g$ is the single original cubic vertex: a diagram with three external fields and only that one interaction vertex has no internal loop. Holomorphy disallows coefficients involving $g^{\ast }$ or $m^{\ast }$, so it cannot conceal the usual nonholomorphic wave-function corrections in $c_1$. Matching the one-vertex term gives $c_1=1/3$. Thus the holomorphy argument for superpotential non-renormalization givesThe infrared qualification matters: a singular one-particle-irreducible expression or the action after eliminating an entire massive multiplet is not the same local Wilsonian effective action for the retained field.

Consequently the holomorphic superpotential and its parameters m and g receive no perturbative vertex renormalization. The Kähler potential does undergo wave-function renormalization. If its quadratic term is $Z{\Phi }^{†}\Phi$, then canonical normalization gives ${\Phi }_c=Z^{1/2}\Phi$ andThese canonically normalized running couplings can therefore be renormalized despite the non-renormalization theorem for the holomorphic superpotential. Loop corrections can also change the effective scalar potential through the corrected Kähler metric. The distinction between holomorphic coefficients and canonically normalized couplings is essential to answering the renormalization question.

For a nonzero common mass,This agrees with supersymmetric mass degeneracy in an unbroken chiral superfield multiplet. The two real scalar modes have the same mass as the two physical polarizations of the massive fermion, as the rest-frame multiplet construction predicts.

At $\Delta =0$ both masses vanish; the numerical ratio is $0/0$ and is undefined, but the equality of masses still expresses the unbroken supersymmetry. Equal zero masses in the exceptional flat, pure-linear breaking model do not prove unbroken supersymmetry: there the nonzero auxiliary expectation and positive energy in part (c) are the decisive tests.

Take canonical charged chiral superfield kinetic terms, gauge coupling $e>0$, and no Fayet–Iliopoulos term, since none is specified. Elimination of the three complex auxiliary fields givesThe Abelian gauge auxiliary field obeys $D=-e(|{\phi }_{+}{|}^2-|{\phi }_{-}{|}^2)$ in this normalization. Thus the F-term scalar potential and the gauge D-term giveAll terms are nonnegative. The normalization of $e$ can be changed together with the vector-field normalization, but the relative charges and the zero-potential conditions cannot. A noncanonical Kähler potential would change the inverse-metric factors in the F-term potential; adding a Fayet–Iliopoulos term would shift $D$ and define a different model. Neither is silently introduced here.

Chirality and the component expansion. A chiral superfield is constrained byUse left Grassmann derivatives. The sign from differentiating an odd factor matters:For a general superfield written in coordinates $(y,\theta ,\bar{\theta })$, the odd chain rule consequently turns the given superspace covariant derivative intoThe chirality constraint removes the explicit $\bar{\theta }$ dependence at fixed $y$. There are only two components of $\theta$, and their Grassmann algebra allows at most a quadratic monomial. Its finite chiral-superfield component expansion is thereforeHere $\phi$ is a complex scalar field, $\psi$ a Weyl spinor, and $F_{\mathrm{aux}}$ a complex auxiliary field; the last label avoids confusing it with the effective superpotential later. The factor $\sqrt{2}$ is the standard canonical component normalization. In ordinary coordinates the same statement isThis translation exponential terminates because its shift is nilpotent; it displays the full component dependence without an unstated convention for the barred spinor square.

A superspace covariant derivative obeys the graded product rule. Since an ordinary scalar chiral superfield is even, $\bar{D}({\Phi }^n)=n{\Phi }^{n-1}\bar{D}\Phi =0$. Linear combinations prove the polynomial claim. More generally, a nonsingular holomorphic function of chiral fields is chiral; inserting conjugate fields generally spoils this holomorphic closure of chiral superfields.

The superspace action. For a real Kähler potential and a holomorphic superpotential, the global chiral-field action has Lagrangian densityFull superspace integration gives a D-term, and chiral superspace integration an F-term. The action is real, and its supersymmetry variations are spacetime total derivatives. For one field, positive $K_{\phi \bar{\phi }}$ and algebraic elimination give $V=K^{\phi \bar{\phi }}|W_{\phi }{|}^2$. The Wess–Zumino model uses the canonical choice $K={\Phi }^{†}\Phi$ at tree level.

Scalar potential and the vertex. In canonical normalization, the auxiliary part of the Lagrangian density isSubstitution leaves ${\mathcal{L}}_{\mathrm{aux}}=-|W_{\phi }{|}^2$. Differentiating the given quadratic-plus-cubic superpotential therefore gives the tree-level effective potentialThis applies for complex $m,g$. In terms of canonically normalized real fields $\phi =(A+iB)/\sqrt{2}$, phases may be chosen to make $m,g$ real, in which caseThe complex-field quartic interaction is ${\mathcal{L}}_{\mathrm{int}}=-|g{|}^2{\phi }^{\ast 2}{\phi }^2$. There are two identical external legs of each field type. Differentiating with respect to those four fields, or counting the Wick attachments to the vertex, produces the factor $2!2!$:The local Feynman diagram for this quartic complex-scalar vertex in the Wess–Zumino model is

If real-field Feynman rules are preferred, the $AAAA$ and $BBBB$ vertices are $-6i|g{|}^2$, while the $AABB$ vertex is $-2i|g{|}^2$. These are the same interaction in a different component basis. They should not be confused with a convention that absorbs $2!2!$ into the coefficient of the complex quartic term.

Spurion symmetries. Treat $m,g$ as chiral spurions. An ordinary $U(1)$ acts on $\Phi ,m,g$ with charges $(1,-2,-3)$ and leaves $\theta$ neutral. An R-symmetry gives $\theta$ charge one and $\Phi ,m,g$ charges $(1,0,-1)$. Thus

Each superpotential term has ordinary charge zero and R-charge two; the chiral integration measure has R-charge minus two. In particular $m$ is neutral under the specified R-symmetry. These are formal transformations of fields and parameters together, not two exact symmetries of a theory with arbitrary fixed nontransforming numerical couplings. This Wess–Zumino spurion charge assignment is useful because it constrains possible quantum terms.

The general holomorphic form. Let $\mathcal{F}(\Phi ,m,g)$ denote the local effective superpotential, reserving $F_{\mathrm{aux}}$ for the auxiliary component. The holomorphy argument for superpotential non-renormalization permits dependence on the chiral spurions, not on their conjugates. The dimensionless combination $z=g\Phi /m$ is neutral under both formal symmetries, whereas $m{\Phi }^2$ has the required dimension and charges. Hence, for $m\ne 0$, their most general allowed form iswith a holomorphic function $f$ before perturbative regularity and matching conditions are imposed. Equivalently, a monomial $m^a{g}^b{\Phi }^c$ must obeySolving gives $a=1-b$, $c=b+2$. Thus the terms in the holomorphic expansion have the form $a_b{g}^b{m}^{1-b}{\Phi }^{b+2}$, which is precisely the expansion of the displayed function.

What is and is not renormalized. Apply the non-renormalization theorem to a local Wilsonian effective action retaining the elementary field and a nonzero infrared cutoff. Perturbative coefficients are regular as $g\to 0$ and $m\to 0$: no massless infrared modes have been integrated all the way to zero momentum. Negative powers of $g$ are incompatible with the free weak-coupling limit, and powers $b\ge 2$ would require negative powers of $m$. Only the quadratic and cubic structures survive this regularity test. Their coefficients cannot acquire a loop correction here: a quadratic term with no $g$ is the free-theory mass term, while a cubic term only linear in $g$ is already the tree interaction. A loop renormalizing that cubic term requires additional interaction insertions. Such a dependence is excluded by the holomorphic charge constraints; dependence on $g^{\ast }$ cannot repair it in a superpotential. Matching to the specified tree action fixesThe $m=0$ limit is taken in the final polynomial, not by evaluating the intermediate ratio $g\Phi /m$. The holomorphic Wilsonian superpotential and its parameters $m,g$ receive no independent perturbative vertex renormalization. Symmetries alone would only give the arbitrary function $f$; regularity and the free/tree matching are necessary to reach the stronger conclusion.

The Kähler potential is not protected by that theorem. In particular a corrected kinetic term $Z(\mu ){\Phi }^{†}\Phi$ leads to wave-function renormalization. Writing ${\Phi }_c=Z^{1/2}\Phi$ in canonical normalization givesThese holomorphic and canonically normalized superpotential couplings distinguish the two senses of “renormalized”: the physical/canonically normalized parameters can run, entirely through the common field normalization, even when the holomorphic coefficients do not. The scalar effective potential can consequently receive quantum corrections through the Kähler potential. Nor does the local perturbative statement automatically apply to infrared-singular one-particle-irreducible actions or to integrating out whole massive fields. Non-renormalization protects the local holomorphic F-term, not the complete quantum action or every physically normalized mass and coupling.

Use the hypercharge normalization $Q_{\mathrm{electric}}=T^3+Y$. All matter chiral superfields are written with left-handed Weyl spinors, so the fields denoted by a superscript $c$ contain the charge conjugates of the usual right-handed Standard Model fermions. The MSSM superfield representations areEach fermion generation contributes the first five chiral superfields. Their complex scalar field partners are squarks for the quarks and sleptons for the leptons. Each Higgs chiral doublet contains a Higgs complex scalar field and a higgsino. Every chiral superfield also has a complex auxiliary field. The vector superfields areThey contain the corresponding gauge bosons, gauginos in the Adjoint representation, and real auxiliary fields. A right-handed neutrino chiral superfield is not part of the minimal field content.

A gauge anomaly is a quantum obstruction to a classical gauge symmetry. For hypercharge, triangle diagrams with left-handed Weyl spinors can violate the Ward identity of the gauge boson; an uncancelled gauge anomaly makes the gauge theory inconsistent. Anomaly cancellation sums over every component, including colour and weak multiplicities. Complex scalar fields do not contribute to these chiral gauge anomalies. For one fermion generation, the cubic hypercharge coefficient isThe other coefficients involving a hypercharge gauge boson vanish too. With the fundamental index $T(\mathbf{N})=1/2$,The last line is the mixed gauge-gravitational anomaly. Coefficients with one non-Abelian generator and two hypercharge generators vanish by tracelessness. For completeness, the purely colour cubic gauge anomaly cancels between the two fundamental quark components and the two antifundamentals; the weak group has no perturbative cubic gauge anomaly. Its four left-handed doublets per fermion generation also avoid the Witten SU(2) anomaly. Thus each family is separately anomaly-free, not merely their sum.

The gauginos do not spoil this result: their hypercharge is zero and their Adjoint representation is real. One extra Higgs chiral doublet is different because its higgsino is chiral. For $H_u$, its contributions areThey have no compensating contribution from the Higgs complex scalar field. The higgsino in $H_d$ supplies precisely the negative of each coefficient. Opposite-hypercharge Higgs chiral doublets restore anomaly cancellation. The same pair restores an even number of weak fermion doublets, so the Witten SU(2) anomaly provides an additional check of the higgsino anomaly cancellation.

The independent reason is the holomorphic need for two Higgs chiral doublets. A superpotential is a holomorphic function of chiral superfields, so it cannot use a conjugate Higgs superfield to generate the missing Yukawa couplings. The ordinary Standard Model can use a Higgs scalar and its conjugate, but the MSSM needs distinct chiral superfields of both hypercharges. For example,where the dot contracts weak indices with the antisymmetric tensor. All three terms are gauge-invariant holomorphic functions. $H_u$ supplies up-type masses, while $H_d$ supplies down-type and charged-lepton masses. Replacing either by the conjugate of the other would violate the holomorphic closure of chiral superfields.

Introduce $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$ and retain the same Grassmann variables. The supersymmetric derivatives in chiral coordinates follow from the left Grassmann derivative chain rule. Differentiating $y$ givesThe minus sign in the second relation comes from moving the odd Grassmann derivative past $\theta$. Hence, as operators on a superfield expressed in $(y,\theta ,\bar{\theta })$,Substitution into the two supersymmetric covariant derivatives adds the two unbarred spacetime terms and cancels the two barred ones:These operator equalities prove both requested actions on $V$. In particular, a chiral superfield becomes independent of $\bar{\theta }$ at fixed $y$. The TeX aid corrupts the second formula by replacing its ordinary barred derivative with a covariant one; the original PDF has the ordinary $-{\bar{\partial }}_{\dot{\alpha }}$.

$W_{\alpha }$ is a fermionic chiral spinor superfield. Its chirality follows directly from the Abelian field-strength chiral projection:There are only two independent barred supersymmetric covariant derivatives, and their equal-chirality anticommutators vanish. Every product of three barred derivatives therefore vanishes. Equivalently, the previous calculation has no independent barred Grassmann variable at fixed $y$. The free undotted index makes this a chiral spinor superfield, rather than a scalar chiral superfield; its lowest component is the odd gaugino ${\lambda }_{\alpha }$.

Use the minimal four-dimensional Super-Poincaré algebra, with $\{Q_{\alpha },Q_{\beta }\}=0$, $\{Q_{\alpha },{\bar{Q}}_{\dot{\beta }}\}=2{\sigma }_{\alpha \dot{\beta }}^{\nu }{P}_{\nu }$ and commuting translations. Lorentz covariance and closure on the existing supercharges permit onlyThe graded Jacobi identity for $P^{\mu },Q_{\alpha },Q_{\beta }$ therefore givesFor example, with ${\sigma }^0=I$, $\mu =0$ and $\alpha =\beta =1$, the coefficient is a nonzero multiple of $A(P_1-i{P}_2)$. Independence of the translation generators forces $A=0$. Its adjoint gives the barred result. Thus every supercharge commutes with four-momentum:Consequently $[Q_{\alpha },P^2]=0$. If $|u⟩$ has a particular mass-shell condition, the nonzero state $Q_{\alpha }|u⟩$ has the same one. This gives supersymmetric mass degeneracy within an unbroken physical supermultiplet. The qualification matters: after spontaneous supersymmetry breaking, the chosen vacuum is not annihilated by the supercharges, and its one-particle excitations need not constitute degenerate supermultiplets.

For the O'Raifeartaigh model, write $X={\varphi }_1$, $Y={\varphi }_2$, $Z={\varphi }_3$. Field phases allow $g,M,m$ to be positive. The canonical Kähler potential and elimination of the auxiliary fields give the F-term scalar potentialA supersymmetric vacuum would require $W_Y=MZ=0$, hence $Z=0$, while $W_X=g(Z^2-m^2)=-g{m}^2$ could not vanish. Thus F-flatness is impossible. More precisely, with $Z=u+iv$,The stated strict inequality makes every nonconstant term nonnegative. The entire classical vacuum family isHere $F_X=-\overline{W_X}=g{m}^2$, while $F_Y=F_Z=0$. The Weyl spinor ${\psi }_X$ is the massless goldstino, and the complex scalar field $X$ is a pseudomodulus with two zero tree-level squared masses. The hierarchy $M\gg m$ separates the massive fields from the breaking scale, but it does not itself select $x=0$ along this flat direction.

At the representative vacuum $x=0$, the chiral-superfield fermion mass matrix isIts physical masses are its singular values: $0,M,M$. The two massive Weyl spinors can be combined into one massive Dirac spinor. For $Y=(y_R+i{y}_I)/\sqrt{2}$, $Z=(z_R+i{z}_I)/\sqrt{2}$, the real scalar squared masses areIn particular, there is no tachyon under the stated inequality. The splitting of the $Z$ masses displays supersymmetry breaking directly.

For completeness, the mass spectrum can be given throughout the classical vacuum family, rather than silently fixing the pseudomodulus. A phase rotation makes $x$ real and nonnegative without changing physical masses. The massive fermion squared masses areThe real and imaginary scalar sectors have two-by-two scalar squared-mass matriceswhose eigenvalues areBoth determinants are $M^2(M^2+2ϵg^2{m}^2)>0$ and their traces are positive, so stability holds at every finite $x$. Quantum corrections can lift a pseudomodulus; these formulas describe the classical, tree-level spectrum requested here.

Define the physical mass supertrace byEach real scalar contributes once and each Weyl spinor twice with a minus sign. At $x=0$ this gives $4M^2-2(2M^2)=0$. At general $x$, ${\sum }_{ϵ,\pm }{b}_{ϵ,\pm }=4M^2+8g^2|x{|}^2=2(f_{+}+f_{-})$. Hence the tree-level mass supertrace vanishes despite broken supersymmetry:This is the tree-level supertrace mass sum rule, not a claim that the individual masses are equal.

For direct breaking in the Minimal supersymmetric Standard Model, the difficulty is both field content and the tree-level spectrum. A linear gauge-invariant superpotential term needs a gauge-singlet chiral superfield, which the minimal model does not contain. Moreover, with canonical kinetic terms and purely neutral F-term breaking, without D-term mass shifts, the MSSM tree-level sfermion mass constraint applies separately to conserved charge sectors. For one electron pair it gives $m_{{\tilde{e}}_1}^2+m_{{\tilde{e}}_2}^2=2m_e^2$: both scalar partners cannot be heavy. This is an illustrative charge-sector consequence under those assumptions, not an unrestricted inference from the total supertrace alone. A separate hidden supersymmetry-breaking sector communicating effective soft supersymmetry breaking through loops or suppressed operators avoids the direct canonical tree-level obstruction.

A four-dimensional $\mathcal{N}=1$ chiral superfield is a complex scalar superfield satisfyingThe supersymmetric covariant derivatives anticommute with the supercharges, so the constraint is preserved by supersymmetry. Use left Grassmann derivatives and the printed convention ${\bar{D}}_{\dot{\alpha }}=-{\bar{\partial }}_{\dot{\alpha }}-i{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }{\partial }_{\mu }$. The conjugate antichiral superfield satisfies $D_{\alpha }{\Phi }^{†}=0$.

Any nonsingular holomorphic function $f({\Phi }_1,\dots ,{\Phi }_n)$ of chiral superfields is itself a chiral superfield. The graded Leibniz rule and the chain rule giveThus sums, products, convergent power series and inverses on patches where the denominator is nonzero preserve chirality. An ordinary function involving ${\Phi }^{†}$ is generally not chiral, because $\bar{D}{\Phi }^{†}$ need not vanish. This is why a superpotential is holomorphic. Spacetime derivatives also commute with $\bar{D}$, so derivatives of chiral superfields remain chiral; chirality by itself does not impose renormalizability.